# NCERT Solutions for Class 11 Physics Chapter 3 - Motion in a straight line

**3.1. In which of the following examples of motion, can the body be considered approximately a point object?**

**(a) a railway carriage moving without jerke 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 railway carriage can be considered as a point object because the distance between the two stations is very large as compared to the size of the railway carriage.

(b) The monkey can be considered as a point object if the cyclist describes a circular track of very large radius. Because in that case the distance covered by the cyclist is very large as compared to the size of the monkey.

(c) The spinning cricket ball cannot be considered as a point object because the size of ball quite appreciable as compared to the distance through which the ball may turn on hitting the ground.

(d) A beaker slipping off the edge of a table cannot be considered as a point object because the size of the beaker is not negligible as compared to the height of the table.

**3.2. 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 3.19. 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).**

**Ans.** (a) Since OP < OQ, so A lives closer to the school than B.

(b) For x = 0, t = 0 for A; while t has some finite value for B. Therefore. A starts from the school earlier than B.

(c) The velocity is equal to the slope of x – t graph and the slope of x – t graph for B is greater than that for A, hence B walks faster than A.

(d) Corresponding to points P and Q, the value of t from x – t graph for A and B is the same, which can be checked by drawing lines through P and Q parallel to t-axis. Thus, both A and B reach home at the same time.

(e) The x - t graph for A and B intersects each other only once. Since B starts from the school afterwards, therefore B overtakes A on the road only once.

**3.3. A woman starts from her home at 9:00 a.m., walks with a speed of 5 km/h on straight road up to her office 2.5 km away, stays at the office up to 5:00 p.m. and returns home by an auto with a speed of 25 km/h. Choose suitable scales and plot the x - t graph of her motion.**

$$\textbf{Ans. }\text{Time taken in reaching office =} \frac{\text{distance}}{\text{speed}}\\ =\frac{2.5}{5} = \text{0.5 h = 30 min} $$

Time taken in returning from office

$$=\frac{2.5}{25} = \text{0.1 h = 6 min} $$

It means that woman reaches the office at 9 : 30 a.m. and returns home at 5 : 06 p.m. The x – t graph of this motion will be shown in the graph below.

**3.4. 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 1s. 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.** The x - t graph for the motion of drunkard is shown in figure below :

It is clear from the graph that he takes 37s to fall in the pit.

Distance of the pit from the start = 13 m

Time taken to move first 5 m = 5 s

5 steps (i.e., 5 m) forward and 3 steps (i.e, 3 m) backward means that net distance covered in one cycle = 5 – 3 = 2 m

and time taken during the process = 5 + 3 = 8 s.

$$\text{Time taken in moving 8 m =}=\frac{8 × 8}{2} = \text{32 s.} $$

Distance of the pit from the drunkard after covering a net distance of 8 m = 13 – 8 = 5 m. Now, in next 5 steps forward, the drunkard will fall into the pit, for which he will take 5 s more i.e., total time taken to fall in the pit = 32 + 5 = 37 s.

**3.5. A jet airplane traveling at the speed of 500 kmh ^{–1} ejects its product of combustion at the speed of 1500 kmh^{–1} relative to the jet plane. What is the speed of the latter with respect to the observer on the ground?**

**Ans.** Let us consider the direction of motion of airplane to be positive direction of x-axis.

Here speed of jet airplane, v_{A} = 500 kmh^{–1}

v_{P} = velocity of the product w.r.t. ground.

Relative speed of product of combustion w.r.t. jet airplane, v_{PA} = – 1500 kmh^{–1}

Relative velocity of the product of combustion w.r.t. jet airplane is v_{PA} = v_{P} – v_{A} = – 1500

⇒ v_{P} = v_{A} – 1500

= 500 – 1500 = – 1000 kmh^{–1}

Here negative sign shows that the direction of products of combustion is opposite to that of the jet airplane,

Thus, the magnitude of relative velocity is 1000 kmh^{–1}.

**3.6. 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

final velocity v = 0, distance covered s = 200 m, acceleration a = ? and time, t = ?

But third equation of motion,

v^{2} = u^{2} + 2as

∴ 0 = (35)^{2} + 2 × a × 200.

$$\Rightarrow \space \space \text{a = -} \frac{(25)^2}{2 × 500} = - \frac{49}{16}= \text{-3.06 m}s^{-2} $$

By 1^{st} equation of motion

∵ v = u + at

∴ 0 = 35 + (– 3.06)t

$$\Rightarrow \space \space \text{t = } \frac{35}{3.06} = \text{11.43 s }$$

**3.7. 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 accelerates 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?**

$$\textbf{Ans. } \text{For train A : u = 72 kmh}^{-1} = \frac{75 × 1000}{60 × 60} = \text{20 m}s^{-1} $$

t = 50 s; a = 0; s = s_{A}

By equation of motion

$$∵ \space \space \text{s = ut + }\frac{1}{2}\text{a}t^2\\ ∴ \space \space s_A = \text{20 × 50 +} \frac{1}{2}\text{×0×}(50)^2 = \text{1000 m}$$

For train B : u = 72 kms^{–1} = 20 ms^{–1};

a = 1 ms^{–2}, t = 50 s, s = s_{B}

Distance covered is,

$$ ∴ \space \space s_B = \text{20 × 50 +} \frac{1}{2}\text{×1×}(50)^2 = \text{2250 m}$$

Taking the guard of the train B is the last compartment of the train B, it follows that

Original distance between the two trains + length of train A + length of train B = S_{B} – S_{A},

⇒ Original distance between the two trains + 400 + 400 = 1250 = 2250 – 1000

Original distance between the two trains

= 1250 – 800 = 450 m

**3.8. On a two-lane road, car A is traveling 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.** Velocity of car A = 36 kmh^{–1} = 10 ms^{–1}

Velocity of car B = 54 kmh^{–1} = 15 ms^{–1}

Velocity of car C = 54 kmh^{–1} = 15 ms^{–1}

Relative velocity of B w.r.t A = 15 – 10 = 5 ms^{–1}

Relative velocity of C w.r.t. A = 15 + 10 = 25 ms^{–1}

∵ AB = AC = 1 km = 1000 m

Time available to B or C for crossing A

$$=\frac{1000}{25} = \text{40 s}$$

If car B acceleration by a, to cross A before car C does, then u = 5 ms^{–1}, t = 40 s, s = 1000 m, a = ?

Using the formula,

$$\text{s = ut +}\frac{1}{2}\text{a}t^2\\ \Rightarrow \space \space \text{1000 = 5 × 40 +} \frac{1}{2} ×a× (40)^2\\ \Rightarrow \space \space \text{1000 – 200 = 800 a} \\ \Rightarrow \space \space \text{a = } \frac{800}{800} = \text{1 m}s^{-2} $$

**3.9. 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 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 minutes in the opposite direction. What is the period of the bus service and with what speed (assumed constant) do the buses ply on the road?**

**Ans.** Let v kmh^{–1} be the constant speed with which the buses ply between the towns A and B. The relative velocity of the bus (for the motion A to B) with respect to the cyclist (i.e. in the direction in which the cyclist is going = (v – 20)kmh^{–1}. The relative velocity of the bus from B to A with respect to the cyclist = (v + 20) kmh^{–1}. The distance travelled by the bus in time T (minutes) = vT.

According to the question,

$$\frac{vT}{v-20} = \text{ 18 ⇒ vT = 18v – 18 × 20 ...(i) }\\ \text{and } \frac{vT}{v-20} \text{= 6 ⇒ vT = 6v + 20 × 6 ...(ii)}$$

Equation (i) and (ii), we get

18v – 18 × 20 = 6v + 20 × 6

⇒ 12v = 20 × 6 + 18 × 20 = 480

$$\Rightarrow \space \space \text{v = } \frac{480}{12} = \text{= 40 km/h }$$

Putting this value of v in (i), we get

40T = (18 × 40 – 18 × 20) = 18 × 20

$$\Rightarrow \space \space \text{T = } \frac{18 × 20}{40} = \text{ 9 minutes }$$

**3.10. A players throws a ball upwards with as 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 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 = 9.8 ms ^{–2} and neglect air resistance).**

**Ans.** (a) Since the ball is moving under the effect of gravity. The direction of acceleration due to gravity is always vertically downwards.

(b) At the highest point, the velocity of the ball becomes zero and acceleration is equal to the acceleration due to gravity (~ 9.8 ms–2) on vertically downward direction.

(c) When the highest point is chosen as the location for x = 0 and t = 0 and vertically downward direction to be the positive directions x-axis and upward direction as negative direction of x-axis.

During upward motion, x = (–) ve, v = velocity = (–)ve, a = acceleration = (+)ve

During downward motion, x = (+) ve, v =

(+)ve and a = +ve

(d) Let t = time taken by the ball to reach the highest point where height from ground to S.

Taking vertical upward motion of the ball, we have

u = – 29.4 ms^{–1}, a = 9.8 ms^{–2}, v= 0, s = S, t = ?

∵ v^{2} – u^{2} = 2as

∴ 0 – (– 29.4)^{2} = 2 × 9.8 × S

$$\Rightarrow \space \space \text{S = } \frac{(29.4)^2}{2×9.8} = \text{ -44.1 m }$$

Here negative (–ve) sign shows that the distance is covered in upward direction.

∵ v = u + at

∴ 0 = – 29.4 + 9.8 + t

$$\Rightarrow \space \space \text{t = } \frac{29.4}{9.8} \text{= 3s }$$

Then, time of ascent is 3 s.

When an object moves under the effect of gravity along, the time of ascent is always equal to the time of descent.

Therefore, the total time after which the ball returns to the player’s hand = 3 + 3 = 6 s.

**3.11 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.**

**Ans.** (a) True, when a body is thrown vertically upwards in the space, then at the highest point, the body has zero speed but has downward acceleration equal to the acceleration due to gravity.

(b) False, because velocity is the speed of the body in a given direction. When speed is zero, the magnitude of velocity (v) = 0, hence velocity is zero.

(c) True, when a particle is moving along a straight line with a constant speed, its velocity remains constant with time. Therefore,

$$\text{\text{acceleration =}} \frac{\text{change in velocity}}{\text{time}} \text{= 0 }$$

(d) False, The velocity of a body at any instant of time t is v = u + at.

If a = + ve, u = – ve which implies that the statement is not correct, at the instant of time taken as origin. Then for all the time before the time for which v vanishes, there is slowing down of the particle, i.e. speed of the particle keeps on decreasing with time. It happens when the body is projected vertically upwards.

However, the given statement is true, if a = –ve and u = +ve at the instant of time as origin, happens when the body is falling vertically downwards.

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

**Ans.** Let us consider vertical downward motion of a ball from a height 90 m.

u = 0, g = 10 m/s^{2}, h = 90 m, t = ?, v = ?

$$\therefore \text{t = } \frac{2h}{g} = \sqrt{\frac{2×90}{10}} = 3\sqrt{2} \text{= 4.24 s}\\ \text{v = } \sqrt{\text{2gs}} = \sqrt{{2 × 10×90 =}} 30 \sqrt{2} \text{m/s}$$

Rebound velocity of ball,

$$\text{u' =} \frac{9}{10}\\ \text{and v =} \frac{9}{10} ×30\sqrt{2} = 2.7\sqrt{2}$$

Time to reach the highest point is calculated as

$$\text{t' =} \frac{u'}{g}\\ \text{ =} \frac{27\sqrt{2}}{10} = 2.7\sqrt{2} \text{= 3.81 s}$$

Total time + t′ = 4.24 + 3.81 = 8.05 s

The ball will take further 3.81 s to fall back to floor, where its velocity before striking the floor is 27 m/s.

$$\text{Velocity of ball after striking the floor =} \frac{9}{10}\text{u'}\\ = \frac{9}{10}×27\sqrt{2} = 24.3\sqrt{2} \text{m/s}$$

Total time elapsed before upward motion of the ball

= 8.05 + 3.81 = 11.86 s

Thus, speed-time graph of this motion will be as shown in the figure below :

**3.13. 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 timer 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.** (a) Magnitude of displacement of a particle in motion for a given time is the shortest distance between the initial and final position of the particle in that time. Whereas the total length of path covered by the particle is the actual path traversed by the particle in the given time.

If the particle goes from A to B and B to C in time t, then magnitude of displacement = distance AC.

Total path length = distance AB + distance AC.

Total path length (AB + AC) > Magnitude of displacement (AC)

If the motion of particle in one dimension (along straight line),

then

magnitude of displacement = Total path length transversed by the particle in the given time.

(b) Magnitude of average velocity

$$= \frac{\text{Magnitude of displacement}}{\text{Time interval}} = \frac{AC}{t}$$

and average speed

$$= \frac{\text{Total path length}}{\text{Time interval}} = \frac{AB+BC}{t}$$

∵ (AB + BC) > AC

∴ average speed > magnitude of average velocity

If the particle is moving along a straight line, then in a given time the magnitude of displacement = Total path length transversed by particle in that time.

∴ Average speed = magnitude of average velocity

**3.14. A man walks on a straight road from his home to a market 2.5 km away with a speed of ****5 km/h. Finding the market closed, be instantly turns and walks back with a speed of 7.5 km/h. 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.]**

**Ans.** Time taken by man to go from his home to market,

$$t_1 = \frac{\text{distance}}{\text{speed}} = \frac{2.5}{5} = \frac{1}{2}h$$

Time taken by man to come from market to his home,

$$t_2 = \frac{2.7}{7.5} = \frac{1}{3}h \\ \therefore \text{Total time = }t_1 + t_2 = \frac{1}{2}+ \frac{1}{3} = \frac{5}{6}h = 50 \text{ minutes}$$

(i) 0 to 30 minutes

$$\text{(a) Average velocity = } = \frac{\text{displacement}}{\text{time}} = \frac{2.5}{1/2} = \text{5 km/h} \\ \text{(b) Average speed } \\ = \frac{\text{distance}}{\text{time}} = {2.5}{1/2} = \text{5 km/h}$$

(ii) 0 to 50 minutes

Total distance travelled = 2.5 × 2.5 = 5 km

Total displacement = 0

$$\text{(a) Average velocity} = \frac{\text{displacement}}{\text{time}} = 0 \\ \text{(b) Average speed} = \frac{\text{distance}}{\text{time}} \\ =\frac{5}{5/6} = \text{6 km/h}$$

(iii) 0 to 40 minutes

Distance moved in 30 minutes (from home to market) = 2.5 km

Distance moved in 10 minutes (from market to home )with speed 7.5 kmh^{-1}

$$= 7.5×\frac{10}{60} = 1.25 \text{ km}$$

So, displacement = 2.5 – 1.25 = 1.25 km

distance traveled = 2.5 + 1.25 = 3.75 km

$$\text{ (a) Average velocity = } \frac{1.25}{(40/60)}= \text{1.875 km/h} \\ \text{(b) Average speed = } \frac{3.75}{(40/60)}= \text{5.625 km/h}$$

**3.15. In exercises 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 magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?**

$$\textbf{Ans. } \text{Instantaneous speed}(v_{rms}) \text{of the particle at an instant} = \frac{dx}{dt}$$

Since in instantaneous speed we take only a small interval of time dt during which the direction of motion of body is not supposed to change, hence there is no difference between the total path length and magnitude of displacement for small interval of time dt. Hence, instantaneous speed is always equal to magnitude of instantaneous velocity.

**3.16. 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 does not represent one-dimensional motion, because at given instant of time, the particle will have two positions, which is not possible in one-dimensional motion.

(b) This graph does not represent one dimensional motion because, at the given instant of time, particle will have velocity in positive as well as in negative direction, which is not possible in one-dimensional motion.

(c) It does not represent one-dimensional motion because this graph shows that the particle can have the negative speed but speed of the particle can never be negative.

(d) It also does not represent one-dimensional motion because this graph shows the total path length decreases after a certain time but the total path length of a moving particle can never decrease with time.

**3.17. Figure 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.**

**Ans.** No, because the x – t graph does not represent the trajectory of the path followed by a particle. From the graph, it is clear, t = 0, x = 0.

The graph can represent the motion of a body falling freely from a tower under gravity.

**3.18. 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.** Muzzle speed of the bullet,

v_{B} = 150 ms^{–1} = 540 kmh^{–1}

Speed of police van, v_{P} = 30 kmh^{–1}

Speed of thief’s car, v_{T} = 192 kmh^{–1}

Since, the bullet is shearing the velocity of the police van, its elective velocity is

v_{B} = v_{B} + v_{P} = 540 + 30 = 570 kmh^{–1}

The speed of the bullet with respect to the thief’s car moving in the same direction

v_{BT} = v_{B} – v_{T} = 570 – 192

= 378 kmh^{–1}

$$= \frac{378×1000}{60×60} = \text{105 m}s^{-1}$$

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

**Ans.** (a) The x - t graph shows that initially x = 0, i.e. at rest, then it increases with time, attains a constant value and again reduces to zero with time, then it increases in opposite direction till it again attains a constant valuc, i.e. comes to rest. The similar physical situation arises when a ball resting on a smooth floor is kicked which rebounds from a wall with reduced speed. It then moves to the opposite wall which stops it.

(b) The velocity changes sign again and again with passage of time and every time some speed is lost. The similar physical situation arises when a ball is thrown up with some velocity, returns back and falls freely. On striking the floor, it rebounds with reduced speed each time it strikes against the floor.

(c) Initially, a body moves with uniform velocity. Its acceleration increases for a short duration and then falls to zero and thereafter the body moves with a constant velocity. The similar physical situation arises when a cricket ball is moving with a uniform speed is hit with a bat for a very short interval of time.

**3.20. 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.**

**Ans.** In the S.H.M. acceleration a = – w^{2} x, where w(i.e. angular frequency) is constant.

(i) At t = 0.3s, x = –ve, the slope of x – t plot is negative, hence position and velocity are negative. Since a = – w^{2}x, hence acceleration is positive.

(ii) At t = 1.2s, x = +ve, the slope of x – t plot is positive, hence position and velocity are positive. Since a = – w^{2}x, hence acceleration is negative.

(iii) At t = – 1.2 s, x = –ve, the slope of x – t plot is also negative but since both x and T are negative, hence velocity is Positive. So the final acceleration is also positive.

**3.21. 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.**

**Ans.** We know that average speed in a small interval of time is equal the slope of x – t graph in that interval of time. The average speed is the greatest in the interval 3 because the slope is greatest and the average speed is least in interval 2 because the slope is least there. The average speed is positive in intervals 1 and 2 because the slope of x – t is positive and average speed is negative in interval 3 because the slope of x – t is negative.

**3.22. Figure gives a speed-time graph of a particle in motion along 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.** We know that average acceleration in a small interval of time = slope of speed-time graph in that interval.

As the slope of speed-time graph is maximum in the interval 2 as compared to other interval 1 and 3, hence the magnitude of average acceleration is greatest in interval 2. The average speed is greatest in interval 3 for obvious reasons. In interval 1, the slope of speed time graph = +ve, hence acceleration = +ve. Obviously speed u = +ve. In interval 2, the slope of speed-time graph = +ve, hence acceleration

= +ve.

Obviously, speed is positive in this interval.

In interval 3, the speed-time graph is parallel to time axis. Therefore, acceleration a is zero but

v = +ve.

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

**3.23. 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 n^{th} second (n = 1, 2, 3, ...) versus n. What do you expect this plot to be during accelerated motion : a straight line or a parabola?**

**Ans.** Here, u = 0; a = 1 ms^{–2}

Distance covered in nth second is calculated as

$$S_n = \text{u} + \frac{a}{2}(2n-1) = 0 + \frac{1}{2}(2n-1) $$

S_{n} = 0.5 (2n – 1) ...(i)

Putting n = 1, 2, 3, ..., we can find the value of Sn. The various values of n and corresponding values of Sn are shown below:

n | S_{n} |

1 | 0.5 |

2 | 1.5 |

3 | 2.5 |

4 | 3.5 |

5 | 4.5 |

6 | 5.5 |

7 | 6.5 |

8 | 7.5 |

9 | 8.5 |

10 | 9.5 |

_{n}and n, we get a straight line AB as shown in figure. From (i), S

_{n}∝ n so the graph is a straight line. After 10 s the graph is a straight line parallel to time axis.

**3.24. A boy standing on a stationary lift (open from above) throws a ball upwards with the ****maximum initial speed he can, equal to 49 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, bow long does the ball take to return to his hands?**

**Ans.** Taking vertical upwards direction as the positive direction of x-axis.

When lift is stationary, consider the motion of the ball going vertically upwards and coming down to the hands of the boy, we have u = 49 ms^{–1},

a = – 9.8 ms^{–1}, t = ?, x – x_{0} = S = 0.

$$\because \space \space \text{S = ut +}\frac{1}{2}\text{a}t^2 \\ \Rightarrow \space \space \text{0 = 49t +}\frac{1}{2}(-9.8)t^2 \\ \Rightarrow \space \space \text{49t = 4.9}t^2\\ \Rightarrow \space \space \text{t =} \frac{49}{4.9} = \text{10\space s}$$

When the list starts moving with uniform speed. As the lift starts moving upwards with uniform speed of 5 ms^{–1}, there is no change in the relative velocity of the ball w.r.t. the boy which remains 49 ms^{–1}. Hence, even in this case, the ball will return to the boy’s hand after 10 seconds.

**3.25. On a long horizontally moving belt (figure), a child runs to and fro with a speed 9 km/h ****(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 km/h. 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.** Consider the direction from left to right to be the positive direction of x-axis

(a) Here velocity of belt, vs = + 4 km/h

Velocity of child w.r.t. belt,

$$v_C = +9 \text{km/h} = \frac{5}{2}\text{m/s}$$

Speed of the child wr.t. stationary observer,

v′_{C} = v_{C} + v_{s} = 9 + 4 = 13 km/h

(b) Here v_{B} = 4 km/h; v_{C} = – 9 km/h

Speed of the child w.r.t. stationary observer

v′_{C} = v_{C} + v_{s} = – 9 + 4 = – 5 km/h

Here negative (–ve) sign indicates that the child will appear to run in a direction opposite to direction of motion of the belt.

(c) Distance between the parents, S = 50 m

Since, parents and child are located on the same bell the speed of the child as observed by stationary observer in either direction (either from mother to father or from father to mother)

Time taken by child in (a) and (b) is

$$t = \frac{50}{5/2} = \frac{50×2}{5} = \text{20 s}$$

The motion is observed by one of the parents, answer to case (a) or case (b) will get altered. It is so because speed of child w.r.t. either of mother or father is 9 km/h. But answer (c) remains unaltered due to the fact that parents and child are on the same belt and all are equally affected by the motion of the belt.

**3.26. 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 figure alongside, 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.** Taking vertical upward motion of the first stone for time t,

We have x_{0} = 200 m, u = 15 m/s;

a = – 10 m/s^{2}, t = t; x = x_{1}

$$\text{As} \space \space \text{x =}x_0 + \text{ut} + \frac{1}{2}\text{a}t^2 \\ \therefore \space \space x_1 = \text{200 + 15t} + \frac{1}{2}(-10)t^2 \\ \Rightarrow \space \space x_1 = \text{200 + 15t -5}t^2 \space \space ...(i) $$

Taking vertical upward motion of the second stone for time t, we have

x_{0} = 200 m, u = 30 m/s, a = – 10 m/s^{2}, t = t; x = x_{2}

$$\text{Then} \space x_2 = \text{200+30t}−\frac{1}{2}×10t^2 $$

= 200 + 30t – 5t^{2} ...(ii)

When the first stone hits the ground, x_{1} = 0,

so t^{2} – 3t – 40 = 0

⇒ (t – 8)(t + 5) = 0

∴ Either t = 8s or t = – 5s

Since, t = 0 corresponds to the instant, when the stone was projected. Hence, negative time has no meaning in this case.

So t = 8s when the second stone hits the ground, x_{2} = 0, so 0 = 200 + 30t – 5t^{2}

⇒ t^{2} – 6t – 40 = 0

⇒ (t – 10)(t + 4) = 0

∴ Either t = 10 s or t = – 4 s

∴ t = 10 s (neglecting –ve value)

Relative position of second stone w.r.t. first stone = x_{2} – x_{1} = 15t

∵ (x_{2} – x_{1}) and t are linearly related, therefore the graph is a straight line till t = 8 s.

For maximum separation, t = 8 s so maximum separation = 15 × 8 = 120 m

After 8 seconds, only 2^{nd} stone would be in motion for 2 seconds.

So the graph in accordance with the quadratic equation x_{2} = 200 + 30t – 5t^{2} for the interval of time 8 seconds to 10 seconds.

**3.27. The speed-time graph of a particle moving along a fixed direction is shown in figure. Obtain the distance traversed by the particle between (a) t = 0 s to t = 10 s, t = 2 s to 6 s.**

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

**Ans.** (a) Distance travelled by the particle between 0 to 10 s will be = area of AOAB, whose base is

$$\text{10 s and height is 12 m/s =} \frac{1}{2}× \text{10×12× = 60 m} \\ \therefore \text{Average speed =} \frac{60}{10} = \text{6 m/s}$$

(b) Let S_{1} and S_{2} be the distance covered by the particle in the time interval t_{1}, = 2 s to 5 s and t_{2} = 5 s to 6 s, then total distance covered in time interval t_{2} = 2 s to 6 s will be

S = S_{1} + S_{2} ...(i)

To find S_{1} : u_{1} = velocity of the particle after 2 s a_{1}, = acceleration of the particle during the time interval (0 – 5 seconds)

Then u = 0, v = 12 m/s, a = a, and 1 = 5 s

$$\text{We have, } a_1 = \frac{v-u}{t} = \frac{12-0}{5} = \frac{12}{5} {= 2.4 m/}s^2$$

∴ u_{1} = u + a_{1}t = 0 + 2.4 × 2 = 4.8 m/s,

t_{1} = 3s, a_{1} = 2.4 m/s^{2}, S_{1} = ?

$$\text{As } S_1 = u_1t_1 + \frac{1}{2}a_1t^2_1 \\ \therefore \space \space S_1 = \text{4.8×3} + \frac{1}{2}×2.4×3^2 = 25.2 \text{m}$$

To find S_{2} : Let a_{2} = acceleration of the particle during the motion,

t_{1} = 5 s to t = 10 s

$$ a_2 = \frac{0 - 12}{10 - 15} = \frac{-12}{5} = -2.4 \text{m/}s^2$$

Taking motion of the particle in time interval t = 5 s to 6 s, we have u_{2} = 12 m/s, a_{2} = – 2.4 m/s^{2}, t_{2} = 1 s, S_{2} = ?

$$\text{As} \space \space S_2 = u_2t_2 + \frac{1}{2}a_2t^2_2\\ \Rightarrow \space \space S_2 = 12 × 1 + \frac{1}{2}(-2.4)1^2 = 10.8\space \text{m}$$

∴ Total distance travelled,

S = 25.2 + 10.8 = 36 m

$$\text{Then, average speed} = \frac{36}{6 - 2}= \frac{36}{4} = 9 \text{m/s}$$

**3.28. The velocity-time graph of a particle in one dimensional motion is shown in figure. Which of the following formulae are correct for describing the motion of the particle over the time interval t _{1} to t_{2} :**

$$\text{(a) }x(t_2) = x(t_1)+v(t_1)(t_2-t_1)+ \frac{1}{2}a(t_2-t_1)^2$$

(b) v(t_{2}) = v(t_{1}) + a(t_{2} – t_{1})

(c) v_{av} = {x(t_{2}) – x(t_{1})}/(t_{2} – t_{1})

(d) a_{av} = {v(t_{2}) – v(t_{1})}/(t_{2} – t_{1})

$$\text{(e) } x(t_2) = x(t_1)+v_{av}(t_2-t_1)+ \frac{1}{2}a_{av}(t_2-t_1)^2$$

(f) x(t_{2}) – x(t_{1}) = are under the v – t curve bounded by the time axis and the dotted line shown.

**Ans.** From the graph we note that the slope is not constant and is not uniform, hence the relations (a), (b) and (e) are not correct, but the relations (c), (d) and (f) are correct.

## NCERT Solutions for Class 11 Physics Chapter 3 Free PDF Download

Please Click on Free PDF Download link to Download the NCERT Solutions for Class 11 Physics Chapter 3 Motion in a straight line