# NCERT Solutions for Class 11 Physics Chapter 3

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

Every aspect of our life is actually governed by a physical law or concept. Physics has explanations of everything from the beginning of the universe to life that is possible today. Students in class 11 are introduced to some fundamental concepts of physics and laws.

A change in the position of an object is referred to as motion. This change also occurs in relation to the object’s surroundings. Everything in the Universe is in motion. Chapter 3 Physics Class 11, in particular, is a part of kinetics. This section of the chapter contains information about rectilinear motion, reference point and frame of reference. In addition, the chapter discusses acceleration, average velocity, instantaneous velocity, and speed.

To aid students with understanding Chapter 3 better, Extramarks offers NCERT Solutions for Class 11 Physics Chapter 3 – Motion in a Straight Line. Students can download these solutions for free from the website and prepare better for the exams.

## NCERT Solutions for Class 11 Physics Chapter 3

Kinetics is introduced in this section. The topics including Rectilinear Motion, Reference Points, Path Lengths, Relative Velocity, Acceleration, and other intriguing concepts are discussed in Chapter 3 of NCERT Class 11 Physics.

It is necessary for you to be familiar with the topics and sub-topics of Chapter 3 Physics Class 11 before you begin your Class 11 Physics preparation. Here’s a complete list of the topics covered in Chapter 3:

 Section Number Title Name 3 Motion in a Straight Line 3.1 Introduction 3.2 Position, Path Length and Displacement 3.3 Average Velocity and Average Speed 3.4 Instantaneous Velocity and Speed 3.5 Acceleration 3.6 Kinematic Equations for Uniformly Accelerated Motion 3.7 Relative Velocity

3.1 Introduction

This section provides a general overview of the subject. The sub-topics include:

3.2 Path length, position, and displacement

This subunit looks at three different aspects of motion. These aspects are, in particular, position, path length, and displacement.

3.3 Average velocity and average speed

Differences in displacement are caused by average velocity divided by time intervals. Furthermore, it is during these time intervals that the displacement occurs. Average speed, on the other hand, describes the rate of motion along the actual path.

3.4 Instantaneous velocity and speed

As the time interval gets infinitely small, the velocity of an instant is the limit of the average velocity. This contrasts sharply with instantaneous speed. The magnitude of velocity is referred to as instantaneous speed.

3.5 Acceleration

This is the rate at which an object’s velocity changes with respect to time.

3.6 Equations of motion for uniformly accelerated motion

This section covers the various aspects of motion through simple equations.

3.7 Relative velocity

Relative velocity is the velocity at which one object moves with respect to the other.

The topics covered in NCERT Solutions Class 11 Physics Chapter 3 are important for students in Class 11 because they lay the groundwork for concepts that will be discussed in Class 12 and higher-level studies. Understanding fundamental topics like the one covered in this chapter is essential for understanding advanced physics concepts.

Class 11 Physics Chapter 3 NCERT Solutions assist students in learning more effectively and comprehending the fundamental concepts of physics.

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

Motion is a topic that gets a lot of attention. It is necessary in every field of science. In addition, the chapter is critical for automobile manufacturers. This is due to the importance of acceleration, speed, and velocity in automobiles. Engineers also frequently employ motion-related concepts.

After finishing the chapter, you should be able to answer the questions about Motion in a Straight Line. You’ll also understand how to use the distance-time graph to solve problems involving the speed and velocity of objects at various points in time. Students will benefit from NCERT Solutions Class 11 Physics Chapter 3 as they work through the NCERT questions.

This chapter is heavily reliant on numerical and derivative questions. Derivations should be practised multiple times to ensure that all concepts are understood. Not only is NCERT Chapter 11 Motion in a Straight Line important for getting good grades in school, but questions from this chapter are also asked in JEE Main, AIIMS, BITSAT, and NEET entrance exams.

NCERT Chapter 3 Marks Distribution

The maximum number of questions from this chapter is two, and the total number of marks you can score in this chapter in your final term examination is five.

### Benefits of Solving CBSE Class 11 Physics Chapter 3

Students can use the NCERT Solutions for Class 11 Physics Chapter 3 to get simple, step-by-step answers to the questions in the textbook. The answers are useful in understanding how to answer questions in the first term exams. Students can save time and speed up their revision, which is ideal for preparing for the term I exam. Here are some benefits of studying Chapter 3 of Class 11 Physics:

Understanding the World

Studying physics can be personally rewarding as students begin to comprehend everyday objects in terms of the concepts that underpin them. Physicists have been able to explain how the world works, all the way back to the beginning of time. Research findings can be used to develop new technology.

Bringing Other Disciplines Together

Physics is not a self-contained discipline. It is used in a wide range of fields. Medical students, for example, must understand basic physics concepts such as pressure, the velocity of flow, and changes in resistance to flow in order to understand how blood and air flow in the body.

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An individual’s ability to think outside the box and apply a variety of approaches to solve a problem is highly valued in Science. Having and honing problem-solving skills is beneficial not only for your studies but also to your job search once you’ve completed your degree. A degree in Physics is an excellent way to develop strong analytical skills, which are highly valued by many employers.

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Q.1 The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29 :

(a) x(t2) = x(t1) + v(t1)t2- t1 + 12at2 – t12(b) v(t2) = v(t1) + at2 – t1(c) vav = {r(t2) – r(t1)}t2 – t1(d) aav = {v(t2) – v(t1)}t2 – t1(e)  x(t2) = x(t1) + vav(t2 – t1) + 12aavt2 – t12(f) x(t2) – x(t1) = area under the v-t curve bounded by the time axis and dotted line shown.

Ans.

The given plot has a non-uniform slope, therefore the relations given in (a), (b) and (e) are not correct. However, relations given in (c), (d) and (f) are correct.

Q.2 The speed-time graph of a particle moving along a fixed direction is as shown in figure 3.28. Obtain the distance travelled by the particle between (a) T=0 s to 10 s, (b) T=2 s to 6 s.

What is the average speed of the particle over the intervals in (a) and (b)?

Ans.

(a) Distance covered by the particle between 0 to 10 s = Area under the given graph Area under the given graph = 1 2 ×10 ms -1 ×12 s = 60 m Average speed of the  particle, v avg = Distance Time v avg = 60 10 ms -1 = 6  ms -1

(b) Let  s 1  and  s 2  be the distances traveled by the particle in time  t 1 = 2 s to 5 s and  t 2 = 5 s to 6 s respectively. ∴Total distance (s) traveled by the particle in time interval t = 2 s to 6 s, s = s 1 + s 2 …(i) To find distance  s 1 : Let us consider  u 1  is the velocity of the particle after 2s and  a 1  is the acceleration of the particle during the time interval 0 to 5 s. As the particle is in uniform acceleration in the interval t = 0 to t = 5 s,  According to the first equation of motion, we obtain: v = u + at  12 ms -1 = 0 + a 1 ×5 s ∴ a 1 = 12 5 ms -2 = 2.4  ms -2 Again, using first equation of motion, we get u 1 = u + a 1 t u 1 = 0 + 2.4   ms -2 ×2 s = 4.8  ms -1 Distance covered by the particle between time 2 s and 5 s i.e. in 3 s s 1 = u 1 t + 1 2 a 1 t 2 s 1 = 4 .8 ms -1 ×3 s+ 1 2 ×2 .4 ms -2 × 3 s 2 = 25.2 m …(ii) To find distance  s 2 Let  a 2  be acceleration of the particle during t = 5 s to t = 10 s, According to the first equation of motion, v = u + at  0 = 12 ms -1 + a 2 ×5 s ∴ a 2 = -12 5 ms -2 = -2.4  ms -2 Distance covered by the particle in during t = 5 s to t = 6 s s 2 = u 2 t + 1 2 a 2 t 2 s 2 = 12 ms -1 ×1 s+ 1 2 -2 .4 ms -2 × 1 s 2 s 2 = 12 m – 1.2 m = 10.8 m …(iii) From equation (i), (ii) and (iii), we obtain: Total distance travelled, s = 25.2 m + 10.8 m = 36 m  ∴Average speed = Total distance Total time = 36 m 4 s = 9  ms -1

Q.3 Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 ms–1 and 30 ms–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 ms–2. Give the equations for the linear and curved parts of the plot.

Ans.

Given, height of the  cliff, x 0 = 200 m

In case of first stone: Initial velocity,  u 1 = 15 ms -1 Acceleration, a = 10  ms -2 According to the equation of motion,

we obtain: x 1 = x 0 + u 1 t + 1 2 at 2 ∴ x 1 = 200 m +  15t – 5t 2 …(i) When the stone reaches the ground,  x 1 = 0 ∴ -5t 2 + 15t + 200 = 0 t 2 – 3t – 40 = 0 ∴t = 8 s or t =-5 s Since, t=0 is the instant, when the stone was projected, hence the negative time has no meaning. ∴t = 8 s In case of second stone, initial velocity,  u 2 = 30  ms -1 Acceleration, a = -10  ms -2 According to the equation of motion: x 2 = x 0 + u 2 t + 1 2 at 2 ∴ x 2 = 200 + 30t – 5t 2 …(ii) When the stone reaches the ground;  x 2 = 0 -5t 2 + 30t + 200 = 0 t 2 – 6t – 40 = 0 ∴t = 10 s or t = -4 s Since, negative sign is meaningless ∴t = 10 s Subtracting eqation (i) from equation (ii), we get x 2 – x 1 = 200 + 30t – 5t 2 – 200 + 15t – 5t 2 x 2 – x 1 = 15t …(iii) Equation (iii) shows the linear path of both the stones. Because of this linear relation between x  and t, the path is a straight line till t = 8 s. For maximum separation between the two stones, t = 8 s ∴ Maximum separation, x 2 – x 1 max = 15 ms -1 ×8 s = 120 m After 8 s, only second stone is in motion for 2 s. Its variation with time is given by the quadratic equation, x 2 – x 1 = 200 + 30  t – 5t 2 (For interval of time 8 s to 10 s) It represents a curved path.

Q.4 On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 kmh–1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 kmh–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? Ans.

(a) In this case, speed of the belt,  v B = 4  kmh -1 Speed of the child w.r.t. the belt,  v C = 9 kmh -1 ∵ The motion of the child is in the direction of motion of the motion of the moving belt. Speed of the child w.r.t. the stationary  observer, v C’ = v C + v B v C’ = 9 kmh -1 + 4 kmh -1 = 13  kmh -1

(b) In this case, speed of the belt,  v B = 4  kmh -1 Speed of the child w.r.t. the belt,  v C = -9  kmh -1 Speed of the child w.r.t. the stationary  observer, v C’ = v C + v B v C’ = -9 kmh -1 + 4 kmh -1 = -5  kmh -1 The negative sign shows that the motion of the child is opposite to the direction of motion of the belt.

(c) Given, distance between the parents, s = 50 m As parents and child are located on the same belt, the speed of the child in either direction as observed  by the parents will remain the same i.e., 9  kmh -1 = 2.5  ms -1 . ∴ Time taken by the child in case (a) and (b) is, t = 50 m 2 .5 ms -1 = 20 s

(d) If motion is observed by any one of the parents, answers obtained in (a) and (b) will get changed.  This is because the child and parents are located on the same belt, therefore, the speed of the child  w.r.t. either of the father or mother is same i.e., 9  kmh -1 .  But, answer (c) remains unchanged, as parents and child are on the same belt and all of them are equally  affected by the motion of the belt.

Q.5 A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 ms–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 ms-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?

Ans.

Given, initial velocity of the ball, u = 49  ms -1 Acceleration, a = -g = -9.8  ms -2 Case 1 : When the lift is stationary The boy throws the ball upwards.  Final velocity of the ball at the highest point = 0 According to the first equation of motion,time of ascent is given as: v = u + at ∴t = v – u a t = -49 ms -1 -9 .8 ms -2 = 5 s But, time of ascent = time of descent ∴Total time taken by the ball to come back to the boy’s hand = 5 s + 5 s = 10 s Case 2: When the lift is moving upwards with uniform velocity of 5  ms -1 In this case, there is no change in the relative velocity of the ball w.r.t. the boy i.e., it remains 49  ms -1 . Thus, in this case also, the ball will come back to the boy’s hand after 10 s.

Q.6 A three wheeler starts from rest, accelerates uniformly with 1 ms-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 the plot to be during accelerated motion: a straight line or a parabola?

Ans.

In a straight line, the distance covered by a particle in nth second is given as: Dnth = Dn – Dn-1 ,Dn = un + 12an2Dn-1 = u(n – 1) + 12a(n – 1)2∴Dnth = u + a22n – 1Given, u = initial velocity = 0Acceleration, a = 1 ms-2∴Dn = 122n – 1Substituting n = 1, 2, 3, …., we can find the value of Dn.The different values of n and corresponding values of Dn are shown below: The graph between n and  D n  will be a straight line as shown below. As the three-wheeler gains the uniform velocity after n = 10 s, the line will be parallel to the time axis after n = 10 s.

Q.7 Figure 3.25 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 ‘u’ and ‘a’ in the three intervals. What are the accelerations at the points A, B, C and D? Ans.

The slope of the speed-time graph gives acceleration. As the slope of the speed-time graph is the greatest in interval 2, hence the magnitude of the average acceleration is the maximum in this interval. The height of the curve from the time axis tells the average speed of the particle. From the given speed-time graph, it is clear that height is the greatest in interval 3. Therefore, during interval 3 the average speed of the particle is the greatest.

In interval 1, the slope of the given speed-time graph is positive. Therefore, acceleration is positive in this interval. In the same way, speed is also positive in this interval of time.

In interval 2, the slope of the given speed-time graph is negative. Therefore, the sign of acceleration is negative in this interval. But, speed is positive because it is a scalar physical quantity.

In interval 3, the slope of the given speed-time graph is zero. Therefore, acceleration is equal to zero in this interval. But, the particle gains some uniform speed and it is positive in this interval.

At points A, B, C and D, the speed-time graph is parallel to the time axis. Therefore, the acceleration of the particle is zero at points A, B, C and D.

Q.8 Figure 3.24 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 in a small interval of time is equal to the slope of x-t graph in that interval of time. From the graph, it is clear that the slope is maximum and minimum in intervals 3 and 2 respectively. Thus, the average speed of the particle is the maximum in interval 3 and the minimum in interval 2. The average speed is positive in both intervals 1 and 2 because the slope of x-t graph is positive there. But, the average speed is negative in interval 3 because the slope of x-t graph is negative in this interval.

Q.9 Suggest a physical situation for each of the following graph. Ans.

(a) The given x-t graph shows that initially x = 0, then it increases with time and attains a constant value for an instant and reduces to zero with time, then it increases in the opposite direction till it again attains a constant value i.e. comes to rest.

A similar physical situation arises when a football is kicked towards a player and he passes it back with a reduced speed but it misses the player and hits the goal finally coming to rest.

(b) In the given graph the velocity changes sign again and again with the passage of time and its magnitude decreases every time. A similar physical situation appears in the case of damping of a to and fro pendulum motion.

(c) The given graph shows that initially, the body moves with uniform velocity. Its acceleration increases for a small duration, which again reduces to zero. It shows that the body again starts moving with constant velocity. A similar situation appears when a hammer moving with a uniform velocity strikes a nail.

Q.10 A police van moving on a highway with a speed of 30 kmh-1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 kmh-1. If the muzzle speed of the bullet is 150 ms-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).

Ans.

Given, speed of police van, vP = 30 kmh-1 = 8.33 ms-1

Muzzle speed of bullet, vB = 150 ms-1

Speed of the car of the thief, vT = 192 kmh-1 = 53.33 ms-1

Since the bullet is fired from the police van, its effective speed is, vB = vB + vP

vB = 150 ms-1 + 8.33 ms-1 = 158.33 ms-1

As, both the vehicles are traveling in the same direction, the speed of bullet w.r.t. thief’s car,

vBT = vB – vT

vBT = 158.33 ms-1 – 53.33 ms-1 = 105 ms-1

Q.11 Figure 3.21 shows x-t plot of 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, because the given x-t graph does not represent the trajectory of the path followed by a particle moving in a straight line as t = 0, x = 0. The given graph can represent the motion of a body with constant acceleration.

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

(a) This graph cannot represent one dimensional motion of a particle, because in a one dimensional motion a particle cannot have two positions at the same instant of time.

(b) This graph cannot represent one dimensional motion of a particle, because a particle cannot have two values of velocity at the same instant of time in one dimensional motion.

(c) This graph cannot represent one dimensional motion of a particle, because according to this graph the particle can have negative speed, but speed being a scalar quantity cannot be negative.

(d) This graph cannot represent one dimensional motion of a particle, because the total path length of a particle cannot decrease with time.

Q.13 Figure 3.23 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.

Ans.

(a) At time t = 0.3 s, x is negative. Therefore, the slope of x-t graph is also negative. Thus, position and velocity are negative. Since, this graph represents a sinusoidal motion; for SHM of a particle, acceleration is given as: a = -ω2x∴Acceleration is positive.

(b) At time t = 1.2 s, x is positive, thus the slope ofx-t graph is also positive. Hence, position and velocity are positive. Since for SHM of a particle, acceleration is given as: a = -ω2x ∴ Acceleration is negative.

(c) At time t = -1.2 s, x is negative, therefore, the slope of x-t graph is also negative. But, as both x and t are negative, v becomes positive. A is also positive applying the SHM equation.

Q.14 In the above questions 3.13 and 3.14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and the magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?

Ans.

Instantaneous speed of a particle at any instant of time is defined as the first derivative of distance w.r.t. time atthat instant of time i.e.vins = limΔt→0ΔxΔtvins = dxdtSince, in instantaneous speed, we take only a small interval of time dt during which the particle is not supposed to change its direction of motion. Therefore, there is no difference between the path length and magnitude of displacement of the particle during this interval of time. Therefore, the value of instantaneous speed is always equal to the magnitude of instantaneous velocity.

Q.15 A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 kmh–1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 kmh–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 itis 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!]

Ans.

Time taken by the man to go to market from home, t 1 = 2.5 km 5 kmh -1 = 1 2 h = 0.5 h = 30 min Time taken by the man to go to home from market, t 2 = 2.5 km 7 .5 kmh -1 = 1 3 h = 20 min ∴Total time taken by the man = 30 min + 20 min = 50 min

(i) 0 to 30 min (a) Average velocity = displacement time = 2.5 km 0.5 h = 5  kmh -1 (b) Average speed = Total distance time = 2.5 km 0.5 h = 5  kmh -1

(ii) 0 to 50 min Total distance = 2.5 km + 2.5 km = 5 km Total displacement = 0 (a) Average velocity = displacement time = 0 (b) Average speed = total distance time = 5 km 5 6  h = 6  kmh -1

(iii) 0 to 40 min Distance moved in first 30 min (from home to market) = 2.5 km Distance moved in next 10 min (from market to home) = 7.5 km× 10 60 = 1.25 km Total distance covered = 2.5 km + 1.25 km = 3.75 km Total displacement = 2.5 km – 1.25 km = 1.25 km (a) Average velocity = displacement time = 1.25 km 40 60 h = 1.875  kmh -1 (b) Average speed = distance time = 3.75 km 40 60 h = 5.625  kmh -1

Q.16 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;
(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 time true? [For simplicity, consider one dimensional motion only].

Ans.

(a) The magnitude of displacement will be the shortest distance between the initial and final positions of the particle over an interval of time.

The total length of the path covered is the actual total path covered by the particle in the given time interval.

For example, a particle moves from point P to point Q and then, comes back to a point R after the total time t, as shown in the given figure. Then the displacement will be the length PR, that is, the distance between P and R while total length of path covered will be PQ + RQ (distance covered).

(b) Magnitude of average velocity = Magnitude of displacementTimeIn case of the given particle, average velocity = PRtAverage speed = Total path lengthTime= PR+RQtAs (PR + RQ) > PR ∴ Average speed is greater than average velocity. For the equality sign in a given time, the particle needs to move along a straight line.

Q.17 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.

Ans.

Given, Height from which the ball is dropped, s = 90 m  Initial velocity of the ball, u = 0Acceleration, a = 9.8 ms-2Let final velocity of the ball be vAccording to second equation of motion, we obtain:s = ut + 12at2∴90 = 0 + 12×9.8t2∴t = 18.38 s2 = 4.29 sAccording to first equation of motion, final velocity (v) is given as:v = u + atv = 0 + 9.8 ms-2×4.29 s  = 42.04 ms-1Rebound velocity of ball is given as: u’ = 910vu’ = 910×42.04 ms-1 = 37.84ms-1Time (t’) taken by the ball to reach the highest point is given by, t’ = u’at’ = 37.84 ms-19.8 ms-2 = 3.86 sTotal time, T = t + t’T = 4.29+3.86 = 8.15 s As, time of ascent = time of descentThe ball takes 3.86 s to fall back on the floor,where,Velocity before striking the floor = 37.84 ms-1Velocity after striking the floor = 910×37.84 ms-1 = 34.05 ms-1Total time taken for second rebound = 8.15 s + 3.86 s = 12.01 sThe speed-time graph of the motion of the ball is shown in the given figure as: Q.18 Read each statement carefully and state with reasons and examples if it is true or false;
(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 may have zero acceleration
(d) With positive value of acceleration must be speeding up.

Ans.

(a) True, as when there is retardation, comes a point of inflection where the speed becomes zero.

(b) False, as the speed of the object at an instant will always be equal to the magnitude of velocity at that instant.

(c) True, since, constant speed can have a constant velocity while acceleration of an object is the rate of change of velocity with respect to time, therefore, acceleration of the object is equal to zero.

(d) The given statement is true if both velocity and acceleration are positive, at the instant of time taken as origin, example, when the body is falling vertically downwards.

Q.19 A player throws a ball upwards with an initial speed of 29.4 ms-1.
(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 and t = 0 be the location and time 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.
(d) To what height does the ball rise and after how long does the ball returns to the player’s hands. (Take g = 9.8 ms-2 and neglect air resistance)

Ans.

(a) As the ball is moving upwards and gravitational forceis in downward direction. The direction of acceleration due to gravity is always vertically downwards.

(b) At the highest point of its motion, the velocity ofthe ball becomes zero and the  acceleration due to gravity remains constant with constant value 9.8  ms -2 .

(c) The sign of position will be positive for both the upward and downward motion of the ball. The sign of velocity is negative during the upward motion and positive during the downward motion. The sign of acceleration is positive for both the upward and downward motion.

(d) Given, initial velocity of the ball, u = 29.4  ms -1 Final velocity of the ball,v = 0 (∵At maximum height, velocity of ball = 0) Acceleration,a = -g = -9.8  ms -2 Let t be the time taken by the ball to reach the highest point, where distance from the ground is S. According to the third equation of motion, v 2 – u 2 = 2gS ∴S = v 2 – u 2 2g = 0 ms -1 2 – 29 .4 ms -1 2 2 -9 .8 ms -2 = 44.1 m According to first equation of motion,time of ascent (t) is given as: v = u + at ∴t = v – u a t = 0 – 29 .4 ms -1 -9 .8 ms -2 = 3 s As, time of ascent = time of descent ∴Total time taken by ball to come back to player’s hands = 3 s + 3 s = 6 s

Q.20 Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 kmh–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?

Ans.

Let the speed of the bus running between two towns A and B be V Given, speed of the cyclist, v = 20  kmh -1 Relative velocity of the bus moving in the direction of the cyclist = V – v = V – 20   kmh -1 Relative velocity of the bus moving in direction of the cyclist = V + v = V + 20   kmh -1 Let distance traveled by the bus in time T minutes be d According to the first condition of the question: d v – 20 = 18 min ∴d = 18v – 18×20…(i)  According to the second condition of the question: d v + 20 = 6 min ∴d = 6v + 20×6…(ii) From equation (i) and (ii), we obtain: 18v – 18×20 = 6v + 20×6 ∴12v = 20×6 + 18×20 = 480 ∴v = 40  kmh -1 Substituting this value of v in equation (i),we obtain: 40 T = 18×40 – 18×20 ∴T = 18×20 40 = 9 min

Q.21 On a two lane road, car A is travelling with a speed of 36 kmh-1. Two cars B and C approach car A in opposite directions with a speed of 54 kmh-1 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 minimum acceleration of car B is required to avoid an accident?

Ans.

Given, velocity of car A, vA = 36 kmh-1 = 10 ms-1Velocity of car B, vB = 54 kmh-1 = 15 ms-1Velocity of car C, vC = 54 kmh-1 = 15 ms-1∴Relative velocity of car B w.r.t. car A is given as:vBA = vB – vAvBA = 15 ms-1 – 10 ms-1 = 5 ms-1Relative velocity of car C w.r.t. car A is given as:vCA = vC – (- vA) = 15 ms-1 + 10 ms-1 = 25 ms-1At a definite instant of time, AB = AC = 1 km = 1000 mTime taken by car C to overtake car A = 1000 m25 ms-1 = 40 s∴To avoid an accident, car B must travel same distance within 40 s.According to second equation of motion, the minimum acceleration (aB) produced by car B can be calculated as:s = ut + 12at2Putting, u = vBA = 5 ms-1; t = 40 s;  s = 1000 m; a = aBin the above relation, we obtain:1000 = 5×40+12×aB×402∴aB = 1 ms-2

Q.22 Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 kmh-1 in the same direction with A ahead of B. The driver of B decides to overtake A and accelerate by 1 ms-2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

Ans.

In case of train A,

Given, initial velocity,  u = 72 kmh -1  = 20  ms -1 Time, t = 50 s As the train is moving with uniform velocity, Acceleration of the train, a = 0 Let distance travelled by train  A be S A According to the second equation of motion, S = ut + 1 2 at 2 ∴ S A = 20 ms -1 ×50 s+ 1 2 ×0 ms -2 × 50 s 2 = 1000 m For train B, Given, initial velocity,  u = 72 kmh -1    = 20 ms -1 Time, t = 50 s Acceleration of the train,  a = 1 ms -2 Let distance travelled by train  B be S B According to the second equation of motion, S = ut + 1 2 at 2 ∴ S B = 20 ms -1 ×50 s + 1 2 ×1 ms -2 × 50 s 2 = 2250 m ∴The original distance between the driver of train A and guard of train B =2250 m – 1000 m = 1250 m

Q.23 A car moving along a straight highway with speed of 126 kmh-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?

Ans.

Given, initial velocity of the car,  u = 126 kmh -1  = 35  ms -1 Final velocity of the car, v = 0 Distance traveled by the car, S = 200 m Let acceleration produced in the car be a According to the third equation of motion,  v 2 – u 2 = 2aS 0 – 35 ms -1 2 = 2a 200 m ∴a = – 35 ms -1 2 2×200 m = -3.06  ms -2 Here, the negative sign shows retardation. According to the first equation of motion, v = u + at ∴0 = 35 + – 49 16 t ∴t = 35×16 49 s = 11.43 s

Q.24 A jet airplane travelling at the speed of 500 km h–1 ejects its products of combustion at the speed of 1500 kmh–1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

Ans.

Given, speed of the jet airplane, vA = 500 kmh-1

Relative speed of the combustion products of jet airplane w.r.t. the jet airplane, vPA = -1500 kmh-1Let relative speed of the combustion products w.r.t. the observer on the ground be vPRelative speed of the products of combustion w.r.t. the jet airplane, vPA = vP – vA∴vP = vA + vPAvP = 500 kmh-1 – 1500 kmh-1 = -1000 kmh-1The negative sign shows that the direction of combustion products is opposite to that of the jet airplane.

Q.25 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 1m 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.

Ans.

Distance travelled in 1 step = 1 m

Time taken to walk 1 step = 1 s

Time taken to walk first 5 m forward = 5 s

Time taken to walk 3 m backward = 3 s

Total distance travelled = 5 m – 3 m = 2 m

Total time taken to travel 2 m = 8 s

Time taken to travel 2 m = 8 s

Time taken to travel 4 m = 16 s

Time taken to travel 6 m = 24 s

Time taken to travel 8 m = 32 s

Now, the drunkard will move 5 steps forward in the next 5 s and cover a distance of 5 m and a total distance of 13 m and fall into the pit.

Therefore, total time taken by the drunkard to cover 13 m = 32 s + 5 s = 37 s Q.26 A woman starts from her home at 9.00 am, walks with a speed of 5 kmh–1 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 kmh–1. Choose suitable scales and plot the x-t graph of her motion.

Ans.

Given, speed of the woman = 5  kmh -1

Distance between home and office = 2.5 km ∴Time taken to reach office = Distance speed = 2.5 km 5  kmh -1 = 0.5 h

Speed of auto = 25  kmh -1 ∴Time taken to return from office = 2.5 km 25 kmh -1 = 0.1 h = 6 min The x-t graph of the motion of the woman is given below. Q.27 The position-time (x-t) graphs for two children A and B returning from their school to their homes P and Q respectively are shown in Fig. 3.19. Choose the correct entries in the brackets below;
(a) (A/B) lives closer to 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). Ans.

(a) In the given graph, distance OP<oq, therefore=””>

(b) In the given graph, x = 0 and t = 0 for A, but t has a finite value for B. Therefore, A starts from the school earlier than B.

(c) The slope of x-t graph is equal to the speed. In the given x-t graph, the slope of B is greater than that of A. Therefore, B walks faster than A.

(d) From the given graph, it can be observed that A and B reach home at the same time.

(e) From the given graph we can deduce that A starts earlier than B, also, B has a greater speed than A. Thus, we can say that B overtakes A on the road once.

Q.28 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 monkey sitting 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.

Ans.

(a) The carriage moving without jerks is considered as a point object because its size is too small compared to the distance between the stations.

(b) The monkey is considered as a point object because the size of the monkey is too small compared to the size of a circular track.

(c) The spinning cricket ball is not considered as a point object as its size is comparable to the distance the ball might turn after hitting the ground.

(d) A beaker slipping off the edge of a table is not considered as a point object as its size is comparable to the height of the table in this case.

## 1. What is the distinction between Distance and Displacement?

Displacement has a directional component, whereas distance does not. Although the magnitude of displacement can be positive or negative, distance is always positive. It doesn’t get any smaller over time.

## 2. What are the different kinds of motion?

Motion can be classified as follows:

1. a) One-dimensional motion occurs when a particle moves in a straight line or along a path. For example, a train moving in a straight line, a freely falling body under gravity, and so on.
2. b) A particle moving in a plane is said to be moving in two dimensions—for example, the motion of a gun shell, carrom board coins, and so on.
3. c) Three-dimensional motion refers to the movement of a particle in space. The motion of a kite in the sky, the motion of an aeroplane, and so on.

## 3. How can questions from past years assist in the preparation of the Class 11 Physics chapter on Motion in a Straight Line?

You can understand the pattern of the question and analyse the trend of the questions asked by looking at past years’ questions. These papers will help you determine the topics that are worth spending more time and effort on. They will also show you the topics that require appropriate notes. Exam question papers from previous years will help you focus your preparation and feel more prepared and confident on exam day. Students can cross-check their answers to these questions with the Class 11 Physics Chapter 3 NCERT Solutions. On the Extramarks website, you can get all of these resources for free.

## 4. In Chapter 3 of the Class 11 Physics textbook, what is the difference between speed and velocity?

The total path length covered by the object is divided by the total time taken to calculate speed. The change in position divided by time intervals is called velocity. A scalar quantity is speed. However, velocity is a vector quantity. Speed is always positive. Velocity, on the other hand, can be positive, negative, or zero depending on the situation. Speed is the rate at which an object moves over time; speed is the rate at which an object moves. The rate and direction of an object’s movement are referred to as velocity.

## 5. How do I master the contents of Chapter 3 for Class 11 Physics?

Motion in a Straight Line is a fundamental yet important chapter in Class 11 Physics. This chapter can be used to generate a plethora of questions. Start by getting familiar with the fundamental terms or definitions of motion, acceleration, speed, velocity, and relative velocity. Understand the various scalar and vector quantities. Practise drawing the distance-time graph to solve various problems.

## 6. What is Chapter 3 of Class 11 Physics about?

Chapter 3 Physics Class 11 examines motion and the various types of motion. We study concepts such as path lengths and displacement. We also learn about velocity and speed, as well as the various numerical values that go with them. The following pointers can be made about the chapter:

• The chapter contains a number of graphs and formulae that explain motion and displacement.
• The chapter also covers instantaneous velocity and speed, as well as their numerical derivations.
• These numerical values are also represented graphically.
• The chapter includes a variety of in-text examples as well as questions for students to answer
• The chapter explains how velocity and speed are related
• The chapter provides two methods for solving these equations for students to choose from
• The chapter concludes with a brief discussion of relative velocity and its formulae

## 7. In the NCERT class 11 Physics, which chapter gets more weightage?

The chapters that get more weightage are – NCERT Chapter 3 Motion in a Straight Line, Chapter 4 Motion in a Plane, and Chapter 7 System of Particles and Rotational Motion from Part I, and Chapter 12 Thermodynamics, Chapter 13 Oscillations, and Chapter 14 Waves from Part II.