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

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line

**Question 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.

(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 table.

**Answer:**

**(a)** The railway carriage moving without jerks between two stations, so the distance between two stations is considered to be large as compared to the size of the train. Therefore the train is considered as a point object.

**(b)** As the distance covered by monkey is large in the reasonable time, so monkey is considered as a point object.

**(c)** A point object does not have a spinning motion. Therefore ball can not be considered as point object.

**(d)** As the beaker is tumbling and then slips off, so the distance covered by it is not large in the reasonable time. Therefore it is not treated as point object.

**Question 2.**

The postion-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) \(\frac { A }{ B } \) lives closer to the school than \( \frac { B }{ A } \)

(b) \(\frac { A }{ B } \) starts from the school earlier than \( \frac { B }{ A } \)

(c) \( \frac { A }{ B } \) walks faster than \( \frac { B }{ A } \)

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

(e) \( \frac { A }{ B } \) overtakes \( \frac { B }{ A } \) on the road (once/twice).

**Answer:**

(a) A lives closer to school than B, because B has to cover higher distances [OP < OQ],

(b) A starts earlier form school than B, because t = 0 for A but for B, t has some finite time.

(c) As slope of B is greater than that of A, thus B walks faster than A.

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

(e) At the point of intersection (i.e., X), B overtakes A on the roads once.

**Question 3.**

A woman starts from her home at 9.00 am, walks with a speed of 5 km h^{_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 km h^{_1}. Choose suitable scales and plot the x-t graph of her motion.

**Answer:**

Distance covered while walking = 2.5 km

Speed while walking = 5 km h^{_1
}Time taken to reach office while walking =\( \frac { 2.5 }{ 5 } \)= \( \frac { 1 }{2} \)

If O is taken as the origin for both time and distance then at t = 9-00 AM, x = 0 and at t 9-30 AM, x = 2.5 km OA is the x-t graph of the motion when the woman walks from her home to office. She stay in the office from 9-30 AM to 5-00 PM and is represented by the straight line AB.

Now time taken to return home = \( \frac { 2.5 }{ 5 } \)= \( \frac { 1 }{ 10 } \)

h = 6 minutes

So at 5. 06 PM, x = 0

This is represented by the line BC in the graph.

**Question 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 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.

**Answer:**

Distance travelled in 5s = 5 m

Distance travelled in 8s=5-3=2m

Distance travelled in 13 s = 2 + 5 = 7m

Distance travelled in l6s = 7- 3 =4m

Distance travelled in 21s = 4 + 5 = 9m

Distance travelled in 24s = 9- 3 =6m

Distance travelled in 29s = 6 + 5 = 1m

Distance travelled in 32s=11-3 = 8m

Distance travelled in 37s=8 + 5 =13m

Since each step requires one second of time therefore total time is 37 second. The graph is as shown.

**Question 5.**

A jet airplane travelling at the speed of 500 km h^{_1} ejects its products of combustion at the speed of 1500 km h^{_1} relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ?

**Answer:**

Velocity of jet =υ_{J} = + 500 km h^{-1} (away from observer)

Relative velocity of ejection with respect to jet = υ_{ej} = – 1500 kmh^{-1}.

If υ_{e }is the velocity of ejected products, then =V_{e}-V_{j
}=> v_{e} =v_{e}. + vi = – 1500 + (500) = – 1000 km h^{-1}

∴ v_{eg} = v_{e}-v_{g} = – 1000 km h^{_1 }(u_{g}=0)

**Question 6.
**A car moving along a straight highway with speed of 126 km h

^{_1}is brought to a stop within a distance 200 m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop ?

**Answer:**

u = 126 km h

^{1}= 126 x\( \frac { 5 }{ 18 } \)=35 ms

^{_1}

^{ }S = 200 m and υ = 0

υ

^{2}– u

^{2}= 2 a S

∴ 0 – (35)

^{2}= 2a x 200

**Question 7.**

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 ?

**Answer:
**

∴ Original distance between trains = (Sb – S

_{A})-((L

_{A}+ L

_{B}) = (2250 – 1000) – (800) = 450 m.

**Question 8.**

On a two-lane road, car A is travelling with a speed of 36 km h^{_1}. Two cars B and C approach car A in opposite directions with a speed of 54 km h^{_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 ?

**Answer:
**

**Question 9.**

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 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 ?

**Answer:**

Let υ_{b} = speed of each bus and υ_{c} = speed of cyclist

Relative speed of buses plying in the same direction of motion of cyclist = υ_{b} – υ_{c
}The buses plying in the direction of motion of the cyclist go past him after every 18 minutes i.e., \( \frac { 18 }{ 60 } \) h

.’. Distance covered by each bus is (υ_{b} – υ_{c}) x \( \frac { 18 }{ 60 } \)

Since a bus leaves after every T minute therefore distance is also equal to υ_{b} x \( \frac { T }{ 60 } \)

.’. (υ_{b} – υ_{c}) x \( \frac { 18 }{ 60 } \) = υ_{b} x \( \frac { T }{ 60 } \)

Relative velocity of the buses plying opposite to the direction of motion of the cyclist is υ_{b} + υ_{c} after every 6 minutes.

.’. Distance covered by each bus is (υ_{b} + υ_{c}) x \( \frac { 6 }{ 60 } \)

**Question 10.**

A player throws a ball upwards with an initial speed of 294 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 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.

(d) To what height does the ball rise and after how long does the ball return to the player’s hands ? (Take g=98 m s^{_2 }and neglect air resistance).

**Answer:**

(a) The ball is under the influence of acceleration due to gravity which always acts vertically downwards.

(b) Velocity at the highest point = zero

Acceleration at highest point = g = 98 ms^{_2 }(vertically downwards)

**Question 11.**

**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.

**Answer:**

**(a) True.** A particle thrown upward has zero speed at the highest point but a = g =9.8 ms^{_2 } in vertically downward direction.

**(b) False.** Because magnitude of velocity = speed. When speed is zero, velocity may not be non-zero.

**(c) True.** Because if the particle rebounds instantly with the same speed, it means it has infinite acceleration which is not possible.

**(d) False.** It can be true only when the chosen positive direction is along the direction of motion.

**Question 12.**

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.

**Answer:
**

**Question 13.**

**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 sign true ? [For simplicity, consider one-dimensional motion only].

**Answer:**

**(a)** Magnitude of displacement over an interval of time may be zero, whereas the total length of the path covered by the particle over the same interval is not zero. For example, consider a particle moving along a straight line from point A to point B distant S from each other and then back to point A in time interval t as shown in fig.

In this case, magnitude of displacement of the particle over an interval of time t = 0.

Total length of the path covered by the particle over the interval of time t = AB + BA = 2 S

**(b)
**

**(c)**In both the cases (a) and (b), second quantity is greater than the first quantity e. Total length of

path > magnitude of displacement and average speed > magnitude of average velocity.If the direction of motion of a particle along a straight line does not change, then magnitude of displacement of the particle over a time interval = Total length of the path covered by the particle over the same time interval and magnitude of velocity = speed of the particle.

**Question 14.**

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^{-1}. 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

(1) 0 to 30 min,

(2) 0 to 50 min,

(3) 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 hot as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]

**Answer:
**

**(3) In 0-40 min .**

In 0 — 30 min. man goes from home to market with a sped of 5 km h’. In next 10 mm, man goes from market towards

home with speed of 75 km h

^{-1}Distance travelled by man in these 10 min = speed x time = 75 km h

^{-1}x \( \frac { 10 }{ 60 } \) h

**Question 15.**

In 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 ?

**Answer:**

When we consider arbitrarily small interval of time, the magnitude of displacement is always equal to the magnitude of distance.

**Question 16.**

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

**Answer:**

None of the four graphs can represent one-dimensional motion of the particle.

(a) A particle cannot have two different positions at the same time.

(b) A particle cannot have two values of velocity at the same time.

(c) Speed is always positive.

(d) Total path length of a particle can never decrease with time. :

**Question 17.**

Figures shows the 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.

**Answer:**

For t < 0, x = 0, so particle is at rest and not moving in a straight line,

For t > 0, particle can move on a parabolic path if its acceleration is constant.

Therefore, it is not correct to say from graph that the particle moves in a straight line for r < 0 and on parabolic path for t > 0.

For the graph, a suitable physical context can be the particle thrown from the top of a tower at the instant t = 0.

**Question 18.**

A police van moving on a highway with a speed of 30 km h’ fires a bullet at a their car speeding away in the same direction with a speed of 192 km h^{-1}. If the muzzle speed of the bullet is 150 ms^{-1}, with what speed does the bullet hit the theirs car? (Note : Obtain that speed which is relevant for damaging the theirs car).

**Answer:
**

**Question 19.**

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

**Answer:**

**(a)** The x-t graph shows that initially x = 0, attains certain value of x, again x becomes zero and then x increases in opposite direction till it settles at a constant x (i.e., comes to rest). Therefore, it may represent a physical situation such as a ball (initially at rest) on being hit moves with constant speed, rebounds from the wall with less rebound speed and then moves to the opposite wall and then stops.

**(b)** From the υ-t graph, it follows that velocity changes sign again and again with the passage of time and every time losing some speed. Therefore, it may represent a physical situation such as a ball falling freely (after thrown up), on striking the ground rebounds with reduced speed after each hit against the ground.

**(c)** The a-t graph shows that the body gets accelerated for a short duration only. Therefore, it may represent a physical situation such as a ball moving with uniform speed is hit with a bat for a very small time interval.

**Question 20.**

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 Chapter 14). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, -1.2 s.

**Answer:
**

(1) At t = 0-3 s, x is -ve. Velocity = slope of x -t graph.

Since slope of x -t is negative, so velocity is negative. In simple harmonic motion, direction of acceleration is opposite to the direction of displacement of the particle, so acceleration is positive.

(2) At t = 1.2 s, x = + ve. v is also +ve as slope of x -t graph is + ve. Acceleration a is -ve

(3) At t = -1.2 s, x =-ve. so a is +ve, v =Δx/Δt = +ve as both Δx and Δt are negative.

**Question 21.**

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.

**Answer:**

Average speed of particle = slope of x-t graph. Slope of x -t is maximum at t = 3, and minimum at t = 2. Therefore, average speed of a particle is greatest at t = 3 add least at t = 2. o > 0 in 1

and 2 and υ< 0 in 3.

**Question 22.**

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 o and a in the three intervals. What are the accelerations at the points A, B, C and D ?

**Answer:**

Acceleration magnitude is greatest in 2 because slope of υ -t graph at this interval is maximum.

Average speed is greatest in 3.

υ>0 in 1, 2 and 3\a > 0 in 1, a < 0 in 2, a = 0 in 3.

Acceleration is zero at A, B, C and D because slope of υ-t graph at these points is zero.

**Question 23.**

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 ?

**Answer:
**

**Question 24.**

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 ?

**Answer:
**

When lift is moving upward with uniform velocity, initial velocity of ball will remain 49 ms

^{-1}only w.r.t lift. Thus the time take up by ball will be 10 s.

**Question 25.**

On a long of horizontally moving belt (Fig.), a child runs to and fro with a speed 9 km h^{_1} (with respect to the belt) between ms father and mother located 50 m apart on the moving belt. The belt moves with 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 ?

**Answer:
**

**Question 26.**

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. 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.

**Answer:
**

For maximum separation, t = 8 s

So maximum separation is 120 m

After 8 second, only the second stone would be in motion. Its motion is described by eqn.(ii) So, the graph is in accordance with the quadratic equation.

**Question 27.**

The speed-time graph of a particle moving along a fixed direction is shown in

Fig. Obtain the distance traversed 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)?

**Answer:**

**(a) **Distance travelled from t = 0 to t = 10 s

= area under speed time graph,

= \( \frac { 1 }{ 2 } \) x 10 x 12 = 60m

**(b)** Average speed over the interval from t = 0 to t = 10 s is \( \frac { 60 }{ 10 } \) =6 ms^{-1}

**(c)** In order to calculate distance from t = 2 s to t = 6 s, let us first determine separately the distance covered from t

= 2 s to r = 5 s and the distance covered form t = 5 s to t = 6 s, then add.

12

**(i)** Acceleration =\( \frac { 12 }{ 5 } \) = 24 ms2

Velocity at the end of 2 s = 24 X 2 = 4.8 ms^{-1}.

**Question 28.**

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

Which of the following formulae are correct for describing the motion of the particle over the time-interval t_{1} to t_{2} :

**Answer:**

**(a)** This formula is not correct as it is applicable only if a is constant. In time interval t_{1} to t_{2}, a is not constant.

**(b)** This formula is not correct as it is applicable only if a is In time-interval t_{1} to t_{2}, a is not constant.

**(c)** and (d) are correct. They represent the definitions of υ_{av} and υ_{av}.

**(d)** This formula is not correct as such formula does not contain u_{av} and a_{av}.

**(e)** This formula is correct, because area under υ-t graph = displacement of a particle.

We hope the NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line, help you. If you have any query regarding .NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line, drop a comment below and we will get back to you at the earliest.

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