case study on kinematics class 11

CBSE Expert

CBSE Case Study Questions Class 11 Physics PDF Download

Are you a Class 11 Physics student looking to enhance your understanding and prepare effectively for your exams? Look no further! In this comprehensive guide, we present a curated collection of CBSE Case Study Questions Class 11 Physics that will help you grasp the core concepts of Physics while reinforcing your problem-solving skills.

CBSE 11th Standard CBSE Physics question papers, important notes, study materials, Previous Year Questions, Syllabus, and exam patterns. Free 11th Standard CBSE Physics books and syllabus online. Important keywords, Case Study Questions, and Solutions.

Class 11 Physics Case Study Questions

CBSE Class 11 Physics question paper will have case study questions too. These case-based questions will be objective type in nature. So, Class 11 Physics students must prepare themselves for such questions. First of all, you should study NCERT Textbooks line by line, and then you should practice as many questions as possible.

Chapter-wise Solved Case Study Questions for Class 11 Physics

• Chapter 1: Physical World
• Chapter 2: Units and Measurements
• Chapter 3: Motion in a Straight Line
• Chapter 4: Motion in a Plane
• Chapter 5: Laws of Motion
• Chapter 6: Work, Energy, and Power
• Chapter 7: System of Particles and Rotational Motion
• Chapter 8: Gravitation
• Chapter 9: Mechanical Properties of Solids
• Chapter 10: Mechanical Properties of Fluids
• Chapter 11: Thermal Properties of Matter
• Chapter 12: Thermodynamics
• Chapter 13: Kinetic Theory
• Chapter 14: Oscillations
• Chapter 15: Waves

Class 11 students should go through important Case Study problems for Physics before the exams. This will help them to understand the type of Case Study questions that can be asked in Grade 11 Physics examinations. Our expert faculty for standard 11 Physics have designed these questions based on the trend of questions that have been asked in last year’s exams. The solutions have been designed in a manner to help the grade 11 students understand the concepts and also easy-to-learn solutions.

Class 11 Books for Boards

Why Case Study Questions Matter

Case study questions are an invaluable resource for Class 11 Physics students. Unlike traditional textbook exercises, these questions simulate real-life scenarios, challenging students to apply theoretical knowledge to practical situations. This approach fosters critical thinking and helps students build a deep understanding of the subject matter.

Let’s delve into the different topics covered in this collection of case study questions:

1. Motion and Gravitation

In this section, we explore questions related to motion, velocity, acceleration, and the force of gravity. These questions are designed to test your grasp of the fundamental principles governing motion and gravitation.

2. Work, Energy, and Power

This set of questions delves into the concepts of work, energy, and power. You will encounter scenarios that require you to calculate work done, potential and kinetic energy, and power in various contexts.

3. Mechanical Properties of Solids and Fluids

This section presents case study questions about the mechanical properties of solids and fluids. From stress and strain calculations to understanding the behavior of fluids in different situations, these questions cover a wide range of applications.

4. Thermodynamics

Thermodynamics can be a challenging topic, but fear not! This part of the guide offers case study questions that will clarify the laws of thermodynamics, heat transfer, and thermal expansion, among other concepts.

5. Oscillations and Waves

Get ready to explore questions related to oscillations, simple harmonic motion, and wave characteristics. These questions will deepen your understanding of wave propagation and the behavior of oscillatory systems.

6. Kinetic Theory and Laws of Motion

Kinetic theory and the laws of motion can be complex, but with our case study questions, you’ll find yourself mastering these topics effortlessly.

Discover a wide array of questions dealing with light, lenses, and mirrors. This section will improve your problem-solving skills in optics and enhance your ability to analyze optical phenomena.

8. Electrical Effects of Current

Electricity and circuits are fundamental to physics. The case study questions in this section will challenge you to apply Ohm’s law, Kirchhoff’s laws, and other principles in various electrical circuits.

9. Magnetic Effects of Current

Delve into the fascinating world of magnets and magnetic fields. This set of questions will strengthen your understanding of magnetic effects and their applications.

10. Electromagnetic Induction

The final section covers electromagnetic induction, Faraday’s law, and Lenz’s law. You’ll be presented with scenarios that test your ability to predict induced electromotive forces and analyze electromagnetic phenomena.

In conclusion, mastering Class 11 Physics requires a thorough understanding of fundamental concepts and their practical applications. The case study questions provided in this guide will undoubtedly assist you in achieving a deeper comprehension of the subject.

Remember, practice is key! Regularly attempt these case study questions to strengthen your problem-solving abilities and boost your confidence for the exams. Happy studying, and may you excel in your Physics journey!

Leave a Comment Cancel reply

Save my name, email, and website in this browser for the next time I comment.

Download India's best Exam Preparation App Now.

Key Features

• Revision Notes
• Important Questions
• Previous Years Questions
• Case-Based Questions
• Assertion and Reason Questions

No thanks, I’m not interested!

Class 11 Physics Case Study Question

Case study question class 11 physics (cbse / ncert board), cbse class 11 physics case study question, leave a reply cancel reply, we have a strong team of experienced teachers who are here to solve all your exam preparation doubts, chhattisgarh class 4 math numbers solution, up scert solutions class 8 english chapter 3 – the right choice, dav class 6 english literature book solution chapter 9 our tree, dav class 6 english literature book solution chapter 8 hanuman and i.

• Motion In One Dimension
• Active page

Kinematics of Class 11

Displacement.

The displacement of a particle is defined as the difference between its final position and its initial position. We represent the displacement as Δx.

Δx = xf− xi

The subscripts iand f refer to be initialand final positions. These are not necessarily the positions from which the particle starts its motion nor where its motion ceases. The i and f designate the particular initial and final positions we are considering out of the entire motion of the object. Notethe order : final position minus initial position . Whenever we calculate “delta” anywhere, we always take the final value minus the initial value.

Average Velocity and Average Speed

The average velocity of an object travelling along the x−axis is defined as the ratio of its displacement to the time taken for that displacement.

The average speed of a particle is defined as the ratio of the total distance travelled to the time taken.

Average speed = Total distance travelled/Δt

Note that velocity and speed have different meanings.

Example : 7.1

A bird flies toward east at 10 m/s for 100 m. It then turns around and flies at 20 m/s for 15 s. Find

(a) its average speed

(b) its average velocity

Δt= Δt 1 + Δt 2 = 25 s

The bird flies 100 m east and then (20 m/s) (15 s) = 300 m west

(a)Average speed = Distance/Δt = 100m + 300 m/25s = 16m/s

(b)The net displacement is

Δx= Δx 1 + Δx 2 = 100 m − 300 m = −200 m

vav= Δx/Δt = -200m/25s = -8  m/s

The negative sign means that vavis directed toward the west.

Example: 7.2

A jogger runs his first 100 m at 4 m/s and the second 100 m at 2 m/s in the same direction. What is the average velocity ?

The first half took

Δt1= (100 m)/(4 m/s) = 25 s,

while the second took

Δt1= (100 m)/(2 m/s) = 50 s,

The total time interval is

Δt= Δt 1 + Δt 2 = 75 s

Therefore, his average velocity is

vav= Δx/Δt = 200m/75s = 2.67 m/s

Since 2.67 ≠ 1/2 (4+2), we see that the average velocity is not, in general, equal to the average of the velocities.

Average Acceleration is defined as the ratio of the change in velocity to the time taken.

Instantaneous Velocity is defined as the value approached by the average velocity when the time interval for measurement becomes closer and closer to zero, i.e. Δt → 0. Mathematically

The instantaneous velocity function is the derivative with respect to the time of the displacement function.

Instantaneous Acceleration is defined analogous to the method for defining instantaneous velocity. That is, instantaneous acceleration is the value approached by the average acceleration as the time interval for the measurement becomes closer and closer to zero.

The Instantaneous acceleration function is the derivative with respect to time of the velocity function

Example: 7.3

The position of a particle is given by

x= 40 − 5t− 5t 2 , where x is in metre and t is in second

(a) Find the average velocity between 1 s and 2 s

(b) Find its instantaneous velocity at 2 s

(c) Find its average acceleration between 1 s and 2 s

(d) Find its instantaneous acceleration at 2 s

(a) Att = 1 s; xi= 30 m

t = 2 s; xf= 10 m

(b) v = dx/dt = -5 -10t

At t= 2 s; v= −5−10(2) = −25 m/s

(c) At t = 1 s; v = −5−10 (1) = −15 m/s

t = 2 s; v = −5−10 (1) = −25 m/s

(d)a= dv/dt = −10 m/s 2

• Graphical Interpretation of Displacement, Velocity And Acceleration
• Motion With Constant Acceleration
• Motion In Two Dimensions
• Projectile Motion
• Circular Motion
• Relative Velocity
• solved question

Talk to Our counsellor

• NCERT Solutions
• NCERT Class 11
• NCERT 11 Physics
• Chapter 3: Motion In A Straight Line

NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line

Ncert solutions class 11 physics chapter 3 – free pdf download.

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 2.

NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line is an essential tool that will help in your exam preparation. They consist of answers to the questions given in the textbook together with important questions from CBSE previous year question papers and CBSE sample papers. Exemplar problems provide solutions that help you in your exam preparation and cross-check your preparation level.

In the universe, motion is common to all objects. Even while we are asleep, air moves in and out of our body and blood flow in our veins and arteries. The change in the position of an object with time is called motion. Class 11 is an important step in a student’s life to obtain a strong knowledge of basic concepts. The topics learnt using the NCERT Solutions for Class 11 Physics Chapter 3, which are updated according to the CBSE Syllabus (2023-24), would help you in your higher levels of education. To help students in this aspect, the subject matter experts at BYJU’S have designed the NCERT Solutions in a PDF format which can be downloaded for free.

carouselExampleControls112

Previous Next

Access answers of NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line

Q1. In which of the following examples of motion can the body be considered approximately a point object? (a) A railway carriage moving without jerks between two stations. (b) A cap on top of a man cycling smoothly on a circular track. (c) A spinning cricket ball that turns sharply on hitting the ground. (d) A tumbling beaker that has slipped off the edge of a table.

(a), (b) The size of the railway carriage and the cap is very small as compared to the distance they’ve travelled, i.e. the distance between the two stations and the length of the race track, respectively. Therefore, the cap and the carriage can be considered as point objects.

The size of the cricket ball is comparable to the distance through which it bounces off after hitting the floor. Thus, the cricket ball cannot be treated as a point object. Likewise, the size of the beaker is comparable to the height of the table from which it drops.  Thus, the beaker cannot be treated as a point object.

Q2. The position-time (x-t) graphs for two children, A and B, returning from their school O to their homes, P and Q, respectively, are shown in Fig. Choose the correct entries in the brackets below: (a) (A/B) lives closer to the school than (B/A) (b) (A/B) starts from the school earlier than (B/A) (c) (A/B) walks faster than (B/A) (d) A and B reach home at the (same/different) time (e) (A/B) overtakes (B/A) on the road (once/twice).

(a) A lives closer to the school than B, because A has to cover shorter distances [OP < OQ]. (b) A starts from school earlier than B, because for x= 0, t = 0 for A but for B, t has some finite time. (c) The slope of B is greater than that of A; therefore, B walks faster than A. (d) Both A and B will reach their home at the same time. (e) At the point of intersection, B overtakes A on the roads once.

Q3. A woman starts from her home at 9.00 am, walks at a speed of 5 km/h on a  straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and  returns home by auto at a speed of 25 km/h. Choose suitable scales and plot the x-t graph of her motion.

Distance to her office = 2.5 km. Walking speed the woman = 5 km/h Time taken to reach office while walking = (2.5/5 ) h=(1/2) h = 30 minutes

Speed of auto = 25 km/h

Time taken to reach home in auto = 2.5/25 = (1/10) h = 0.1 h = 6 minutes

In the graph, O is taken as the origin of the distance and the time, then at  t = 9.00 am, x = 0 and at t = 9.30 am, x = 2.5 km

OA is the portion on the x-t graph that represents her walk from home to the office. AB represents her time of stay in the office from 9.30 to 5. Her return journey is represented by BC.

Q4. A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backwards,  followed again by 5 steps forward and 3 steps backwards, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically  and otherwise how long the drunkard takes to fall in a pit 13 m away from the  start.

The time taken to go one step is 1 second. In 5 s, he moves forward through a distance of 5 m, and then in the next 3s, he comes back by 3 m. Therefore,  in 8 s, he covers 2 m. So, to cover a distance of 8 m, he takes 32 s. He must take another 5 steps forward to fall into the pit. So, the total time taken is 32 s + 5 s = 37 s to fall into a pit 13 m away.

Q5. A jet aeroplane travelling at the speed of 500 km/h ejects its products of combustion at the speed of 1500 km/h relative to the jet plane. What is the speed of the  latter with respect to an observer on the ground?

Q6. A car moving along a straight highway with a speed of 126 km h –1 is brought to a  stop within a distance of 200 m. What is the retardation of the car (assumed  uniform), and how long does it take for the car to stop?

The initial velocity of the car = u

The final velocity of the car = v

Distance covered by the car before coming to rest = 200 m

Using the equation,

t = (v – u)/a = 11.44 sec.

Therefore, it takes 11.44 sec for the car to stop.

Q7. Two trains, A and B, of length 400 m each, are moving on two parallel tracks with a uniform speed of 72 km h –1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s –2 . If, after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

Length of trains A and B = 400 m

Speed of both the trains = 72 km/h = 72 x (5/18) = 20m/s

Using the relation, s = ut + (1/2)at 2

Distance covered by the train B

S B = u B t + (1/2)at 2

Acceleration, a = 1 m/s

Time = 50 s

S B = (20 x 50) + (1/2) x 1 x (50) 2

Distance covered by the train A

S A = u A t + (1/2)at 2

Acceleration, a = 0

S A = u A t  = 20 x 50 = 1000 m

Therefore, the original distance between the two trains = S B – S A = 2250 – 1000 = 1250 m

Q8. On a two-lane road, car A is travelling at a speed of 36 km/h. Two cars, B and C, approach car A in opposite directions with a speed of 54 km/h each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What is the minimum acceleration of car B required to avoid an accident?

The speed of car A = 36 km/h = 36 x (5/8) = 10 m/s

Speed of car B = 54 km/h = 54 x (5/18) = 15 m/s

Speed of car C = – 54 km/h = -54 x (5/18) = -15 m/s (negative sign shows B and C are in opposite directions)

Relative speed of A w.r.t. C, V AC = V A – V B = 10 – (-15) = 25 m/s

Relative speed of B w.r.t. A, V BA = V B – V A = 15 – 10 = 5 m/s

Distance between AB = Distance between AC = 1 km = 1000 m

Time taken by the car C to cover the distance AC, t = 1000/V AC = 1000/ 25 = 40 s

If a is the acceleration, then

s = ut + (1/2) at 2

1000 = (5 x 40) + (1/2) a (40) 2

a = (1000 – 200)/ 800 = 1 m/s 2

Thus, the minimum acceleration of car B required to avoid an accident is 1 m/s 2

Q9.   Two towns, A and B, are connected by regular bus service, with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h –1 in the direction A to B notices that a bus goes past him every 18 min in the direction of  his motion and every 6 min in the opposite direction. What is the period T of the bus service, and with what speed (assumed constant) do the buses ply on the road?

Speed of each bus = V b

Speed of the cyclist = V c  = 20 km/h

The relative velocity of the buses plying in the direction of motion of the cyclist is V b – V c. The buses playing in the direction of motion of the cyclist go past him after every 18 minutes, i.e. (18/60) s.

Distance covered = (V b – V c  ) x 18/60

Since the buses are leaving every T minutes. Therefore, the distance is equal to V b x (T/60)

(V b – V c  ) x 18/60 = V b x (T/60) ——(1)

The relative velocity of the buses plying in the direction opposite to the motion of the cyclist is V b  + V c The buses go past the cyclist every 6 minutes, i.e. (6/60) s.

Distance covered = (V b  + V c  ) x 6/60

Therefore, (V b  +V c  ) x 6/60 = V b x (T/60)——(2)

Dividing (2) by (1)

(V b – V c  ) 18/(V b  +V c  ) 6 = 1

(V b – V c  )3 = (V b  +V c  )

Substituting  the value of V c

(V b – 20 )3= (V b  + 20 )

3V b – 60 = V b + 20

V b = 80/2 = 40 km/h

To find the value of T, substitute the values of V b and V c in equation (1)

(V b – V c  ) x 18/60 = V b x (T/60)

(40 – 20) x (18/60) = 40 x (T/60)

T = (20 x 18) /40 = 9 minutes

Q10. A player throws a ball upwards with an initial speed of 29.4 m/s. (a) What is the direction of acceleration during the upward motion of the ball? (b) What are the velocity and acceleration of the ball at the highest point of its  motion? (c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its  highest point, vertically downward direction to be the positive direction of  the x-axis, and give the signs of position, velocity and acceleration of the ball during its upward and downward motion. (d) To what height does the ball rise, and after how long does the ball return to the player’s hands? (Take g = 9.8 m s –2 and neglect air resistance).

(a) The acceleration due to gravity always acts downwards towards the centre of the Earth. (b) At the highest point of its motion, the velocity of the ball will be zero, but the acceleration due to gravity will be 9.8 m s –2    acting vertically downward. (c) If we consider the highest point of ball motion as x = 0, t = 0, and vertically downward direction to be +ve direction of the x-axis, then (i) During the upward motion of the ball before reaching the highest point position, x = +ve, velocity, v = -ve and acceleration, a = +ve. (ii) During the downward motion of the ball after reaching the highest point position, velocity and acceleration, all three quantities are positive. (d) Initial speed of the ball, u= -29.4 m/s

The final velocity of the ball, v = 0

Acceleration = 9.8 m/s 2

Applying in the equation v 2 – u 2 = 2gs

0 – (-29.4) 2 = 2 (9.8) s

s = – 864.36/19.6 = – 44.1

Height to which the ball rises = – 44.1 m (negative sign represents upward direction)

Considering the equation of motion

0 = (-29.4) + 9.8t

t = 29.4/9.8 = 3 seconds

Therefore, the total time taken for the ball to return to the player’s hands is 3 +3 = 6 s.

Q11. Read each statement below carefully and state with reasons and examples, if it is  true or false;  A particle in one-dimensional motion (a) with zero speed at an instant may have non-zero acceleration at that instant (b) with zero speed may have non-zero velocity (c) with constant speed must have zero acceleration (d) with positive value of acceleration must be speeding up

(c) True (if the particle rebounds instantly with the same speed, it implies infinite acceleration, which is unphysical)

(d) False (true only when the chosen positive direction is along the direction of motion)

Q12. A ball is dropped from a height of 90 m on a floor. At each collision with the floor,  the ball loses one-tenth of its speed. Plot the speed-time graph of its motion  between t = 0 to 12 s.

Height from which the ball is dropped = 90 m

The initial velocity of the ball, u = 0

Let v be the final velocity of the ball

Using the equation

v 2 – u 2 = 2as ——–(1)

v 1 2 – 0 = 2 x 10 x 90

v 1 = 42.43 m/s

Time taken for the first collision can be given by the equation

42.43 = 0 + (10) t

t 1 = 4.24 s

The ball losses one-tenth of the velocity at collision. So, the rebound velocity of the ball is

v 2 = v – (1/10)v

v 2 = (9/10) v

v 2 = (9/10) (42.43)

= 38.19 m/s

Time taken to reach maximum height after the first collision is

38.19 = 0 + (10)t 2

t 2 = 3.819 s

The total time taken by the ball to reach the maximum height is

T = t 1 + t 2

T = 4.24 + 3.819  = 8.05 s

Now the ball will travel back to the ground in the same time as it took to reach the maximum height = 3.819 s

Total time taken will be, T = 4.24 + 3.819 + 3.819 = 11.86

Velocity after the second collision

v 3 = (9/10) (38.19)

v 3 = 34.37 m/s

Using the above information, the speed time graph can be plotted

Q13. Provide clear explanations and examples to distinguish between:

(a) The total length of a path covered by a particle and the magnitude of displacement over the same interval of time.

(b) The magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval].

In (a) and (b), compare and find which of the two quantities is greater.

When can the given quantities be equal? [For simplicity, consider one-dimensional motion only].      

( a ) Let us consider an example of a football; it is passed to player B by player A and then instantly kicked back to player A along the same path. Now, the magnitude of displacement of the ball is 0 because it has returned to its initial position. However, the total length of the path covered by the ball = AB +BA = 2AB. Hence, it is clear that the first quantity is greater than the second.

( b ) Taking the above example, let us assume that football takes t seconds to cover the total distance. Then,

The magnitude of the average velocity of the ball over time interval t = Magnitude of displacement/time interval

= 0 / t = 0.

The average speed of the ball over the same interval = total length of the path/time interval

Thus, the second quantity is greater than the first.

The above quantities are equal if the ball moves only in one direction from one player to another (considering one-dimensional motion).

Q14. A man walks on a straight road from his home to a market 2.5 km away at a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h –1 . What is the (a) Magnitude of average velocity, and (b) Average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time and not as the magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]

Distance to the market = 2.5 km = 2500 m

Speed of the man  walking to the market= 5 km/h = 5 x (5/18) = 1.388 m/s

Speed of the man walking when he returns back home = 7.5 km/h = 7.5  x (5/18) = 2.08 m/s

(a) Magnitude of the average speed is zero since the displacement is zero

(i) Time taken to reach the market = Distance/Speed = 2500/1.388 = 1800 seconds = 30 minutes

So, the average speed over 0 to 30 minutes is 5 km/h or  1.388 m/s

(ii) Time taken to reach back home = Distance/Speed = 2500/2.08 = 1200 seconds = 20 minutes

So, the average speed is

Average speed over a interval of 50 minutes= distance covered/time taken = (2500 + 2500)/3000 = 5000/3000 = 5/3 = 1.66 m/s

(ii) Average speed over an interval of 0-40 minutes = distance covered/ time taken = (2500+ 1250)/2400 = 1.5625 seconds = 5.6 km/h

Q15. In Exercises 3.13 and 3.14, we have carefully distinguished between average speed and the magnitude of average velocity. No such distinction is necessary when we consider the instantaneous speed and the magnitude of velocity. The instantaneous speed  is always equal to the magnitude of instantaneous velocity. Why?

Instantaneous velocity and instantaneous speed are equal for a small interval of time because the magnitude of the displacement is effectively equal to the distance travelled by the particle.

Q16. Look at the graphs (a) to (d) carefully and state, with reasons, which of these cannot possibly represent the one-dimensional motion of a particle? 

None of the four graphs shows a one-dimensional motion. 

(a) Shows two positions at the same time, which is not possible. 

(b) A particle cannot have velocity in two directions at the same time.

(c) Graph shows negative speed, which is impossible. Speed is always positive.

(d) Path length decreases in the graph; this is also not possible.

Q17. The figure shows the x-t plot of the one-dimensional motion of a particle. Is it correct to say from the graph that shows that the particle moves in a straight line for t < 0 and on a parabolic path for t >0? If not, suggest a suitable physical context for this graph.

It is not correct to say that the particle moves in a straight line for t < 0 (i.e., -ve) and on a parabolic path for t > 0 (i.e., + ve) because the x-t graph does not represent the path of the particle.

A suitable physical context for the graph can be the particle is dropped from the top of a tower at t =0.

Q18. A police van moving on a highway with a speed of 30 km h –1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h –1 . If the muzzle speed of the bullet is 150 m s –1 , with what speed does the bullet hit the thief’s car? (Note: Obtain that speed which is relevant for damaging the thief’s car).

Speed of the police van = 30 km/h = 30 x (5/18) = 25/3 m/s

Speed of a thief’s car = 192 km/h = 192 x (5/18) = 160/3 m/s

Muzzle Speed of  the bullet = 150 m/s

Speed of the bullet = speed of the police van + muzzle speed of  the bullet

= (25/3)+ 150 = 475/3 m/s

The relative velocity of the bullet w.r.t. the thief’s car is

v = Speed of the bullet – Speed of a thief’s car

= (475/3) – (160/3) = 105 m/s

The bullet hits the thief’s car at a speed of 105 m/s

Q19. Suggest a suitable physical situation for each of the following graphs.

(a)The graph is similar to kicking a ball, and then it hits the wall and rebounds with a reduced speed. The ball then moves in the opposite direction and hits the opposite wall, which stops the ball. (b) The graph shows a continuous change in the velocity of the object, and at some instant, it losses some velocity. Therefore, it may represent a situation where a ball falls on the ground from a certain height and rebounds with a reduced speed. (c) A cricket ball moving with a uniform speed is hit by a bat for a very short time interval.

Q20. The following figure gives the x-t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.

In S.H.M., acceleration, a = – ω 2 x , ω is the angular frequency —-(1) (i) At t = 0.3 s, x < 0, i.e. position is negative. Moreover, as x becomes more negative with time, it shows that velocity is negative (i.e., v < 0). However, using equation (1), acceleration will be positive. (ii) At t = 1.2 s,  Positions and velocity will be positive. Acceleration will be negative. (iii) At t = -1.2 s, Position, x is negative.  Velocity and acceleration will be positive.

Q21. The figure gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.

Interval 3 is the greatest, and 2 is the least. The average velocity is positive for intervals 1 and 2, and it is negative for interval 3.

Q22. The following figure gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion gives the signs of v and a in the three intervals. What are the accelerations at points A, B, C and D?

The change in the speed with time is maximum in interval 2. Therefore, the average acceleration is greatest in magnitude in interval 2.

The average speed is maximum in interval 3.

The sign of velocity is positive in intervals 1, 2 and 3. The acceleration depends on the slope. The acceleration is positive in intervals 1 and interval 3, as the slope is positive. The acceleration is negative in interval 2, as the slope is negative.

Acceleration at A, B, C and D is zero since the slope is parallel to the time axis at these instants.

Q23. A three-wheeler starts from rest, accelerates uniformly with 1 m s –2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1,2,3….) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?

For a straight line, the distance covered by a body in nth  second is: S N  = u + a (2n – 1) / 2     . . . . . . . . ( 1 ) Where,

a = Acceleration

u = Initial velocity n = Time = 1, 2, 3, . . . . . , n In the above  case, a = 1 m/s 2 and u = 0. ∴ S N  = (2n – 1)  /  2     . . . . . . . . . . .( 2 ) This relation shows that: S N  ∝ n                             . . . . .. . . . . ( 3 )

Now substituting different values of n in equation ( 2 ) we get:

Q24. A boy, standing on a stationary lift (open from above), throws a ball upwards with the maximum initial speed he can, equal to 49 m s -1 . How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m s -1  and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?

The initial velocity of the ball, u = 49 m/s

The boy throws the ball upwards when the lift is stationary. The vertically upward direction is taken as the positive direction. The displacement of the ball is zero.

s=ut + (1/2)at 2

0 = (49)t + (1/2) (-9.8)t 2

t = (49 x 2)/9.8 = 98/9.8 = 10 sec

As the lift starts moving with a speed of 5 m/s, the initial speed of the ball will be 49 m/s + 5 m/s = 54 m/s

The displacement of the ball will be s =  5t’

Therefore, the time taken can be calculated using the formula

5t’ = (54) t’ + (1/2)(-9.8) t’ 2

t’ = 2(54 – 5)/9.8 = 10 sec

The time taken will remain the same in both cases.

Q25. On a long, horizontally moving belt figure, a child runs to and fro with a speed 9 km h –1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves at a speed of 4 km h –1 . For an observer on a  stationary platform outside, what is the (a) speed of the child running in the direction of motion of the belt? (b) speed of the child running opposite to the direction of motion of the belt? (c) time taken by the child in (a) and (b)? Which of the answers alter if motion is viewed by one of the parents?

Speed of child = 9 km h -1

Speed of belt = 4 km h -1 (a) When the boy runs in the direction of motion of the belt, then his speed as observed by the stationary observer = (9 + 4) km h -1  = 13 km h -1 .

(b) When the boy runs opposite to the direction of motion of the belt, then speed of the child as observed by the stationary observer = (9 – 4) km h -1  = 5 km h -1

(c) Distance between the two parents = 50 m = 0.05 km

The speed of the boy, as observed by both the parents, is 9 km h -1

Time taken by the boy to move towards one of the parents =0.05 km/9k h -1 =0.0056 h =20 S

Q26. Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m s –1 and 30 m s –1 . Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 m s –2 . Give the equations for the linear and curved parts of the plot.

For the first stone:

Acceleration, a = –g = – 10 m/s 2 Initial velocity, u I  = 15 m/s

Now, we know s 1  = s 0  + u 1 t + (1/2)at 2 Given, the height of the tree, s 0  = 200 m s 1  = 200 + 15t – 5t 2       . . . . . . . . .  . ( 1 ) When this stone hits the jungle floor, s 1  = 0 ∴– 5t 2  + 15t + 200 = 0 t 2  – 3t – 40 = 0 t 2  – 8t + 5t – 40 = 0 t (t – 8) + 5 (t – 8) = 0 t = 8 s or t = – 5 s Since the stone was thrown at time t = 0, the negative sign is not possible ∴t = 8 s For the second stone:

Acceleration, a = – g = – 10 m/s 2 Initial velocity, u II  = 30 m/s

We know, s 2  = s 0  + u II t + (1/2)at 2 = 200 + 30t – 5t 2 . . . . . . . . .  . . .  . ( 2 ) when this stone hits the jungle floor; s 2  = 0 – 5t 2  + 30 t + 200 = 0 t 2  – 6t – 40 = 0 t 2  – 10t + 4t + 40 = 0 t (t – 10) + 4 (t – 10) = 0 (t – 10) (t + 4) = 0 t = 10 s or t = – 4 s Here again, the negative sign is not possible ∴ t = 10 s Subtracting equation (1) from equation (2), we get s 2  – s 1  = (200 + 30t -5t 2 ) – (200 + 15t -5t 2 ) s 2  – s 1  =15t                                           . . .  . . . . . . . . . .. . . . . . ( 3 ) Equation (3) represents the linear trajectory of the two stones because to this linear relation between (s 2  – s 1 ) and t,  the projection is a straight line till 8 s. The maximum distance between the two stones is at t = 8 s. (s 2  – s 1 ) max  = 15× 8 = 120 m This value has been depicted correctly in the above graph. After 8 s, only the second stone is in motion, whose variation with time is given by the quadratic equation: s 2  – s 1  = 200 + 30t – 5t 2 Therefore, the equation of linear and curved path is given by: s 2  – s 1  = 15t (Linear path)

s 2  ­– s 1  = 200 + 30t – 5t 2  (Curved path)

Q27. The speed-time graph of a particle moving along a fixed direction is shown in the figure. Obtain the distance traversed by the particle between

(a) t = 0 s to 10 s, (b) t = 2 s to 6 s.

(a) Distance traversed by the particle between t = 0 s and t = 10 s

= area of the triangle = (1/2) x base x height

= (1/2) x 10 x x12 = 60 m

The average speed of the particle is 60 m/ 10 s = 6 m/s

(b) The distance travelled by the particle between t = 2 s and t = 6 s

= Let S 1 be the distance travelled by the particle in time 2 to 5 s and S 2 be the distance travelled by the particle in time 5 to 6 s.

For the motion from 0 sec to 5 sec

Now, u = 0 , t = 5 , v = 12 m/s

From the equation v = u + at, we get

a = (v – u)/t = 12/ 5 = 2.4 m/s 2

Distance covered from 2 to 5 s, S1 = distance covered in 5 sec – distance covered in 2 sec

= (1/2) a (5) 2 – (1/2) a (2) 2 = (1/2) x 2.4 x (25 – 4) = 1.2 x 21 = 25.2 m

For the motion from 5 sec to 10 sec, u = 12 m/s and a = -2.4 m/s 2

and t = 5 sec to t = 6 sec means n = 1 for this motion

Distance covered in the 6 the sec is S 2 = u + (1/2) a (2n – 1)

= 12 – (2.4/2) (2 x 1 – 1) = 10.8 m

Therefore, the total distance covered from t = 2 s to 6 s = S 1 + S 2

= 25.2 + 10.8 = 36 m

Q28. The velocity-time graph of a particle in one-dimensional motion is shown in the figure.

(a) x (t 2 ) = x (t 1 ) + v (t 1 ) (t 2 – t 1 ) + (1/2) a(t 2 – t 1 )2

(b) v(t 2 ) = v(t 1 ) + a(t 2 – t 1 )

(c) V average =  [ x(t 2 ) – x (t 1 )] /(t 2 – t 1 )

(d)  a average =  [ v(t 2 ) – v (t 1 )] /(t 2 – t 1 )

(e) x (t 2 ) = x (t 1 ) + v av (t 2 – t 1 ) + (1/2) a av (t 2 – t 1 ) 2

(f) x(t 2 ) – x (t 1 ) = Area under the v-t curve bounded by t- axis and the dotted lines.

The graph has a non-uniform slope between the intervals t 1 and t 2 (since the graph is not a straight line). The equations (a), (b) and (e) does not describe the motion of the particle. Only the relations (c), (d) and (f) are correct.

For students of Class 11, the topics that come under the NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line are important as they lay out the foundation for the topics to follow in the future. In order to understand advanced concepts of Physics, getting a grasp on fundamental topics like the one in this chapter is very crucial.

This chapter of NCERT Class 11 Physics covers simple but essential topics such as comparing objects as point objects. We will be plotting graphs and determining values according to them. When you walk at a certain speed and then climb down a bus at a certain distance, then what will be the (x-t) graph of your motion? Check out the NCERT Solutions provided below on how to plot the graph.

Have you ever seen a drunk guy walking some steps forward and a few steps backwards; have you wondered what will be the (x-t) motion graph of his motion? We will be finding the speed of a supercar which will go in front of you and the time taken by a biker to stop when he is stopped by a policeman; know the answer to these questions here. Are you afraid of snakes? Want to know who will catch its prey first among two anacondas racing for their prey? Find it by accessing the NCERT Solutions for Class 11  for free.

Do you love basketball? Want to know in what direction the ball is accelerating when it is thrown upwards by the referee and also find out here what will be the ball’s acceleration and velocity at the highest point. Let’s see what statements are true regarding the one-dimensional motion of an object. Have you seen a rubber ball bounce back when you throw it down the ground? What will be the speed-time graph of that? See the answers below. You can check out NCERT Solutions for Class 11 Physics for more chapter-wise solutions.

We will be providing the explanation for the difference between the total length of the path covered by a particle and the magnitude of displacement in the same interval of time. We will also find out the difference between average velocity and average speed in the same time interval. Do you know there is no difference between instantaneous speed and velocity? Find out why here. This chapter has a vast number of questions on every important topic in this chapter.

Subtopics of NCERT Class 11 Physics Chapter 3 Motion in a Straight Line

• Introduction
• Position, path length and displacement
• Average velocity and average speed
• Instantaneous velocity and speed
• Acceleration
• Kinematic equations for uniformly accelerated motion
• Relative velocity

Here, we have provided students with NCERT Solutions for Class 11 Physics Chapter 3 PDF to help them learn more effectively and understand the basic concepts of Physics. Solving these questions would ensure better clarity.

NCERT Solutions for Class 11 can be accessed anytime and can be downloaded easily. BYJU’S presents the best study materials, notes, study material, books, previous year question papers, sample papers, videos and animation to help the students in their all-important Class 11 exams and entrance examinations.

Disclaimer – 

Dropped Topics – 

Frequently Asked Questions on NCERT Solutions for Class 11 Physics Chapter 3

Why should i use the ncert solutions for class 11 physics chapter 3 motion in a straight line while preparing for the exams, what are the topics covered under chapter 3 of ncert solutions for class 11 physics, what is the meaning of acceleration in chapter 3 of ncert solutions for class 11 physics.

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

Kinematics - Physics For Class 11 - Class 11 - Notes, Videos & Tests

Part of the course

Kinematics Study Material

Notes for Kinematics - Physics For Class 11 | Class 11

Online test for kinematics - physics for class 11 | class 11, other chapters in physics for class 11.

Top Courses for Class 11

Importance of Kinematics Class 11

Kinematics notes free pdf download, important questions for kinematics, kinematics practice questions.

Welcome Back

Create your account for free.

Forgot Password

Unattempted tests, change country, practice & revise.

Kinematics Class 11 Physics | Notes

Kinematics:.

The branch of physics which deals with the study of motion as the function of time is known as kinematics. It does not give any information about force that causes it to move.

An object is said to be in rest if it does not changes its position with respect to a reference point.

An object is said to be in motion if it changes its position with respect to a reference point.

Motion in straight line

The total length of actual path moved by an object from its initial position to final position is known as distance. It is a scalar quantity.

The distance of a moving body can be never be zero. Its unit is metre (m).

Displacement:

The shortest distance between any two points in specified direction. (i.e. initial to final) is known as displacement. It is vector quantity and denoted by ($\overrightarrow{S}$)

The displacement of a moving body may be zero. Its unit is metre (m).

Note: The ratio of distance to magnitude of displacement is greater than or equal to 1

Q. An object moves 3.5 times around a circular path of radius 100m. Calculate its total distance and total displacement. 

Radius, r = 100 m

Distance, S = 2$\pi $r × n = 2×$\frac{\text{22}}{\text{7}}$×100×3.5 = 2200m

Displacement, ($\overrightarrow{S}$) = 2r = 2×100= 200m

Q. A car is moving on a circular track of radius 70m. Find the distance and displacement made when the car reaches to opposite end of the track.

Radius, r = 70m.

Distance, S = $\frac{\text{2 }\!\!\pi\!\!\text{ }\text{r}}{\text{2}}$

= $\frac{\text{22}}{\text{7}}$×70

Displacement, ($\overrightarrow{S}$)= 2r

It is defined as the rate of change of distance with respect to time.

Mathematically, speed= $\frac{\text{distance(s)}}{\text{time(t)}}$

Its unit is ms –1 . It is a scalar quantity.

Types of Speed:

(i) Average speed:

It is defined as the ratio of total distance covered to the total time taken.

Mathematically, average speed = $\frac{\text{total distance}}{\text{total time}}$

Avg. speed = $\frac{\text{total distance}}{\text{total time}}$ = $\frac{{{\text{S}}_{\text{1}}}\text{+}{{\text{S}}_{\text{2}}}\text{+}{{\text{S}}_{\text{3}}}}{{{\text{t}}_{\text{1}}}\text{+}{{\text{t}}_{\text{2}}}\text{+}{{\text{t}}_{\text{3}}}}$= $\frac{\sum S}{\sum t}$

(ii) Uniform speed:

The speed of an object is said to be uniform if it covers equal distances in equal interval of time. 

(iii) Non-uniform speed:

The speed of an object is said to be non-uniform if it covers unequal distances in equal interval of time.

(iv) Instantaneous speed:

The speed of an object calculated for very small interval of time is called instantaneous speed.

Mathematically, V ins =  $\underset{\Delta t\to 0}{\mathop{\lim }}\,\text{ }\frac{\Delta s}{\Delta t}$

The rate of change of displacement with respect to time is known as velocity.

Mathematically, Velocity = $\frac{\text{displacement}}{\text{time}}$

It’s unit is ms –1 . It is a vector quantity.

Types of velocity:

(i) average velocity:.

It is defined as the ratio of total displacement covered to total time taken.

Mathematically, average velocity = $\frac{\text{total displacement}}{\text{total time}}$                                                        

(ii) Uniform velocity:

The velocity of an object is said to be uniform if it covers equal displacements in equal interval of time.

(iii) Non-uniform velocity:

The velocity of an object is said to be non-uniform if it covers unequal displacements in equal interval of time.

(iv) Instantaneous velocity:

The velocity of an object calculated for very small interval of time is called instantaneous velocity.

Mathematically, $\overrightarrow{{{V}_{ins}}}$= $\underset{\Delta t\to 0}{\mathop{\lim }}\,\text{ }\frac{\overrightarrow{\Delta s}}{\Delta t}$=$\frac{\overrightarrow{ds}}{dt}$

Note: (i) If time is equally divided then the average speed is V =  $\frac{{{\text{v}}_{\text{1}}}\text{+}{{\text{v}}_{\text{2}}}}{\text{2}}$

         (ii) If distance is equally divided, then the average speed is     V =  $\frac{\text{2}{{\text{v}}_{\text{1}}}{{\text{v}}_{\text{2}}}}{{{\text{v}}_{\text{1}}}\text{+}{{\text{v}}_{\text{2}}}}$

         (iii) For uniform motion i.e. constant velocity, $\overrightarrow{{{V}_{avg}}}$=  $\overrightarrow{{{V}_{ins}}}$

Facts about uniform velocity:

(i)         The object always moves in a straight line and in the same direction.

(ii)        The magnitude of velocity is equal to speed.

(iii)       The average velocity is equal to instantaneous velocity.

(iv)       Since velocity is constant, acceleration is zero and hence net force (resultant force) is also zero.

Acceleration and retardation (or deceleration):

The rate of change of velocity with respect to time is known as acceleration. It is denoted by ‘a’ and given by

                                     a = $\frac{v-u}{t}$    i.e. v = u+at

It is a vector quantity retardation. The negative value of acceleration is called retardation (or deceleration).

For uniform velocity, acceleration is zero.

Equations related to uniform acceleration.

1.      v = u+at

2.      S = ut + $\frac{1}{2}$at 2

3.      V 2 = u 2 + 2as

Types of acceleration

(a) Average acceleration,

$\overrightarrow{{{a}_{avg}}}$= $\frac{\overrightarrow{\Delta v}}{\Delta t}$

(b) Instantaneous acceleration,

$\overrightarrow{{{a}_{ins}}}$= $\underset{\Delta t\to 0}{\mathop{\lim }}\,\text{ }\frac{\overrightarrow{\Delta v}}{\Delta t}$=$\frac{\overrightarrow{dv}}{dt}$

Graphical treatment:

(i)         displacement- time graph for uniform motion.

(i.e. velocity, V = constant and acceleration, a = 0)

Slope = $\frac{BC}{AC}$ = velocity

(ii)        Displacement- time graph for non-uniform motion

Slope at P = $\frac{BC}{AC}$= Velocity at point P

(iii)       Velocity – time graph for uniform motion

(iv) Velocity – time graph for constant acceleration (for u = 0)

(v) Velocity – time graph for constant acceleration (for u > 0)

(vi) Velocity – time graph for non-uniform acceleration.

Slope at P = $\frac{BC}{AC}$ = acceleration at point P

Displacement-time graph (s-t graph):

 Fig: Displacement-time graph for uniform motion

Let us consider a body initially at S o is moving with constant velocity for time ‘t’ where it covers distance ‘S’. Let this motion is represented by line AB in a s-t graph.

Draw BM perpendicular to t-axis and AN perpendicular to BM.

OA = MN = S o

OM = AN = t

BN = BM – MN = S – S o

Slope of S-t graph  = Slope of line AB

=  $\frac{BN}{AN}$

= $\frac{S-{{S}_{o}}}{t}$ = velocity         

$\therefore $  Slope of S-t graph = Velocity of a body

Physical significance of s-t graph

• The slope of (s-t) graph gives velocity   i.e.$\overrightarrow{V}$= $\frac{\overrightarrow{\Delta s}}{\Delta t}$
• It gives position of a body at any instant.
• It gives distance covered by a body in any interval of time.  

(i) Derivation of   S = S o + vt for a body moving with constant velocity by graphical method.

We have, Slope of S-t graph = velocity

Slope of line AB = V

  $\frac{BN}{AN}$ = V

$\frac{S-{{S}_{o}}}{t}$ = v

$\therefore $ S = S o + vt ……….is the required expression for displacement.

Velocity-time graph (v-t graph):

Let us consider a body moving initially with velocity ‘u’ is accelerated for time ‘t’ and velocity increases to ‘v’. Let this motion is represented by line AB in v-t graph.

Draw perpendicular BM to t-axis AN perpendicular to BM.

OA = MN = u

BN = BM – MN = v -u

Slope of v-t graph = slope of line AB

= $\frac{BN}{AN}$  =  $\frac{v-u}{t}$ = a

$\therefore $ Slope of v-t graph  = acceleration

Also, Area of v-t graph with t- axis = Area of trapezium OABM

= Area of rectangle OANM + Area of triangle ABN

= OA× OM + $\frac{1}{2}$ AN ×BN

=  ut + $\frac{1}{2}$ t × at =  displacement (S)

$\therefore $ Area of v-t graph with t- axis = displacement (S)

Physical significance of v-t graph are

• The slope of (v – t) graph gives acceleration   i.e.$\overrightarrow{a}$= $\frac{\overrightarrow{\Delta v}}{\Delta t}$
• The area under (v – t) graph with time – axis gives displacement .
• It gives velocity at any instant of time.

Kinematical equations by graphical method:

(i) Derive v = u + at for a body with constant acceleration by graphical method.

Draw perpendicular BM to t-axis and AN perpendicular to BM.

Slope of v-t graph = acceleration

i.e. slope of line AB = a

Or,    $\frac{BN}{AN}$ = a

Or,   $\frac{v-u}{t}$ = a

$\therefore $ v = u + at  ………This is the required expression.

(ii) Derivation of S = ut + $\frac{\text{1}}{\text{2}}$ at 2 for a uniformly accelerating body by graphical method.

Or, BN = at

Now, Area of v-t graph with t- axis = distance

Or, Area of trapezium OABM = S

Or, Area of rectangle OANM + Area of triangle ABN = S

Or, OA× OM + $\frac{1}{2}$ AN ×BN = S

Or, ut + $\frac{1}{2}$ t × at = S

$\therefore $ S = ut + $\frac{\text{1}}{\text{2}}$ at 2 ………This is the required expression.

(iii)  Derivation of  v 2 = u 2 + 2as for a uniformly accelerating body by graphical method.

BN = BM – MN = v – u

i.e, slope of line AB = a

Or, $\frac{BN}{AN}$ = a

Or, AN = $\frac{v-u}{a}$ ………………. (i)

i.e, Area of trapezium OABM = S

Or, $\frac{1}{2}$ (OA + BM) × AN = S

Or, $\frac{1}{2}$ ( u + v)  $\frac{v-u}{a}$ = S

Or, $\frac{{{v}^{2}}-{{u}^{2}}}{2a}$ = S                 

Or, v 2 – u 2 = 2as

$\therefore $ v 2 = u 2 + 2as ………..This is the required expression.

Distance covered by a body in n th second ( S n th formula):

The distance covered by a body in the interval of ‘n’ sec and (n-1) sec is called S n th formula.

It is given by S n th = S n – S n-1

i.e, S n th   = u + $\frac{a}{2}$ (2n – 1)

Derivation:

Let us consider a body moving initially with velocity ‘u’ is accelerated to distances S n and S n-1 in ‘n’ sec and (n-1) sec respectively.

As we know,    S = ut + $\frac{1}{2}$ at 2

So,    S n = u × n + $\frac{1}{2}$ an 2 ……………..(i)

and   S n-1 = u(n-1) + $\frac{1}{2}$ a(n-1) 2 ……………….(ii)

Now, distance covered in n th second = S n – S n-1

Or,       S n th   = un + $\frac{1}{2}$ an 2 – [u(n-1) + $\frac{1}{2}$ a(n-1) 2 )]

Or,       S n th = un + $\frac{1}{2}$ an 2 – [un-u + $\frac{1}{2}$ a(n 2 – 2n+1)]

Or,       S n th = un + $\frac{1}{2}$ an 2 – [un – u + $\frac{1}{2}$ an 2 – an +$\frac{1}{2}$ a]

Or,       S n th = un + $\frac{1}{2}$ an 2 – un + u – $\frac{1}{2}$ an 2 + an – $\frac{1}{2}$ a

Or,      S n th = u + a( n- $\frac{1}{2}$ )

$\therefore $ S n th = u + $\frac{\text{a}}{\text{2}}$ (2n – 1)…………………is the required expression.

Projectile: 

An object which is thrown in space and falls under the action of gravity alone is known as projectile. We neglect air resistance during the projectile motion. The path followed by projectile is known as trajectory.

Examples of projectile:

i. A stone thrown into space.

ii. A ball thrown towards a player in a playground. 

iii. A bomb dropped from an airplane.

Assumptions for the study of projectile motion 

1. The acceleration along y–axis is only considered, i.e. the projected object falls under the action of gravity alone.

2. The acceleration along x–axis is zero (a x = 0). The horizontal velocity does not change at all throughout the projectile motion, i.e. u x = V x  

3. Air resistance is neglected.

Case I: When an object is projected making an angle ‘$\theta $’ with the horizontal.

Let us consider an object projected from ground with an initial speed ‘u’ making an angle ‘$\theta $’ with horizontal (ground) as shown in figure.

Resolving the initial velocity ‘u’ into its rectangular components, we get, 

u x = u cos$\theta $ and u y = u sin$\theta $ along horizontal and vertical direction respectively.

1. Path of the projectile:

Let S x = x and S y = y be the components of displacement along horizontal and vertical direction respectively at any point.

We know, S = ut + $\frac{1}{2}$at 2 …………(i)

Taking horizontal component, equation (i) becomes,

 S x = u x t + $\frac{1}{2}$a x t 2  

Or, x = u cos$\theta $.t    $\because $ a x = 0

Or, t = $\frac{x}{u\cos \theta }$………….(ii)

Taking vertical component, equation (i) becomes,

 S y = u y t + $\frac{1}{2}$a y t 2  

Or, y = u sin$\theta $$\frac{x}{u\cos \theta }$ + $\frac{1}{2}$(–g) ${{\left( \frac{x}{u\cos \theta } \right)}^{2}}$

Or, y = tan$\theta $.x – $\frac{g}{2{{u}^{2}}{{\cos }^{2}}\theta }$x 2

This equation is of the parabolic form y = ax + bx 2 ,

where, a = tan$\theta $ and b = $\frac{-g}{2{{u}^{2}}{{\cos }^{2}}\theta }$

2. Maximum Height (H max ):

It is the maximum vertical displacement attained by the projectile.

We know, v 2 = u 2 + 2as ….(iii)

Taking vertical component, equation (iii) becomes,

v y 2 = u y 2 + 2 a y s y

For maximum height, s y = H max and v y = 0

$\therefore $ 0 = u 2 sin 2 $\theta $ + 2 (-g) H max

Or, 2 g H max = u 2 sin 2 $\theta $

$\therefore $ H max = $\frac{{{u}^{2}}si{{n}^{2}}\theta }{2g}$ 

This is the required expression for maximum height.

3. Time of flight (T):

It is the total time spent by projectile in air.

We know, S = ut + $\frac{1}{2}$at 2 …………(iv)

Taking vertical component, equation (iv) becomes,

For time of flight (total time), t = T and S y = 0

Or, 0 = u sin$\theta $. T + $\frac{1}{2}$(–g) T 2

Or, $\frac{1}{2}$g T 2 = u sin$\theta $. T

$\therefore $T = $\frac{2u\sin \theta }{g}$

This is the required expression for time of flight.

Time to reach the greatest height / time of ascend (t asc ) / time of descend (t des ):

T 1/2 = t asc = t des =  $\frac{T}{2}$= $\frac{u\sin \theta }{g}$

4. Horizontal Range (R):

It is the total horizontal distance covered by the projectile.

We know, S = ut + $\frac{1}{2}$at 2 …………(v)

Taking horizontal component, equation (v) becomes,

Or, S x = u x t + a x t 2  

Or, S x = u cos$\theta $. t $\because $ a x = 0

For horizontal range (total horizontal displacement), S x = R and t = T, then 

R = u cos$\theta $.T

Or, R = u cos$\theta $.$\frac{2u\sin \theta }{g}$

 $\therefore $ R = $\frac{{{u}^{2}}\sin 2\theta }{g}$

This is the required expression for horizontal range.

Condition for maximum horizontal range (R max ):

We have, R = $\frac{{{u}^{2}}\sin 2\theta }{g}$

For maximum value of R, $\operatorname{sin}2\theta $ must be maximum.

i.e. $\operatorname{sin}2\theta $ = 1

Or, $\operatorname{sin}2\theta $ = Sin90 o

$\therefore $ $\theta $ = 45 o

The horizontal range is maximum for$\theta $ = 45 o .

5. Velocity of the projectile at any instant:

Let $\overrightarrow{V}$ be the velocity of the projectile at any instant. Let ‘V x ’ and ‘V y ’ be the components of velocity $\overrightarrow{V}$along x-axis and y-axis respectively and ‘$\alpha $’ be the angle made by the velocity with the horizontal.

We have V = u + at   ……. (vi)

Taking horizontal component, equation (vi) becomes,

V x = u x + a x t  

Or, V x = ucos$\theta $  {$\because $ a x = 0} 

Similarly taking vertical component, equation (vi) becomes,

V y = u y + a y t  

Or, V y = usin$\theta $ – gt 

Now, magnitude of velocity 

V = $\sqrt{{{V}_{x}}^{2}+{{V}_{y}}^{2}}$= $\sqrt{{{u}^{2}}{{\cos }^{2}}\theta +{{\left( u\sin \theta -gt \right)}^{2}}}$

Or, V =$\sqrt{{{u}^{2}}{{\cos }^{2}}\theta +{{\left( u\sin \theta -gt \right)}^{2}}}$

and direction, $\tan \alpha $= $\frac{{{V}_{y}}}{{{V}_{x}}}$

Or, $\alpha $= tan –1 $\left( \frac{usin~\theta gt}{u\cos \theta } \right)$

For the velocity with which the projectile hits the ground, we replace the time ‘t’ by time of flight ‘T’ and angle ($\alpha $) by the striking angle ($\beta $)  then magnitude and direction become

Magnitude, V =$\sqrt{{{u}^{2}}{{\cos }^{2}}\theta +{{\left( u\sin \theta -gT \right)}^{2}}}$

and direction, $\beta $= tan -1 $\left( \frac{usin~\theta gT}{u\cos \theta } \right)$

6. Two angles of projection for same horizontal range:

For the angle of projection ‘$\theta $’, the horizontal range is given by,

R 1 = $\frac{{{u}^{2}}\sin 2\theta }{g}$

Again, for the angle of projection (90 –$\theta $), the horizontal range is given by

R 2 = $\frac{{{u}^{2}}\sin 2(90-\theta )}{g}$  

=  $\frac{{{u}^{2}}\sin (180-2\theta )}{g}$

=  $\frac{{{u}^{2}}\sin 2\theta }{g}$

From above equations

Hence, two angles of projection for the same horizontal range are $\theta $ and (90 –$\theta $ ).

Case II: When an object is projected horizontally from height (h) above the ground.

Let us consider an object projected horizontally from a height ‘h’ above the ground with initial velocity ‘u’. Here angle made by initial velocity with horizontal is zero (i.e. $\theta $= 0). In projectile motion, vertical velocity changes continuously but the horizontal velocity remains constant throughout the motion because the acceleration due to gravity doesn’t act in horizontal direction. Let ‘V x ’ and ‘V y ’ be the rectangular components of velocity ‘V’ after time ‘t’, u y = usin0 o = 0.

1. Path of the projectile: 

Let S x = x and S y = y be two rectangular components of displacement along horizontal and vertical direction respectively at any point.

Or, x = u cos$\theta $.t    {$\because $ a x = 0}

Or, x = u cos0 o .t      {$\because $$\theta $   = 0}

Or, t = $\frac{x}{u}$………….(ii)

Or, y = u sin$\theta $.t + $\frac{1}{2}$.g ${{\left( \frac{x}{u} \right)}^{2}}$

Or, y = u sin0 o .t +$\frac{g}{2{{u}^{2}}}$x 2

Or, y = $\frac{g}{2{{u}^{2}}}$x 2

This equation is of the parabolic form y = ax 2 ,

where, a = $\frac{g}{2{{u}^{2}}}$

2. Time of flight / time to reach the ground (T):

The total time of a projectile spent in air is known as time of flight. When the projectile reaches the ground.

We know, S = ut + $\frac{1}{2}$at 2 …………(iii)

For total time, S y = h and t = T

Or, h = u sin0 o .t +$\frac{1}{2}$g T 2

Or, h = $\frac{1}{2}$g T 2  

This is the required expression for height. 

From above equation

T 2 = $\frac{2h}{g}$

$\therefore $T = $\sqrt{\frac{2h}{g}}$

This is required expression for time of flight.

3. Horizontal Range (R):

Taking horizontal component, equation (iv) becomes,

 S x = u x t + $\frac{1}{2}$a x t 2 

S x = u x t {$\because $ a x = 0}

For horizontal range (total horizontal displacement) S x = R and t = T

Or, R = u cos0 o .T      {$\because $$\theta $   = 0}

Or, R = u  $\sqrt{\frac{2h}{g}}$

4. Velocity with which the projectile hits the ground.

Let $\overrightarrow{V}$ be the velocity of the projectile at any instant. Let ‘V x ’ and ‘V y ’ be the rectangular components of velocity $\overrightarrow{V}$along x-axis and y-axis respectively and ‘$\alpha $’ be the angle made by the velocity with the horizontal.

We have, V = u + at  …….  (v)

Or, V x = ucos$\theta $  {$\because $ a x = 0}

 Or, V x = ucos0 o = u 

Similarly taking vertical component, equation (v) becomes,

Or, V y = usin$\theta $ + gt 

Or, V y = gt {$\because $$\theta $= 0 o }

V = $\sqrt{{{V}_{x}}^{2}+{{V}_{y}}^{2}}$= $\sqrt{{{u}^{2}}+{{g}^{2}}{{t}^{2}}}$

Or, V =  $\sqrt{{{u}^{2}}+{{g}^{2}}{{t}^{2}}}$

And direction, $\tan \alpha $= $\frac{{{V}_{y}}}{{{V}_{x}}}$

Or, $\alpha $= tan –1 $\left( \frac{gt}{u} \right)$

For the velocity with which the projectile hits the ground, we replace the time ‘t’ by time of flight ‘T’ and angle ($\alpha $) by the striking angle ($\beta $) then magnitude and direction becomes

Magnitude, V =$\sqrt{{{u}^{2}}+{{g}^{2}}{{T}^{2}}}$

And direction, $\beta $= tan –1 $\left( \frac{gT}{u} \right)$

Crossing of River:

Let, width of river = AB

Velocity of flow of river = V R

Velocity of swimmer in still water = V S  

1. For shortest route: 

(a) Direction:

Sin$\theta $ = $\frac{p}{h}$= ,

$\theta $ = sin –1$\left( \frac{{{V}_{R}}}{{{V}_{S}}} \right)$

(b) Velocity along AB, V AB = Vs cos$\theta $

(c) Time to cross the river (t) = $\frac{AB}{Vscos\theta }$

2. For least time to cross the river:

We have, 

Time to cross the river, t = $\frac{AB}{{{V}_{s}}cos\theta }$

For t min , cos$\theta $ = max m  

i.e. cos$\theta $ = 1

Or, cos$\theta $ =  cos0 o

Or, $\theta $ = 0 o

Then, t min = $\frac{AB}{{{V}_{S}}}$

From fig (ii), 

Tan$\theta $ = $\frac{{{V}_{R}}}{{{V}_{s}}}$= $\frac{BC}{AB}$

Distance apart from opposite end

BC = $\frac{{{V}_{R}}}{{{V}_{s}}}$AB

$\therefore $Distance apart from the opposite end (BC) = $\frac{{{V}_{R}}}{{{V}_{s}}}$× width of river {OR, BC Tan$\theta $ × width of river} 

When a swimmer is swimming in river (Downstream or Upstream):

Let V S be the velocity of swimmer in still water and V R   be the velocity of flow of river.

V total = V S + V R V total = V S – V R

In downstream resultant velocity increases while in upstream it decreases.

Relative velocity:

The rate of change of displacement of an object with respect to another object when both are in motion is known as relative velocity of one object w.r.t. another object.

The relative velocity of ‘A’ w.r.t. ‘B’ is 

  $\overrightarrow{{{V}_{AB}}}$=$\overrightarrow{{{V}_{A}}}$ –$\overrightarrow{{{V}_{B}}}$

Let us consider two objects A and B moving with velocities at $\overrightarrow{{{V}_{A}}}$and $\overrightarrow{{{V}_{B}}}$an angle ‘$\theta $’. To find the relative velocity of ‘A’ w.r.t. ‘B’, the velocity of ‘B’ is supposed to be zero. For this, draw a negative vector of  $\overrightarrow{{{V}_{B}}}$ and complete a parallelogram as shown in figure.

The resultant of $\overrightarrow{{{V}_{A}}}$and –$\overrightarrow{{{V}_{B}}}$gives the relative velocity of ‘A’ w.r.t. ‘B’ i.e. $\overrightarrow{{{V}_{AB}}}$

In fig, ‘β’ is the dir n of$\overrightarrow{R}$ (=$\overrightarrow{{{V}_{AB}}}$) with $\overrightarrow{{{V}_{A}}}$

Now, the magnitude of $\overrightarrow{{{V}_{AB}}}$is 

V AB =$\sqrt{{{V}_{A}}^{2}+2{{V}_{A}}{{V}_{B}}\cos (180-\theta )+{{V}_{B}}^{2}}$

= $\sqrt{{{V}_{A}}^{2}-2{{V}_{A}}{{V}_{B}}\cos \theta +{{V}_{B}}^{2}}$

the direction of $\overrightarrow{{{V}_{AB}}}$ with $\overrightarrow{{{V}_{A}}}$is

α = tan –1 $\left( \frac{{{V}_{B}}\sin (180-\theta )}{{{V}_{A}}+{{V}_{B}}\cos (180-\theta )} \right)$

α = tan –1 $\left( \frac{{{V}_{B}}\sin \theta }{{{V}_{A}}-{{V}_{B}}\cos \theta } \right)$

Special Cases :

(i) When two objects are moving in the same direction i.e. $\theta $=0° then,

V AB = $\sqrt{{{V}_{A}}^{2}-2{{V}_{A}}{{V}_{B}}\cos {{0}^{o}}+{{V}_{B}}^{2}}$

  = $\sqrt{{{V}_{A}}^{2}-2{{V}_{A}}{{V}_{B}}+{{V}_{B}}^{2}}$

= $\sqrt{{{({{V}_{A}}-{{V}_{B}})}^{2}}}$

$\therefore $V AB = V A –V B

(ii)  When two objects are moving in opposite direction i.e. $\theta $=180° then, 

V AB = $\sqrt{{{V}_{A}}^{2}-2{{V}_{A}}{{V}_{B}}\cos {{180}^{o}}+{{V}_{B}}^{2}}$

= $\sqrt{{{V}_{A}}^{2}+2{{V}_{A}}{{V}_{B}}+{{V}_{B}}^{2}}$

= $\sqrt{{{({{V}_{A}}+{{V}_{B}})}^{2}}}$

$\therefore $V AB = V A +V B

(iii) When two objects are moving perpendicular to each direction i.e. $\theta $=90°then, 

V AB = $\sqrt{{{V}_{A}}^{2}-2{{V}_{A}}{{V}_{B}}cos90{}^\circ +{{V}_{B}}2}$

$\therefore $V AB = $\sqrt{{{V}_{A}}^{2}+{{V}_{B}}^{2}}$

(iv) When two objects are moving with same magnitude velocities:

Let,$\left| \overrightarrow{{{V}_{A}}} \right|$= $\left| \overrightarrow{{{V}_{B}}} \right|$ = V 

then, 

V AB =$\sqrt{{{V}_{A}}^{2}-2{{V}_{A}}{{V}_{B}}\cos \theta +{{V}_{B}}^{2}}$

= $\sqrt{{{V}^{2}}-2{{V}^{2}}\cos \theta +{{V}^{2}}}$

= $\sqrt{2{{V}^{2}}-2{{V}^{2}}cos\theta }$

= $\sqrt{2{{V}^{2}}(1-\theta cos)}$

= $\sqrt{2{{V}^{2}}.2si{{n}^{2}}\frac{\theta }{2}}$

$\therefore $V AB = 2Vsin$\frac{\theta }{2}$

Examples of Relative Velocity:

1. The front windscreen of a moving car gets wet in rain while the behind screen remains dry.

2. Rain drops hitting the side of windows of a car in motion often leave diagonal streaks.

3. Person has to incline the umbrella forward with vertical while running in rain. etc.

Q. The front windscreen of a moving car gets wet in rain while the behind screen remains dry. Why?

This is due to the relative velocity of rain with respect to the car. 

Let $\overrightarrow{{{V}_{C}}}$ the velocity of the car along the horizontal direction and $\overrightarrow{{{V}_{R}}}$ be the velocity of rain vertically downward. From the above figure, the resultant of $\overrightarrow{{{V}_{R}}}$and –$\overrightarrow{{{V}_{C}}}$gives the relative velocity of rain with respect to the car as shown in the figure above.

Here, Tan$\theta $ = $\frac{p}{b}$= $\frac{AY}{OY}$= $\frac{{{V}_{C}}}{{{V}_{R}}}$

$\therefore $ $\theta $= tan –1 $\left( \frac{{{V}_{C}}}{{{V}_{R}}} \right)$

So, the raindrop strikes the front windscreen making an angle $\theta $= tan –1$\left( \frac{{{V}_{C}}}{{{V}_{R}}} \right)$ with vertical direction. Thus, the front windscreen of a moving car gets wet in rain while behind the screen remains dry.

Numerical Problems ( Kinematics)

Take acceleration due to gravity, g = 10 ms –2

Q.1 A car travelling with a speed of 15m/s is braked and it slows down with uniform retardation. It covers a distance of 88 m as its velocity reduces to 7m/s. If the car continues to slow down with the same rate, after what further distance will it be brought to rest?  Ans: 24.5 m

Q.2 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the height of tower.   Ans: 45 m

Q.3 An object is dropped from the top of the tower of height 156.8 m. and at the same time another object is   thrown vertically upward with the velocity of 78.1 m/s from   the foot of the tower, when and where the object meet?            Ans: 2 sec. and 20 m below from top

Q.4 An object is dropped from the top of the tower of height 100m and after 2 sec another object is thrown vertically upward with velocity 25 m/s from the foot of the tower. When and where the objects meet?

Ans: after 3 sec of 1 st object dropped, 45m below the top.

Q.5 A swimmer’s speed along the river (down-stream) is 20 km/h and can swim up-stream at 8 km/h. Calculate the velocity of the stream and the swimmer’s possible speed in still water.          Ans: 14 km/hr, 6 km/hr

Q.6 A swimmer’s speed along the river is 20 kmph and up-stream is 8 kmph. Calculate the velocity of the stream and the swimmer’s possible speed in still water.     Ans: 14 km/hr, 6 krnihr

Q.7 A man wishes to swim across a river 600m wide. If he can swim at the rate of 4km/h in still water and the river flows at 2km/h. Then in what direction must he swim to reach a point exactly opposite to the starting point and when will he reach it?         Ans: 120° with water and 10.4 min

Q.8 A projectile is fired from ground level with a velocity 500 ms -1 at 30 o to the horizontal. Find the horizontal range, the greatest height and the time to reach the greatest height.     Ans:  21651m, 3125m, 25 sec

Q.9 A baseball is thrown towards a player with an initial velocity 20 ms -1 at 45 o with the horizontal. At the moment, the ball is thrown the player is 50 m from the thrower. At what is speed and in what direction must he run to catch the ball at the same height at which it is released?          Ans: 3.5 ms -1 towards the thrower

Q.10 A body is projected upwards making an angle $\theta $ with the horizontal with a velocity of 300 m/s. Find the value of $\theta $ so that the horizontal range will be maximum. Find its range and time of flight.  Ans: 45°, 9000 m

Q.11 A stone on the top of a vertical cliff is kicked horizontally so that its initial velocity is 9 ms -1 . If the cliff is 200 m high, calculate

(i) taken by the stone to reach the ground.

(ii) How far from the bottom of the cliff, the stone will hit the ground?       Ans: (i) 6.32 sec (ii) 56.92 m

Q.12 A stone on the edge of a vertical cliff is kicked so that its initial velocity is 9 m/s horizontally. If the cliff is 200m high, calculate: i. time taken by stone to reach the ground. ii. How far from the cliff the stone will hit the ground?                  Ans: (i) 6.32 sec (ii) 56.92 m

Q.13 A bullet is fired with a velocity of 100 m/s from the ground at an angle of 60° with the horizontal. Calculate the horizontal range covered by the bullet. Also calculate the maximum height attained.                Ans: 866m; 375 m

Q.14 A projectile is fired from the ground level with velocity 150 m/s at 30° to the horizontal. Find its   horizontal range. What will be the least speed with which it can.be projected to achieve the same horizontal range?       Ans: 1949 m, 140 m/s

Q.15 A projectile is fired from the ground level with velocity of 500 m/s at 30° to horizon. Find the horizontal range, and greatest vertical height to which it rises. What is   the least speed with which it could be projected in order to   achieve the same horizontal range? [g = 10 N kg -1 ]     Ans: 21651 m, 3125 m, 464 ms -1

Q.16 A body is projected horizontally from the top of   a tower 100 m high with a velocity of 9.8 ms -1 . Find the velocity with which it hits the ground.      Ans: 45.82 m/s at 77.9°

Q.17 A batter hits a baseball so that it leaves the bat with an initial speed 37m/s at an angle of 53°. Find the position of the ball and the magnitude and direction of its velocity after 2 seconds. Treat the baseball as a projectile.         Ans: 24.23 m/sec, 23.21°

Q.18 A stone is projected horizontally with 20m/s from top of a tall building. Calculate its position and velocity after 3 sec neglecting the air resistance.        Ans: 36 m/sec; 56.3°

Q.19 A projectile is launched with an initial velocity of 30 m/s at an angle of 60° above the horizontal. Calculate the magnitude and direction of its velocity 5s after launch. Ans: 28.3 m/s and 58° C from horizontal

Q.20 An airplane is flying with a velocity of 90 m/s at an angle of 23.0° above the horizontal. When the plane is 114 m directly above a dog that is standing on level ground, a suitcase drops out of luggage compartment far from the dog will the suitcase land? You can ignore it resistance.            Ans: 778.7 m from dog

Q.21 To a cyclist riding due west with a speed of 4 m/s, a wind appears to blow from south west. The wind appears to blow from south to a man, running at 2ms -1 in the same direction. What is the actual velocity of the wind?  Ans: 2$\sqrt2$ ms -1 from south east

Q.22 To a person going due east in a car with a velocity 25 km/hr, a train appear to move due north with a velocity of 25$\sqrt3$ km/hr. What is the actual velocity and direction of motion of train?

Ans: 50 km/hr along a direction 30° east of north.

Q.23 Two ships A and B are moving in a sea. If A moves with uniform velocity of 8 kmhr –1 due east B moves with uniform velocity 6 kmhr –1 due south. Calculate the relative velocity of ship A with respect to ship B.

Also Read: Vector Notes Class 11

14 thoughts on “Kinematics Class 11 Physics | Notes”

Excellent Notes, just a complete concept on kinematics. Thanks for your hard work!

Thank you very much for this. I really really need this one. again thank you.

very helpful

yes very helpful

thank u very much this too much helpful

Thanks alot sir

These notes are really very helpful all the topics are covered loved them thankyou

THANK YOU SO MUCH. THIS WAS EXACTLY WHAT I WAS LOOKING FOR.

Great Notes. Motion in a plane is also required . Kindly help ..

Thankyou AJ sir for this notes it helped me a lot 🙂

Nice sir 🙏🏾🙏🏾🙏🏾🙏🏾🙏🏾🙏🏾🙏🏾 God bless you लेकिन भगवान करे कि जो आपने notes बनाएं हैं वो एक एक question syllabus में हो।

Great note . Best teacher of uniglobe SS college AJ sir. Thank you very helpful note.

Excellent explanation

This notes were amazing👍

Leave a Comment Cancel Reply

Your email address will not be published. Required fields are marked *

Save my name, email, and website in this browser for the next time I comment.

Talk to our experts

1800-120-456-456

NCERT Solutions for Class 11 Physics Chapter 2 Motion in a Straight Line

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line - FREE PDF Download

NCERT for Chapter 2 Motion In A Straight Line Class 11 Solutions by Vedantu, introduces the basics of kinematics, which deals with the motion of objects. This chapter explains how objects move along a straight path, providing the essential concepts required for further studies in physics. In this chapter, you will study motion and the different kinds of motion. Also read about concepts like path lengths, displacement, velocity, and speed and their various numerical associated. Through these Class 11 Physics NCERT Solutions , we aim to simplify complex concepts into easy-to-understand explanations, helping you with each topic effectively. With Vedantu's NCERT Solutions, you will find step-by-step explanations of all the questions in your textbook, ensuring that you understand the concepts thoroughly. 

Note: ➤ Unlock your dream college possibilities with our NEET College Predictor !

Glance on Physics Chapter 2 Class 11 - Motion in a Straight Line

An object is said to be in motion if its position changes with time. The position of the object can be specified with reference to a conveniently chosen origin. For motion in a straight fine. position to the right of the origin is taken as positive and to the left as negative.

The path to the right of the origin is taken as positive and to the left is taken as negative when motion is in a straight line.

Average Speed - The total path length travelled in the total given time interval is known as average speed.

The path length is defined as the total length of the path traversed by an object.

An object is said to be in uniform motion in a straight line if its displacement is equal in equal intervals of time. Otherwise, the motion is said to be nonuniform.

This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 2 - Motion In A Straight Line, which you can download as PDFs.

There are 18 fully solved questions in class 11th physics chapter 2 Motion In A Straight Line.

Access NCERT Solutions for Class 11 Physics Chapter 2 – Motion in a Straight Line

1. In which of the following examples of motion, can the body be considered approximately a point object:

(a) a railway carriage moving without jerks between two stations.

Ans: As the size of a carriage is very small as compared to the distance between two stations, the carriage can be treated as a point sized object.

(b) a monkey sitting on top of a man cycling smoothly on a circular track.

Ans: As the size of a monkey is very small as compared to the size of a circular track, the monkey can be considered as a point sized object on the track.

(c) a spinning cricket ball that turns sharply on hitting the ground.

Ans: As the size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground, the cricket ball cannot be considered as a point object.

(d) a tumbling beaker that has slipped off the edge of a table.

Ans: As the size of a beaker is comparable to the height of the table from which it slipped, the beaker cannot be considered as a point object.

2. The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in figure. Choose the correct entries in the brackets below:

a. (A/B) lives closer to the school than (B/A)

Ans: From the graph, it is clear that $OP<OQ$. Therefore, A lives closer to the school than B.

b. (A/B) starts from the school earlier than (B/A)

Ans: From the graph it is clear that for \[x\text{ }=\text{ }0,\text{ }t\text{ }=\text{ }0\] for A and t has some finite value for B. Hence, A starts from the school earlier than B.

c. (A/B) walks faster than (B/A)

Ans: As the velocity is equal to slope of x-t graph, in case of uniform motion and slope of x-t graph for B is greater than that for A. Thus B walks faster than A.

d. A and B reach home at the (same/different) time

Ans: From the graph it is clear that both A and B reach their respective homes at the same time.

e.  (A/B) overtakes (B/A) on the road (once/twice).

Ans: As B moves later than A and his/her speed is greater than that of A. From the graph, it is clear that B overtakes A only once on the road.

3. A woman starts from her home at 9.00 am, walks with a speed of \[5\text{ }km/hr\] on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of \[25\text{ }km/hr\]. Choose suitable scales and plot the x-t graph of her motion

Ans: In the above question it is given that:

Speed of the woman \[=\text{ }5\text{ }km/h\].

Distance between her office and home \[=\text{ }2.5\text{ }km\].

\[Time\text{ }taken\text{ }=\text{ }Distance\text{ }/\text{ }Speed\]

\[=\text{ }2.5\text{ }/\text{ }5\text{ }=\text{ }0.5\text{ }h\text{ }=\text{ }30\text{ }min\]

It is given that she covers the same distance in the evening by an auto.

Now, \[speed\text{ }of\text{ }the\text{ }auto\text{ }=\text{ }25\text{ }km/h\].

\[=\text{ }2.5\text{ }/\text{ }25\text{ }=\text{ }1\text{ }/\text{ }10\text{ }=\text{ }0.1\text{ }h\text{ }=\text{ }6\text{ }min\]

The suitable x-t graph of the motion of the woman is shown in the given figure.

4. A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

Distance covered with 1 step \[=\text{ }1\text{ }m\]

Time taken \[=\text{ }1\text{ }s\]

Time taken to move first \[5\text{ }m\] forward \[=\text{ }5\text{ }s\]

Time taken to move \[3\text{ }m\]backward \[=\text{ }3\text{ }s\]

\[Net\text{ }distance\text{ }covered\text{ }=\text{ }5\text{ }\text{ }3\text{ }=\text{ }2\text{ }m\]

Net time taken to cover \[2\text{ }m\text{ }=\text{ }8\text{ }s\].

Drunkard covers \[2\text{ }m\]in \[8\text{ }s\].

Drunkard covered \[4\text{ }m\]in \[16\text{ }s\].

Drunkard covered \[6\text{ }m\]in $2\text{4 s}$.

Drunkard covered \[8\text{ }m\]in \[32\text{ }s\].

In the next \[5\text{ }s\], the drunkard will cover a distance of \[5\text{ }m\] and a total distance of \[13\text{ }m\] and then fall into the pit.

Net time taken by the drunkard to cover $km$\[13\text{ }m\text{ }=\text{ }32\text{ }+\text{ }5\text{ }=\text{ }37\text{ }s\]

The x-t graph of the drunkard’s motion can be shown below:

5. A car moving along a straight highway with speed of \[126\text{ }km/hr\] is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

Initial velocity of the car is \[u=126\text{ }km/hr=35m/s\]. 

Final velocity of the car is $v=0km/hr$.

Distance covered by the car before coming to rest is \[200\text{ }m\].

Consider retardation produced in the car $=a$

From third equation of motion, ${{v}^{2}}-{{u}^{2}}=2as$

Therefore, $-{{35}^{2}}=2a\left( 200 \right)$

$a=-3.0625m/{{s}^{2}}$

From first equation of motion, time (t) taken by the car to stop can be obtained as:

$0=35-\left( 3.065 \right)t$

Hence, the retardation of the car (assumed uniform) is $3.0625m/{{s}^{2}}$, and it takes $11.44s$ for the car to stop.

6. A player throws a ball upwards with an initial speed of \[29.4\text{ }m/s\].

a. What is the direction of acceleration during the upward motion of the ball?

Ans: Acceleration of the ball (which is actually acceleration due to gravity) always acts in the downward direction towards the centre of the Earth, irrespective of the direction of the motion of the ball.

b. What are the velocity and acceleration of the ball at the highest point of its motion?

Ans: Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e.,n\[g\text{ }=\text{ }9.8\text{ }m/{{s}^{2}}\]. At maximum height, velocity of the ball becomes zero.

c. Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.

Ans: The sign of position is positive, sign of velocity is negative, and sign of acceleration is positive during upward motion. During downward motion, the signs of position, velocity, and acceleration are all positive.

d. To what height does the ball rise and after how long does the ball return to the player’s hands? (Take \[g\text{ }=\text{ }9.8\text{ }m/{{s}^{2}}\] and neglect air resistance).

Ans: Initial velocity of the ball, \[u\text{ }=\text{ }29.4\text{ }m/s\].

Final velocity of the ball, \[v\text{ }=\text{ }0\] (At maximum height, the velocity of the ball becomes zero)

Acceleration, \[a=g\text{ }=9.8\text{ }m/{{s}^{2}}\]

From first equation of motion, 

$0=-29.4+\left( 9.8 \right)t$

$t=\frac{29.4}{9.8}=3s$

\[Time\text{ }of\text{ }ascent\text{ }=\text{ }Time\text{ }of\text{ }descent\]

Hence, the total time taken by the ball to return to the player’s hands\[=\text{ }3\text{ }+\text{ }3\text{ }=\text{ }6\text{ }s\].

7. Read each statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion

a. with zero speed at an instant may have non-zero acceleration at that instant

Ans: The above statement is true. When an object is thrown vertically up in the air, its speed becomes zero at maximum height. It has acceleration equal to the acceleration due to gravity (g) Which acts in the downward direction at that point.

b. with zero speed may have non-zero velocity,

Ans: The above statement is false as speed is the magnitude of velocity. If speed is zero, the magnitude of velocity along with the velocity is zero.

c. with constant speed must have zero acceleration,

Ans: The above statement is true. If a car is moving on a straight highway with constant speed, it will have constant velocity. Acceleration is defined as the rate of change of velocity. Hence, the acceleration of the car is also zero.

d. with a positive value of acceleration must be speeding up.

Ans: The above statement is false. If acceleration is positive and velocity is negative at the instant time is taken as origin. Thus, for all the time before velocity becomes zero, there is slowing down of the particle. This case occurs when a particle is projected upwards. This statement will be true when both velocity and acceleration are positive, at that instant time taken as origin. This case happens when a particle is moving with positive acceleration or falling vertically downwards from a height.

8. A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between \[\mathbf{t}\text{ }=\text{ }\mathbf{0}\text{ }\mathbf{to}\text{ }\mathbf{12}\text{ }\mathbf{s}\].

Ball is dropped from a height is \[s\text{ }=\text{ }90\text{ }m\].

Initial velocity of the ball is \[u\text{ }=\text{ }0\].

Acceleration is $a=g=9.8m/{{s}^{2}}$.

Final velocity of the ball to be v.

Using second equation of motion, time (t) taken by the ball to hit the ground can be obtained

$s=ut+\left( 1/2 \right)a{{t}^{2}}$

$90=0+\left( 1/2 \right)9.8{{t}^{2}}$

$t=\sqrt{18.38}=4.29s$

Using, first equation of motion, final velocity is given as:

$v=0+9.8\left( 4.29 \right)=42.04m/s$

Rebound velocity of the ball is calculated as:

${{u}_{r}}=9v/10=9\left( \frac{42.04}{10} \right)=37.84m/s$

Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:

$v={{u}_{r}}+at'$

$0=37.84+\left( -9.8 \right)t'$

$t'=37.84/9.8=3.86s$

Total time taken by the ball will be $t+t'=4.29+3.86=8.15s$.

As the time of ascent is equal to the time of descent, the ball takes \[3.86\text{ }s\] to strike back on the floor for the second time.

The velocity with which the ball rebounds from the floor will be ${{u}_{r}}=9v/10=9\left( \frac{37.84}{10} \right)=34.05m/s$

Total time taken by the ball for second rebound will be $t+t'=8.15+3.86=12.01s$

The speed-time graph of the ball is represented in the given figure as:

9. Explain clearly, with examples, the distinction between:

a. magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;

Ans: The shortest distance (which is a straight line) between the initial and final positions of the particle gives the magnitude of displacement over an interval of time. The total path length of a particle is the actual path length covered by the particle in a given interval of time. For example, suppose a particle moves from point A to point B and then comes back to a point, C taking a total time t, as shown below. Then, the magnitude of displacement of the particle is AC.

Whereas, total path length \[=\text{ }AB\text{ }+\text{ }BC\]

We know that the magnitude of displacement can never be greater than the total path length. But, in some cases, both quantities are equal to each other.

b. magnitude of average velocity over an interval of time, and the average speed over the same interval. Average speed of a particle over an interval of time is defined as the total path length divided by the time interval. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? (For simplicity, consider one-dimensional motion only).

Ans: We know that: 

\[Magnitude\text{ }of\text{ }average\text{ }velocity\text{ }=\text{ }Magnitude\text{ }of\text{ }displacement\text{ }/\text{ }Time\text{ }interval\]

Hence, for the given particle,

Average velocity $=AC/t$

\[Average\text{ }speed\text{ }=\text{ }Total\text{ }path\text{ }length\text{ }/\text{ }Time\text{ }interval\]

\[=\left( AB+BC \right)/t\]

Since, \[AB+BC>AC\], average speed is greater than the magnitude of average velocity. The two quantities will be equal if the particle continues to move along a straight line.

10. A man walks on a straight road from his home to a market 2.5 km away with a speed of \[5\text{ }km/hr\]. Finding the market closed, he instantly turns and walks back home with a speed of \[7.5\text{ }km/hr\].What is the

(a) magnitude of average velocity, and

(b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ? (Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero)

Time taken by the man to reach the market from home is ${{t}_{1}}=2.5/5=1/2hr=30\min $.

Time taken by the man to reach home from the market is ${{t}_{2}}=2.5/7.5=1/3hr=20\min $. 

Total time taken in the whole journey \[=\text{ }30\text{ }+\text{ }20\text{ }=\text{ }50\text{ }min\].

0 to 30 min

\[Average\text{ }velocity\text{ }=\text{ }Displacement/Time\]

\[Average\text{ }speed\text{ }=\text{ }Distance/Time\]

0 to 50 min

\[Time\text{ }=\text{ }50\text{ }min\text{ }=\text{ }50/60\text{ }=\text{ }5/6\text{ }h\]

\[Net\text{ }displacement\text{ }=\text{ }0\]

\[Total\text{ }distance\text{ }=\text{ }2.5\text{ }+\text{ }2.5\text{ }=\text{ }5\text{ }km\]

\[Average\text{ }velocity\text{ }=\text{ }Displacement\text{ }/\text{ }Time\text{ }=\text{ }0\]

\[Average\text{ }speed\text{ }=\text{ }Distance\text{ }/\text{ }Time\text{ }=\text{ }5/\left( 5/6 \right)\text{ }=\text{ }6\text{ }km/h\]

0 to 40 min

\[Speed\text{ }of\text{ }the\text{ }man\text{ }=\text{ }7.5\text{ }km/h\]

\[Distance\text{ }travelled\text{ }in\text{ }first\text{ }30\text{ }min\text{ }=\text{ }2.5\text{ }km\]

Distance travelled by the man (from market to home) in the next 10 min

\[=\text{ }7.5\text{ }\times \text{ }10/60\text{ }=\text{ }1.25\text{ }km\]

\[Net\text{ }displacement\text{ }=\text{ }2.5\text{ }\text{ }1.25\text{ }=\text{ }1.25\text{ }km\]

\[Total\text{ }distance\text{ }travelled\text{ }=\text{ }2.5\text{ }+\text{ }1.25\text{ }=\text{ }3.75\text{ }km\]

\[Average\text{ }velocity\text{ }=\text{ }Displacement\text{ }/\text{ }Time\text{ }=\text{ }1.25\text{ }/\text{ }\left( 40/60 \right)\text{ }=\text{ }1.875\text{ }km/h\]

\[Average\text{ }speed\text{ }=\text{ }Distance\text{ }/\text{ }Time\text{ }=\text{ }3.75\text{ }/\text{ }\left( 40/60 \right)\text{ }=\text{ }5.625\text{ }km/h\].

11. In Exercises 13 and 14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?

Ans: We know that instantaneous velocity is the first derivative of distance with respect to time.

Here, the time interval is so small that it is assumed that the particle does not change its direction of motion. Therefore, both the total path length and magnitude of displacement become equal in this interval of time. Thus, instantaneous speed is always equal to instantaneous velocity.

12. Look at the graphs (a) to (d) (figure) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

a. Consider the x-t graph, given in fig (a). It does not represent the one-dimensional motion of the particle. This is because a particle cannot have two positions at the same instant of time.

b. Consider the x-t graph, given in fig (b). It does not represent the one-dimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time.

c. Consider the x-t graph, given in fig (c). It does not represent the one-dimensional motion of the particle. This is because speed being a scalar quantity cannot be negative.

d. Consider the x-t graph, given in fig (d). It does not represent the one-dimensional motion of the particle. This is because the total path length travelled by the particle cannot decrease with time.

13. Figure shows the x-t plot of the one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for $t<0$ and on a parabolic path for $t>0$? If not, suggest a suitable physical context for this graph.

Ans: No, this is because the x-t graph does not represent the trajectory of the path followed by a particle. Also from the graph, it is clear that at \[t=0,\text{ }x=0\].

14. A police van moving on a highway with a speed of \[30\text{ }km/hr\] fires a bullet at a thief’s car speeding away in the same direction with a speed of \[192\text{ }km/hr\]. If the muzzle speed of the bullet is \[150\text{ }m/s\], with what speed does the bullet hit the thief’s car? (Note: Obtain that speed which is relevant for damaging the thief’s car).

Speed of the police van is \[{{v}_{p}}=30\text{ }km/hr=8.33m/s\].

Muzzle speed of the bullet is \[{{v}_{b}}=150\text{ }m/s\].

Speed of the thief’s car is \[{{v}_{t}}=192\text{ }km/hr=53.33m/s\].

As the bullet is fired from a moving van, its resultant speed will be: $150+8.33=158.33m/s$.

Since both the vehicles are moving in the same direction, the velocity with which the bullet hits the thief’s car can be obtained as:

${{v}_{bt}}={{v}_{b}}-{{v}_{t}}=158.33-53.33=105m/s$

15. Suggest a suitable physical situation for each of the following graphs (figure):

Ans: Consider fig 3.22 given in the question:

a. From the x-t graph given it is clear that initially a body was at rest. Further, its velocity increases with time and attains an instantaneous constant value. The velocity then reduces to zero with an increase in time. Further, its velocity increases with time in the opposite direction and acquires a constant value. A similar physical situation arises when a football (initially kept at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced. Then, it passes from the player who has kicked it and ultimately stops after sometime.

b. From the given v-t graph it is clear that the sign of velocity changes and its magnitude decreases with a passage of time. This type of situation arises when a ball is dropped on the hard floor from a height. It strikes the floor with some velocity and upon rebound, its velocity decreases by a factor. This continues till the velocity of the ball eventually becomes zero.

c. From the given a-t graph it is clear that initially the body is moving with a certain uniform velocity. Its acceleration increases for a short interval of time, which again drops to zero. This shows that the body again starts moving with the same constant velocity. This type of physical situation arises when a hammer moving with a uniform velocity strikes a nail.

16. Figure gives the x-t plot of a particle executing one-dimensional simple harmonic motion.

(You will learn about this motion in more detail in Chapter14). Give the signs of position, velocity and acceleration variables of the particle at $t=0.3s,1.2s,-1.2s$.

Ans: Negative, Negative, Positive

Positive, Positive, Negative

Negative, Positive, Positive

When a particle executes simple harmonic motion (SHM), acceleration (a) is given by the relation: $a=-{{\omega }^{2}}x$

$\omega $ is the angular frequency ... 

i. \[~t\text{ }=\text{ }0.3\text{ }s\]

For this time interval, x is negative. Hence, the slope of the x-t plot will be negative. Thus, both position and velocity are negative. However, using equation (i), acceleration of the particle will be positive.

ii. \[~t\text{ }=1.2s\]

For this time interval, x is positive. Hence, the slope of the x-t plot will be positive. Thus, both position and velocity are positive. However, using equation (i), acceleration of the particle comes to be negative.

iii. \[\text{t }=-1.2s\]

For this time interval, x is negative. Hence, the slope of the x-t plot will be negative. Thus, both x and t are negative, the velocity comes to be positive. From equation (i), it can be interpreted that the acceleration of the particle will be positive.

17. Figure gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.

Ans: The average speed is greatest in interval 3 and least in interval 2. It is positive in intervals 1 & 2 and negative in interval 3.

The average speed of a particle shown in the x-t graph is given by the slope of the graph in a particular interval of time.

From the graph it is clear that the slope is maximum and minimum restively in intervals 3 and 2 respectively. Thus, the average speed of the particle is the greatest in interval 3 and is the least in interval 2. The sign of average velocity is positive in both intervals 1 and 2 as the slope is positive in these intervals. However, it is negative in interval 3 because the slope is negative in this interval.

18. Figure gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D?

Ans: From the graph given in the question,

Average acceleration is greatest in interval \[2\].

Average speed is greatest in intervals of $3$.

v is positive for intervals \[1,\text{ }2\], and $3$.

a is positive for intervals $1$ and $3$ and negative in interval $2$

\[a\text{ }=\text{ }0\] at A, B, C, D

Acceleration is calculated as the slope of the speed-time graph. In the given case, it is given by the slope of the speed-time graph within the given interval of time.

As the slope of the given speed-time graph is maximum in interval \[2\], average acceleration will be the greatest in this interval.

From the time-axis, height of the curve gives the average speed of the particle. It is clear that the height is the greatest in interval 3. Thus, average speed of the particle is the greatest in the interval 3.

For interval 1:

The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.

In interval 2:

As slope of the speed-time graph is negative, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.

In interval 3:

As the slope of the speed-time graph is zero, acceleration is zero in this interval.

However, here the particle acquires some uniform speed. It is positive in this interval.

Points A, B, C, and D are all parallel to the time-axis. Thus, the slope is zero at these points.

Therefore, at points A, B, C, and D, acceleration of the particle is zero.

Motion in a Straight Line Chapter Summary - Class 11 NCERT Solutions

Path length is defined as the total length of the path traversed by an object.

Displacement is the change in position: $\Delta x=x_{2}-x_{1} $ Path length is greater or equal to the magnitude of the displacement between the same points.

An object is said to be in uniform motion in a straight line if its displacement is equal in equal intervals of time. Otherwise, the motion is said to be nonuniform

Average velocity is the displacement divided by the time Interval in which the displacement occurs: $ v\bar{}=\frac{\Delta x}{\Delta t}

On an x-t graph. The average velocity over a time Interval is the slope of the line connecting the initial and final positions corresponding to that interval.

Average Speed is the ratio of total path length traversed and the corresponding time Interval.

Instantaneous velocity or simply velocity is defined as the limit of the average velocity as the time interval $\Delta t$ becomes infinitesimally small $v=\lim_{\Delta t\rightarrow 0}v\bar{}=\lim_{\Delta t\rightarrow 0}\frac{\Delta x}{\Delta t}=\frac{dx}{dt}$

The velocity at a particular instant is equal to the slope of the tangent drawn on position-time graph at that instant.

Average acceleration is the change in velocity divided by the time interval during which the change occurs: $a\vec{}=\frac{\Delta v}{\Delta t}$

Instantaneous acceleration is defined as the limit of the average acceleration as the time interval at goes to zero: $a=\lim_{\Delta \rightarrow t}a\vec{}=\lim_{\Delta t\rightarrow 0}\frac{\Delta t}{\Delta v}=\frac{dv}{dt}$

The acceleration of an object at a particular time is the slope of the velocity-time graph at that instant of time. For uniform motion. acceleration is zero and the x-t graph is a straight line inclined to the time axis and the v-t graph is a straight line parallel to the time axis. For motion with uniform acceleration. x - t graph is a parabola while the v-t graph is a straight line inclined to the time axis.

The area under the velocity-time curve between times t 1 and t 2 is equal to the displacement of the object during that interval of time.

For objects in uniformly accelerated rectilinear motion. the five quantities, displacement x, time taken t. Initial velocity v 0,  final velocity v and acceleration a are related by a set of simple equations called kinematic equations of motion: v=v 0 +at x=v 0 t+$\frac{1}{2}$at 2 v 2 =$v_{0}^{2}$+2ax

For Uniform Motion

For Non-Uniform Motion

Overview of Deleted Syllabus for CBSE Class 11 Physics Motion in a Straight Line

NCERT Class 11 Physics Chapter 2 Exercise Solutions on Motion In A Straight Line provided by Vedantu are comprehensive, and helps students understand motion and the different kinds of motion. This chapter explores different types of motion, including uniform motion, non-uniform motion, and uniformly accelerated motion. In Ch 2 physics class 11 ncert solutions we learn how to solve problems involving these concepts using various equations relating to distance, time, speed, velocity, and acceleration. These concepts are essential for mastering the topic and are often tested in exams. From previous year's question papers, typically around 2–3 questions are asked from this chapter. These questions test students' grasp of theoretical concepts as well as their problem-solving skills.

Other Study Material for CBSE Class 11 Physics Chapter 2

Chapter-Specific NCERT Solutions for Class 11 Physics

Given below are the chapter-wise NCERT Solutions for Class 11 Physics. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 11 Physics Chapter 2 Motion in a Straight Line

1. If I'm not a Non-Medical Student, do I Still Need to Learn and Take Physics as One of the Subjects?

The answer to this question is both yes and no, you can opt for Physics even if you are not a non-medical student, but the only other branch which offers Physics as one of its subjects is medical. In that branch, you don't have maths as the main subject, but you can opt for maths as well. On the other hand, if you are from commerce or arts, you don't choose Physics. You can't opt for Physics as it's one of the core subjects and isn't easy. In the medical stream, you can opt for Physics.

2. If an Object has One Dimension Motion with a Positive Acceleration Value and Negative Velocity, is the Object Speeding up?

No, in this case, the object present in the starting point is not speeding up. As the velocity is negative, we can see that the particle is slowing down. This type of scenario happens only when the object is thrown upwards in the air. We can say the object is in acceleration if both the acceleration and the object's velocity have a positive value, or the object is falling from a height.

3. What are the contents of Chapter 3 of the Class 11 Physics textbook?

The theme of the chapter is Kinematics, motion in a straight line. Mechanics is a scientific term that refers to the study of the movement of physical objects. The chapter explains all the relevant terms in a simple and easy-to-understand language. The chapter distinguishes between rest and motion. You will also come across the terms position, distance, and displacement as well as the difference between speed and velocity. The chapter explains acceleration. The chapter also contains the units and mathematical representations.

4. What is the difference between distance and displacement according to Chapter 3 of the Class 11 Physics textbook?

Distance means the actual path traversed by the object whereas displacement refers to the difference between initial and final positions. The former is the scalar quantity while the latter is the vector quantity. The distance covered by an object is always positive; it can never be negative or zero. However, the displacement of an object can be either positive, negative, or even zero. The distance is dependent on the path traversed by the object but displacement is not dependent on the path.

5. What is the difference between speed and velocity Chapter 3 of the Class 11 Physics textbook?

Speed is the total path length covered by the object divided by the total time taken. Velocity is the change in position divided by time intervals. Speed is a scalar quantity. Velocity, however, is a vector quantity. Speed is always positive. Velocity, on the other hand, can be positive, negative, or zero. Speed is the time rate at which an object moves; speed is about how fast an object moves. Velocity is the rate and direction of the movement of an object.  

6. What are the types of motion mentioned in Chapter 3 of the Class 11 Physics textbook?

The motion refers to changes in the position or orientation of an object with the change in time. Motion can be of two types- Uniform motion or Non-Uniform motion. Uniform motion exists when the object completes equal distance in equal time intervals. Non-Uniform motion, on the other hand, exists when an object completes unequal distance in equal intervals of time. During a uniform motion, the velocity of the object remains the same. When an object undergoes non-uniform motion, the magnitude of the velocity increases or decreases with time.

7. How can previous years’ questions help in preparation of Motion in a straight line chapter of Class 11 Physics?

Previous years’ questions enable you to understand the pattern of the question and analyse the trend of the questions asked. These papers will give you an idea regarding the topics that you should invest more time and effort on. These question papers should also motivate you to make notes accordingly. The previous years’ question papers will lend a direction to your preparation and you will feel more prepared and confident on the day of the exam. These papers are available on the official website of  Vedantu  or the Vedantu app. All the resources are free of cost.

8. In Motion In A Straight Line Class 11 NCERT Solutions, which motion travels in a straight line?

In Class 11 Physics Chapter 2, Motion In A Straight Line refers to the movement of an object along a straight path. This type of motion is characterized by having only one spatial dimension. Examples include a car driving on a straight road, a ball rolling down a straight hill, or an object falling freely under gravity (neglecting air resistance).

9. What are the two motions of a straight line mentioned in Physics Class 11 Chapter 2?

The two types of motion in a straight line are:

Uniform Motion : When an object moves with a constant speed in a straight line, covering equal distances in equal intervals of time.

Non-uniform Motion : When an object's speed varies as it moves along a straight line, covering unequal distances in equal intervals of time.

10. What are the important things in Motion In A Straight Line Class 11 Questions?

Important aspects of motion in a straight line include:

Displacement : The change in position of an object, considering direction.

Velocity : The rate of change of displacement with time.

Speed : The rate at which an object covers distance.

Acceleration : The rate of change of velocity with time.

Kinematic Equations : These equations relate displacement, velocity, acceleration, and time.

11. In Class 11 Motion In A Straight Line NCERT Solutions, Is motion always in a straight line?

No, motion is not always in a straight line. While this chapter focuses on straight-line motion, objects can move in various types of paths, such as circular motion, projectile motion, or any arbitrary path in two or three dimensions.

12. What is another name for a straight line motion in Motion In Straight Line Class 11 NCERT Solutions?

Another name for straight-line motion is rectilinear motion. This term is often used in physics to describe motion that occurs along a straight path.

13. What is the other name of motion in a straight line in Class 11 Physics Chapter 2 Exercise Solutions?

The other name for motion in a straight line is linear motion. This term highlights the one-dimensional aspect of the motion, indicating that it occurs along a single straight axis.

NCERT Solutions for Class 11 Physics

Class 11 Physics Case Study Questions Chapter 2 Units and Measurements

• Post author: studyrate
• Post published:
• Post category: Class 11
• Post comments: 0 Comments

In Class 11 Final Exams there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Downloads of CBSE Class 11 Physics Chapter 2 Case Study and Passage-Based Questions with Answers were Prepared Based on the Latest Exam Pattern. Students can solve Class 11 Physics Case Study Questions Units and Measurements  to know their preparation level.

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

In CBSE Class 11 Physics Paper, There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Units and Measurements Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 11 Physics  Chapter 2 Units and Measurements

Case Study/Passage-Based Questions

Case Study 1: The rules for determining the uncertainty or error in the measured quantity in arithmetic operations can be understood from the following examples.

a.)  If the length and breadth of a thin rectangular sheet are measured, using a meter scale as 16.2 cm and, 10.1 cm respectively, there are three significant figures in each measurement. It means that the length L may be written as L = 16.2 ± 0.1 cm   = 16.2 cm ± 0.6 %.

Similarly, the breadth b may be written as b = 10.1 ± 0.1 cm = 10.1 cm ± 1 %

Then, the error of the product of two (or more) experimental values, using the combination of errors rule, will be L*b = 163.62 cm2 + 1.6% = 163.62 + 2.6 cm2

This leads us to quote the final result as L*b = 164 + 3 cm 2 . Here 3 cm 2  is the uncertainty or error in the estimation of the area of the rectangular sheet.

b.)  If a set of experimental data is specified to n significant figures a result obtained by combining the data will also be valid to n significant figures. However, if data are subtracted, the number of significant figures can be reduced. For example, 12.9 g – 7.06 g, both specified to three significant figures, cannot properly be evaluated as 5.84 g but only as 5.8 g, as uncertainties in subtraction or addition combine in a different fashion (smallest number of decimal places rather than the number of significant figures in any of the number added or subtracted).

c.)  The relative error of a value of a number specified to significant figures depends not only on n but also on the number itself. For example, the accuracy in the measurement of mass 1.02 g is ± 0.01 g whereas another measurement of 9.89 g is also accurate to ± 0.01 g. The relative error in 1.02 g is

= (± 0.01/1.02) × 100 % = ± 1%

Similarly, the relative error in 9.89 g is = (± 0.01/9.89) × 100 % = ± 0.1 %

Finally, remember that intermediate results in a multi-step computation should be calculated to one more significant figure in every measurement than the number of digits in the least precise measurement.

d.)  The nature of a physical quantity is described by its dimensions. All the physical quantities represented by derived units can be expressed in terms of some combination of seven fundamental or base quantities. We shall call these base quantities the seven dimensions of the physical world, which are denoted with square brackets [ ]. Thus, length has the dimension [L], mass [M], time [T], electric current [A], thermodynamic temperature [K], luminous intensity [cd], and amount of substance [mol]. The dimensions of a physical quantity are the powers (or exponents) to which the base quantities are raised to represent that quantity. Note that using the square brackets [ ] around a quantity means that we are dealing with ‘the dimensions of’ the quantity. In mechanics, all the physical quantities can be written in terms of the dimensions [L], [M], and [T]. For example, the volume occupied by an object is expressed as the product of length, breadth, and height, or three lengths. Hence the dimensions of volume are [L] × [L] × [L] = [L 3 ].

When measuring the length and breadth of a rectangular sheet as 16.2 cm and 10.1 cm respectively, what is the uncertainty in the length? a) 0.6% b) 0.1% c) 1% d) 0.6 cm

Answer: a) 0.6%

In the calculation of the area of a rectangular sheet, the final result is expressed as 164 + 3 cm². What does the 3 cm² represent? a) The length of the rectangular sheet b) The breadth of the rectangular sheet c) The uncertainty or error in the estimation d) The percentage error in the calculation

Answer: c) The uncertainty or error in the estimation

When dealing with subtraction of numbers like 12.9 g – 7.06 g, how should the result be properly evaluated? a) 5.84 g b) 5.8 g c) 5.9 g d) 6.0 g

Answer: b) 5.8 g

Which of the following accurately represents the relative error in the measurement of 9.89 g? a) ± 1% b) ± 0.1% c) ± 0.01% d) ± 0.9%

Answer: b) ± 0.1%

In a multi-step computation, intermediate results should be calculated to how many more significant figures than the least precise measurement? a) Two more significant figures b) One more significant figure c) The same number of significant figures d) Three more significant figures

Answer: b) One more significant figure

Which of the following represents the dimensions of volume? a) [L] b) [L^2] c) [L3] d) [L * M * T]

Answer: c) [L3]

How many fundamental or base quantities are used to express all physical quantities in derived units? a) Five b) Six c) Seven d) Eight

Answer: c) Seven

Case Study 2: Measurement of Physical Quantity : All engineering phenomena deal with definite and measured quantities and so depend on the making of the measurement. We must be clear and precise in making these measurements. To make a measurement, the magnitude of the physical quantity (unknown) is compared. The record of a measurement consists of three parts, i.e. the dimension of the quantity, the unit which represents a standard quantity, and a number which is the ratio of the measured quantity to the standard quantity.

(i) A device which is used for measurement of length to an accuracy of about 10 -5 m, is (a) screw gauge (b) spherometer (c) vernier calipers (d) Either (a) or (b)

Answer: (d) Either (a) or (b)

(ii) Which of the technique is not used for measuring time intervals? (a) Electrical oscillator (b) Atomic clock (c) Spring oscillator (d) Decay of elementary particles

Answer: (c) Spring oscillator

(iii) The mean length of an object is 5 cm. Which of the following measurements is most accurate? (a) 4.9 cm (b) 4.805 cm (c) 5.25 cm (d) 5.4 cm

Answer: (a) 4.9 cm

(iv) If the length of rectangle l = 10.5 cm, breadth b = 2.1 cm and minimum possible measurement by scale = 0.1 cm, then the area is (a) 22.0 cm 2 (b) 21.0 cm 2 (c) 22.5 cm 2 (d) 21.5 cm 2

Answer: (a) 22.0 cm2

(v) Age of the universe is about 10 10  yr, whereas the mankind has existed for 10 6  yr. For how many seconds would the man have existed, if age of universe were 1 day? (a) 9.2 s (b) 10.2 s (c) 8.6 s (d) 10.5 s

Answer: (c) 8.6 s

Hope the information shed above regarding Case Study and Passage Based Questions for Class 11 Physics Chapter 2 Units and Measurements with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 11 Physics Units and Measurements Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible. By Team Study Rate

You Might Also Like

Mcq questions class 11 business studies chapter 9 small business with answers, mcq questions class 11 english hornbill chapter 8 silk road with answers, mcq questions class 11 accountancy chapter 13 computerised accounting system with answers, leave a reply cancel reply.

Save my name, email, and website in this browser for the next time I comment.

This site uses Akismet to reduce spam. Learn how your comment data is processed .

The Topper Combo Flashcards

• Contains the Latest NCERT in just 350 flashcards.
• Colourful and Interactive
• Summarised Important reactions according to the latest PYQs of NEET(UG) and JEE

No thanks, I’m not interested!

myCBSEguide

• Mathematics
• Class 11 Mathematics Case...

Class 11 Mathematics Case Study Questions

Table of Contents

myCBSEguide App

Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

If you’re seeking a comprehensive and dependable study resource with Class 11 mathematics case study questions for CBSE, myCBSEguide is the place to be. It has a wide range of study notes, case study questions, previous year question papers, and practice questions to help you ace your examinations. Furthermore, it is routinely updated to bring you up to speed with the newest CBSE syllabus. So, why delay? Begin your path to success with myCBSEguide now!

The rationale behind teaching Mathematics

The general rationale to teach Mathematics at the senior secondary level is to assist students:

• In knowledge acquisition and cognitive understanding of basic ideas, words, principles, symbols, and mastery of underlying processes and abilities, notably through motivation and visualization.
• To experience the flow of arguments while demonstrating a point or addressing an issue.
• To use the information and skills gained to address issues using several methods wherever possible.
• To cultivate a good mentality in order to think, evaluate, and explain coherently.
• To spark interest in the subject by taking part in relevant tournaments.
• To familiarise pupils with many areas of mathematics utilized in daily life.
• To pique students’ interest in studying mathematics as a discipline.

Case studies in Class 11 Mathematics

A case study in mathematics is a comprehensive examination of a specific mathematical topic or scenario. Case studies are frequently used to investigate the link between theory and practise, as well as the connections between different fields of mathematics. A case study will frequently focus on a specific topic or circumstance and will investigate it using a range of methodologies. These approaches may incorporate algebraic, geometric, and/or statistical analysis.

Sample Class 11 Mathematics case study questions

When it comes to preparing for Class 11 Mathematics, one of the best things Class 11 Mathematics students can do is to look at some Class 11 Mathematics sample case study questions. Class 11 Mathematics sample case study questions will give you a good idea of the types of Class 11 Mathematics sample case study questions that will be asked in the exam and help you to prepare more effectively.

Looking at sample questions is also a good way to identify any areas of weakness in your knowledge. If you find that you struggle with a particular topic, you can then focus your revision on that area.

myCBSEguide offers ample Class 11 Mathematics case study questions, so there is no excuse. With a little bit of preparation, Class 11 Mathematics students can boost their chances of getting the grade they deserve.

Some samples of Class 11 Mathematics case study questions are as follows:

Class 11 Mathematics case study question 1

• 9 km and 13 km
• 9.8 km and 13.8 km
• 9.5 km and 13.5 km
• 10 km and 14 km
• x  ≤   −1913
• x <  −1613
• −1613  < x <  −1913
• There are no solution.
• y  ≤   12 x+2
• y >  12 x+2
• y  ≥   12 x+2
• y <  12 x+2

Answer Key:

• (b) 9.8 km and 13.8 km
• (a) −1913   ≤  x 
• (b)  y >  12 x+2
• (d) (-5, 5)

Class 11 Mathematics case study question 2

• 2 C 1 × 13 C 10
• 2 C 1 × 10 C 13
• 1 C 2 × 13 C 10
• 2 C 10 × 13 C 10
• 6 C 2​ × 3 C 4   × 11 C 5 ​
• 6 C 2​ × 3 C 4   × 11 C 5
• 6 C 2​ × 3 C 5 × 11 C 4 ​
• 6 C 2 ​  ×   3 C 1 ​  × 11 C 5 ​
• (b) (13) 4  ways
• (c) 2860 ways.

Class 11 Mathematics case study question 3

Read the Case study given below and attempt any 4 sub parts: Father of Ashok is a builder, He planned a 12 story building in Gurgaon sector 5. For this, he bought a plot of 500 square yards at the rate of Rs 1000 /yard². The builder planned ground floor of 5 m height, first floor of 4.75 m and so on each floor is 0.25 m less than its previous floor.

Class 11 Mathematics case study question 4

Read the Case study given below and attempt any 4 sub parts: villages of Shanu and Arun’s are 50km apart and are situated on Delhi Agra highway as shown in the following picture. Another highway YY’ crosses Agra Delhi highway at O(0,0). A small local road PQ crosses both the highways at pints A and B such that OA=10 km and OB =12 km. Also, the villages of Barun and Jeetu are on the smaller high way YY’. Barun’s village B is 12km from O and that of Jeetu is 15 km from O.

Now answer the following questions:

• 5x + 6y = 60
• 6x + 5y = 60
• (a) (10, 0)
• (b) 6x + 5y = 60
• (b) 60/√ 61 km
• (d) 2√61 km

A peek at the Class 11 Mathematics curriculum

The Mathematics Syllabus has evolved over time in response to the subject’s expansion and developing societal requirements. The Senior Secondary stage serves as a springboard for students to pursue higher academic education in Mathematics or professional subjects such as Engineering, Physical and Biological Science, Commerce, or Computer Applications. The current updated curriculum has been prepared in compliance with the National Curriculum Framework 2005 and the instructions provided by the Focus Group on Teaching Mathematics 2005 in order to satisfy the rising demands of all student groups. Greater focus has been placed on the application of various principles by motivating the themes from real-life events and other subject areas.

Class 11 Mathematics (Code No. 041)

Design of Class 11 Mathematics exam paper

CBSE Class 11 mathematics question paper is designed to assess students’ understanding of the subject’s essential concepts. Class 11 mathematics question paper will assess their problem-solving and analytical abilities. Before beginning their test preparations, students in Class 11 maths should properly review the question paper format. This will assist Class 11 mathematics students in better understanding the paper and achieving optimum scores. Refer to the Class 11 Mathematics question paper design provided.

 Class 11 Mathematics Question Paper Design

• No chapter-wise weightage. Care to be taken to cover all the chapters.
• Suitable internal variations may be made for generating various templates keeping the overall weightage to different forms of questions and typology of questions the same.  

Choice(s): There will be no overall choice in the question paper. However, 33% of internal choices will be given in all the sections.

  Prescribed Books:

• Mathematics Textbook for Class XI, NCERT Publications
• Mathematics Exemplar Problem for Class XI, Published by NCERT
• Mathematics Lab Manual class XI, published by NCERT

myCBSEguide guarantees LHS=RHS

With myCBSEguide it will always be LHS=RHS, Hence Proved!

myCBSEguide is a prominent streaming resource for students studying for CBSE examinations. The site offers extensive study material, practice papers, case study questions, online examinations, and other resources at the push of a click. myCBSEguide also has a mobile app for learning on the move. There’s no excuse not to try myCBSEguide with so much convenience and simplicity of access! Hence Proved!

Test Generator

Create question paper PDF and online tests with your own name & logo in minutes.

Question Bank, Mock Tests, Exam Papers, NCERT Solutions, Sample Papers, Notes

Related Posts

• Competency Based Learning in CBSE Schools
• Class 11 Physical Education Case Study Questions
• Class 11 Sociology Case Study Questions
• Class 12 Applied Mathematics Case Study Questions
• Class 11 Applied Mathematics Case Study Questions
• Class 11 Biology Case Study Questions
• Class 12 Physical Education Case Study Questions
• Class 12 Computer Science Case Study Questions

1 thought on “Class 11 Mathematics Case Study Questions”

teri meri meri teri prem kahani hai muskil dolabjo main sayana hop jaye ek ladka aur ek ladki ki prem kahani hai muskil

Leave a Comment

Save my name, email, and website in this browser for the next time I comment.

• DOI: 10.69593/ajahe.v4i03.84
• Corpus ID: 271079373

Assessing English Language Proficiency: A Case Study of English Version Students of Higher Secondary Level of Bangladesh

• Alia Rawshan Banu
• Published in ACADEMIC JOURNAL ON ARTS &amp… 8 July 2024
• Education, Linguistics
• ACADEMIC JOURNAL ON ARTS &amp; HUMANITIES EDUCATION

Related Papers

Showing 1 through 3 of 0 Related Papers

Gurukul of Excellence

Classes for Physics, Chemistry and Mathematics by IITians

Join our Telegram Channel for Free PDF Download

Download Case Study Questions for Class 11 Maths

• Last modified on: 1 year ago
• Reading Time: 9 Minutes

[PDF] Download Case Study Questions for Class 11 Maths

Here we are providing case study questions for Class 11 Maths. In this article, we are sharing links for Class 11 Maths All Chapters. All case study questions of Class 11 Maths are solved so that students can check their solutions after attempting questions.

Click on the chapter to view.

Class 11 Maths Chapters

Chapter 1 Sets Chapter 2 Relations and Functions Chapter 3 Trigonometric Functions Chapter 4 Principle of Mathematical Induction Chapter 5 Complex Numbers and Quadratic Equations Chapter 6 Linear Inequalities Chapter 7 Permutation and Combinations Chapter 8 Binomial Theorem Chapter 9 Sequences and Series Chapter 10 Straight Lines Chapter 11 Conic Sections Chapter 12 Introduction to Three-Dimensional Geometry Chapter 13 Limits and Derivatives Chapter 14 Mathematical Reasoning Chapter 15 Statistics Chapter 16 Probability

What is meant by Case Study Question?

In the context of CBSE (Central Board of Secondary Education), a case study question is a type of question that requires students to analyze a given scenario or situation and apply their knowledge and skills to solve a problem or answer a question related to the case study.

Case study questions typically involve a real-world situation that requires students to identify the problem or issue, analyze the relevant information, and apply their understanding of the relevant concepts to propose a solution or answer a question. These questions may involve multiple steps and require students to think critically, apply their problem-solving skills, and communicate their reasoning effectively.

Importance of Solving Case Study Questions for Class 11 Maths

Case study questions are an important aspect of mathematics education at the Class 11 level. These questions require students to apply their knowledge and skills to real-world scenarios, helping them develop critical thinking, problem-solving, and analytical skills. Here are some reasons why case study questions are important in Class 11 maths education:

• Real-world application: Case study questions allow students to see how the concepts they are learning in mathematics can be applied in real-life situations. This helps students understand the relevance and importance of mathematics in their daily lives.
• Higher-order thinking: Case study questions require students to think critically, analyze data, and make connections between different concepts. This helps develop higher-order thinking skills, which are essential for success in both academics and real-life situations.
• Collaborative learning: Case study questions often require students to work in groups, which promotes collaborative learning and helps students develop communication and teamwork skills.
• Problem-solving skills: Case study questions require students to apply their knowledge and skills to solve complex problems. This helps develop problem-solving skills, which are essential in many careers and in everyday life.
• Exam preparation: Case study questions are included in exams and tests, so practicing them can help students prepare for these assessments.

Overall, case study questions are an important component of Class 11 mathematics education, as they help students develop critical thinking, problem-solving, and analytical skills, which are essential for success in both academics and real-life situations.

Feature of Case Study Questions on This Website

Here are some features of a Class 11 Maths Case Study Questions Booklet:

Many Case Study Questions: This website contains many case study questions, each with a unique scenario and problem statement.

Different types of problems: The booklet includes different types of problems, such as optimization problems, application problems, and interpretation problems, to test students’ understanding of various mathematical concepts and their ability to apply them to real-world situations.

Multiple-choice questions: Questions contains multiple-choice questions to assess students’ knowledge, understanding, and critical thinking skills.

Focus on problem-solving skills: The questions are designed to test students’ problem-solving skills, requiring them to identify the problem, select appropriate mathematical tools, and analyze and interpret the results.

Emphasis on practical applications: The case studies in the booklet focus on practical applications of mathematical concepts, allowing students to develop an understanding of how mathematics is used in real-life situations.

Comprehensive answer key: The booklet includes a comprehensive answer key that provides detailed explanations and step-by-step solutions for all the questions, helping students to understand the concepts and methods used to solve each problem.

Download CBSE Books

Exam Special Series:

• Sample Question Paper for CBSE Class 10 Science (for 2024)
• Sample Question Paper for CBSE Class 10 Maths (for 2024)
• CBSE Most Repeated Questions for Class 10 Science Board Exams
• CBSE Important Diagram Based Questions Class 10 Physics Board Exams
• CBSE Important Numericals Class 10 Physics Board Exams
• CBSE Practical Based Questions for Class 10 Science Board Exams
• CBSE Important “Differentiate Between” Based Questions Class 10 Social Science
• Sample Question Papers for CBSE Class 12 Physics (for 2024)
• Sample Question Papers for CBSE Class 12 Chemistry (for 2024)
• Sample Question Papers for CBSE Class 12 Maths (for 2024)
• Sample Question Papers for CBSE Class 12 Biology (for 2024)
• CBSE Important Diagrams & Graphs Asked in Board Exams Class 12 Physics
• Master Organic Conversions CBSE Class 12 Chemistry Board Exams
• CBSE Important Numericals Class 12 Physics Board Exams
• CBSE Important Definitions Class 12 Physics Board Exams
• CBSE Important Laws & Principles Class 12 Physics Board Exams
• 10 Years CBSE Class 12 Chemistry Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Physics Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Maths Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Biology Previous Year-Wise Solved Papers (2023-2024)
• ICSE Important Numericals Class 10 Physics BOARD Exams (215 Numericals)
• ICSE Important Figure Based Questions Class 10 Physics BOARD Exams (230 Questions)
• ICSE Mole Concept and Stoichiometry Numericals Class 10 Chemistry (65 Numericals)
• ICSE Reasoning Based Questions Class 10 Chemistry BOARD Exams (150 Qs)
• ICSE Important Functions and Locations Based Questions Class 10 Biology
• ICSE Reasoning Based Questions Class 10 Biology BOARD Exams (100 Qs)

✨ Join our Online NEET Test Series for 499/- Only for 1 Year

Leave a Reply Cancel reply

Editable Study Materials for Your Institute - CBSE, ICSE, State Boards (Maharashtra & Karnataka), JEE, NEET, FOUNDATION, OLYMPIADS, PPTs

Discover more from Gurukul of Excellence

Subscribe now to keep reading and get access to the full archive.

Type your email…

Continue reading