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

Question 2.1

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 a table.

Solution:

a, b

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

b. The size of a monkey is very small as compared to the size of a circular track. Therefore, the monkey can be considered as a point sized object on the track.

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

d. The size of a beaker is comparable to the height of the table from which it slipped. Hence, the beaker cannot be considered as a point object.

Question 2.2

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 Fig. 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).

Solution:

a. A lives closer to school than B.
b. A starts from school earlier than B.
c. B walks faster than A.
d. A and B reach home at the same time.
e. B overtakes A once on the road.

Explanation:
a. In the given $x\text{–}t$ graph, it can be observed that distance $\text{OP} < \text{OQ}$. Hence, the distance of school from A’s home is less than that from B’s home.
b. In the given graph, it can be observed that for $x = 0$, $t = 0$ for A, whereas for $x = 0$, $t$ has some finite value for B. Thus, A starts his journey from school earlier than B.
c. In the given $x\text{–}t$ graph, it can be observed that the slope of B is greater than that of A. Since the slope of the $x\text{–}t$ graph gives the speed, a greater slope means that the speed of B is greater than the speed of A.
d. It is clear from the given graph that both A and B reach their respective homes at the same time.
e. B moves later than A and his/her speed is greater than that of A. From the graph, it is clear that B overtakes A only once on the road.

Question 2.3

A woman starts from her home at $9.00\text{ am}$, walks with a speed of $5\text{ km h}^{-1}$ on a straight road up to her office $2.5\text{ km}$ away, stays at the office up to $5.00\text{ pm}$, and returns home by an auto with a speed of $25\text{ km h}^{-1}$. Choose suitable scales and plot the $x\text{-}t$ graph of her motion.

Solution:

$$\text{Distance covered while walking} = 2.5\text{ km}$$
$$\text{Speed while walking} = 5\text{ km/h}$$
$$\text{Time taken to reach office while walking} = \frac{2.5}{5}\text{ h} = \frac{1}{2}\text{ h}$$

If O is regarded as the origin for both time and distance, then at $t = 9.00\text{ am}$, $x = 0$ and at $t = 9.30\text{ am}$, $x = 2.5\text{ km}$.
OA is the $x\text{-}t$ graph of the motion when the woman walks from her home to office. Her stay in the office from $9.30\text{ am}$ to $5.00\text{ pm}$ is represented by the straight line AB in the graph.
$$\text{Time taken to return home by an auto} = \frac{2.5}{25}\text{ h} = \frac{1}{10}\text{ h} = 6\text{ minutes}$$
So, at $t = 5.06\text{ pm}$, $x = 0$. This motion is represented by the straight line BC in the graph. While drawing the $x\text{-}t$ graph, the scales chosen are as under:
– Along time-axis, one division equals $1\text{ hour}$.
– Along position-axis, one division equals $0.5\text{ km}$.

Question 2.4

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

Solution:

Distance covered with $1\text{ step} = 1\text{ m}$, Time taken $= 1\text{ s}$
Time taken to move first $5\text{ m}$ forward $= 5\text{ s}$
Time taken to move $3\text{ m}$ backward $= 3\text{ s}$
$$\text{Net distance covered} = 5 – 3 = 2\text{ m}$$
$$\text{Net time taken to cover } 2\text{ m} = 5 + 3 = 8\text{ s}$$
– The drunkard covers $2\text{ m}$ in $8\text{ s}$.
– The drunkard covers $4\text{ m}$ in $16\text{ s}$.
– The drunkard covers $6\text{ m}$ in $24\text{ s}$.
– The drunkard covers $8\text{ m}$ in $32\text{ s}$.
In the next $5\text{ s}$, the drunkard will cover a distance of $5\text{ m}$ (reaching a total of $13\text{ m}$) and falls into the pit without stepping back.
$$\text{Net time taken by the drunkard to cover } 13\text{ m} = 32 + 5 = 37\text{ s}.$$

Question 2.5

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

Solution:

$$\text{Initial velocity } u = 126\text{ km h}^{-1} = 126 \times \frac{5}{18} = 35\text{ m s}^{-1}$$
$$\text{Final velocity } v = 0$$
$$\text{Distance covered } s = 200\text{ m}$$
Using the third equation of motion ($v^2 = u^2 + 2as$):
$$0 = (35)^2 + 2a(200) \implies 400a = -1225 \implies a = -3.0625\text{ m s}^{-2}$$
Retardation of the car is $3.06\text{ m s}^{-2}$.
Using the first equation of motion ($v = u + at$):
$$t = \frac{v – u}{a} = \frac{0 – 35}{-3.0625} \approx 11.43\text{ s}$$
Therefore, it takes $11.43\text{ s}$ for the car to stop.

Question 2.6

A player throws a ball upwards with an initial speed of $29.4\text{ m s}^{-1}$.
(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 $x = 0\text{ m}$ and $t = 0\text{ s}$ 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 $g = 9.8\text{ m s}^{-2}$ and neglect air resistance).

Solution:

(a) When the ball moves upward, its acceleration is vertically downwards (due to gravity).

(b) At the highest point of the ball’s ascent, the velocity of the ball is $0$ and its acceleration is $a = g = 9.8\text{ m s}^{-2}$ in the vertically downward direction.

(c) With the given sign conventions:
– During upward motion ($x < 0$): Position $x$ is negative, velocity $v$ is negative, acceleration $a$ is positive ($+g$).
– During downward motion ($x > 0$): Position $x$ is positive, velocity $v$ is positive, acceleration $a$ is positive ($+g$).

(d) Initial velocity $u = 29.4\text{ m s}^{-1}$, final velocity $v = 0$, $a = -9.8\text{ m s}^{-2}$.
Using $v^2 = u^2 + 2as$:
$$0 = (29.4)^2 + 2(-9.8)s \implies 19.6s = 864.36 \implies s = 44.1\text{ m}$$
The ball attains a maximum height of $44.1\text{ m}$.
Time of ascent $t = \frac{v – u}{a} = \frac{0 – 29.4}{-9.8} = 3\text{ s}$.
Total time taken to return to the player’s hands (time of flight) $= 2 \times 3 = 6\text{ seconds}$.

Question 2.7

Read each statement below carefully and state with reasons and examples, if it is true or false;
A particle in one-dimensional motion
a. with zero speed at an instant may have non-zero acceleration at that instant
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.

Solution:

a. True
b. False
c. True
d. False

Explanation:
a. When an object is thrown vertically up in the air, its speed becomes zero at maximum height. However, it has an acceleration equal to the acceleration due to gravity ($g$) acting downward at that instant.
b. Speed is the magnitude of velocity. When speed is zero, the magnitude of velocity is zero, so velocity must also be zero.
c. A car moving on a straight highway with constant speed has constant velocity. Since acceleration is the rate of change of velocity, its acceleration is zero.
d. This statement is false when acceleration is positive and velocity is negative (e.g., a particle projected upwards experiences downward acceleration while moving up, so it slows down). It is true only when both velocity and acceleration have the same sign.

Question 2.8

A ball is dropped from a height of $90\text{ m}$ 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 $12\text{ s}$.

Solution:

Height $s = 90\text{ m}$, initial velocity $u = 0$, $a = g = 9.8\text{ m s}^{-2}$.
Using $s = ut + \frac{1}{2}at^2$:
$$90 = 0 + \frac{1}{2}(9.8)t^2 \implies t^2 = \frac{180}{9.8} \implies t \approx 4.29\text{ s (time to hit floor)}$$
Velocity just before hitting the floor: $v = u + at = 0 + 9.8 \times 4.29 = 42.04\text{ m s}^{-1}$.
Rebound velocity $u_R = \frac{9}{10}v = 0.9 \times 42.04 = 37.84\text{ m s}^{-1}$.
Time of ascent to maximum height after first bounce: $t’ = \frac{0 – 37.84}{-9.8} \approx 3.86\text{ s}$.
Total time for first bounce cycle $= 4.29 + 3.86 = 8.15\text{ s}$.
Second impact occurs at $8.15 + 3.86 = 12.01\text{ s}$.

Question 2.9

Explain clearly, with examples, the distinction between:
a. magnitude of displacement 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.
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?

Solution:

a. Consider a football passed from player A to player B and instantly kicked back to player A along the same path. The magnitude of displacement is $0$ (returns to start), but total path length is $\text{AB} + \text{BA} = 2\text{AB}$. Thus, path length $\ge$ magnitude of displacement.

b. Average velocity magnitude $= \frac{\text{Displacement}}{\text{Time}} = \frac{0}{t} = 0$, whereas average speed $= \frac{\text{Total Path Length}}{\text{Time}} = \frac{2\text{AB}}{t} > 0$.
The equality sign holds true only if the particle moves in a single fixed direction along a straight line without turning back.

Question 2.10

A man walks on a straight road from his home to a market $2.5\text{ km}$ away with a speed of $5\text{ km h}^{-1}$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5\text{ km h}^{-1}$. What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) $0$ to $30\text{ min}$, (ii) $0$ to $50\text{ min}$, (iii) $0$ to $40\text{ min}$?

Solution:

– Time taken to reach market $= \frac{2.5\text{ km}}{5\text{ km h}^{-1}} = 0.5\text{ h} = 30\text{ min}$.
– Speed on return $= 7.5\text{ km h}^{-1} = \frac{7.5}{60}\text{ km min}^{-1} = 0.125\text{ km min}^{-1}$.

(a) Magnitude of average velocity over the whole journey ($0$ to $50\text{ min}$) is $0$ because net displacement is $0$.

(b) Average speeds:
(i) $0$ to $30\text{ min}$: $\text{Distance} = 2.5\text{ km}$, $\text{Time} = 0.5\text{ h} \implies \text{Average speed} = 5\text{ km h}^{-1}$.
(ii) $0$ to $50\text{ min}$: $\text{Total distance} = 2.5 + (20 \times 0.125) = 5\text{ km}$, $\text{Time} = \frac{50}{60}\text{ h} \implies \text{Average speed} = \frac{5}{50/60} = 6\text{ km h}^{-1}$.
(iii) $0$ to $40\text{ min}$: $\text{Total distance} = 2.5 + (10 \times 0.125) = 3.75\text{ km}$, $\text{Time} = \frac{40}{60}\text{ h} \implies \text{Average speed} = \frac{3.75}{40/60} = 5.625\text{ km h}^{-1}$.

Question 2.11

In Exercises 2.9 and 2.10, 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?

Solution:

Instantaneous velocity is defined over an infinitesimally small time interval ($dt \rightarrow 0$). In such a short interval, the moving particle does not change its direction of motion. As a result, the magnitude of displacement becomes exactly equal to the actual path length covered, making instantaneous speed equal to the magnitude of instantaneous velocity.

Question 2.12

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

a.

b.

c.

d.

Solution:

a. Cannot represent motion because a particle cannot have two different positions at the same instant of time.
b. Cannot represent motion because a particle cannot possess two different velocities at the same instant of time.
c. Cannot represent motion because speed being a scalar quantity can never be negative.
d. Cannot represent motion because total path length can never decrease with time.

Question 2.13

Figure 2.21 shows the $x\text{-}t$ 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.

Solution:

No.
An $x\text{-}t$ graph represents position versus time for one-dimensional motion, not the physical trajectory (shape of the path) of the particle. A physical situation that resembles this graph is a freely falling body dropped from a height at $t = 0$.

Question 2.14

A police van moving on a highway with a speed of $30\text{ km h}^{-1}$ fires a bullet at a thief’s car speeding away in the same direction with a speed of $192\text{ km h}^{-1}$. If the muzzle speed of the bullet is $150\text{ m s}^{-1}$, with what speed does the bullet hit the thief’s car?

Solution:

$$\text{Speed of police van } v_p = 30\text{ km h}^{-1} = 30 \times \frac{5}{18} = 8.33\text{ m s}^{-1}$$
$$\text{Muzzle speed of bullet } v_b = 150\text{ m s}^{-1}$$
$$\text{Speed of thief’s car } v_t = 192\text{ km h}^{-1} = 192 \times \frac{5}{18} = 53.33\text{ m s}^{-1}$$
$$\text{Actual speed of bullet w.r.t. ground} = 150 + 8.33 = 158.33\text{ m s}^{-1}$$
$$\text{Relative speed hitting the thief’s car} = 158.33 – 53.33 = 105\text{ m s}^{-1}.$$

Question 2.15

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

b. 

c. 

Solution:

a. The given x-t graph shows that initially a body was at rest. Then, its velocity increases with time and attains an instantaneous constant value. The velocity then reduces to zero with an increase in time. Then, its velocity increases with time in the opposite direction and acquires a constant value. A similar physical situation arises when a football (initially kept at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced. Then, it passes from the player who has kicked it and ultimately gets stopped after sometime.
b. In the given v-tgraph, the sign of velocity changes and its magnitude decreases with a passage of time. A similar situation arises when a ball is dropped on the hard floor from a height. It strikes the floor with some velocity and upon rebound, its velocity decreases by a factor. This continues till the velocity of the ball eventually becomes zero.
c. The given a-t graph reveals that initially the body is moving with a certain uniform velocity. Its acceleration increases for a short interval of time, which again drops to zero. This indicates that the body again starts moving with the same constant velocity. A similar physical situation arises when a hammer moving with a uniform velocity strikes a nail.

Question 2.16

Figure 2.23 gives the $x\text{-}t$ plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particle at $t = 0.3\text{ s}$, $1.2\text{ s}$, $-1.2\text{ s}$.

Solution:

– At $t = 0.3\text{ s}$: Position ($x$) is negative, velocity ($v$) is negative, acceleration ($a$) is positive.
– At $t = 1.2\text{ s}$: Position ($x$) is positive, velocity ($v$) is positive, acceleration ($a$) is negative.
– At $t = -1.2\text{ s}$: Position ($x$) is negative, velocity ($v$) is positive, acceleration ($a$) is positive.

Question 2.17

Figure 2.24 gives the $x\text{-}t$ 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.

Solution:

– Average speed is greatest in **Interval 3** and least in **Interval 2** (determined by the magnitude of slope).
– Sign of average velocity: Positive in Intervals 1 and 2, Negative in Interval 3.

Question 2.18

Figure 2.25 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 $v$ and $a$ in the three intervals. What are the accelerations at the points A, B, C and D?

Solution:

Average acceleration is greatest in interval 2
Average speed is greatest in interval 3
v is positive in intervals 1, 2, and 3
a is positive in intervals 1 and 3 and negative in interval 2
a = 0 at A, B, C, D
Acceleration is given by the slope of the speed-time graph. In the given case, it is given by the slope of the speed-time graph within the given interval of time.
Since the slope of the given speed-time graph is maximum in interval 2, average acceleration will be the greatest in this interval.
Height of the curve from the time-axis gives the average speed of the particle. It is clear that the height is the greatest in interval 3. Hence, average speed of the particle is the greatest in interval 3.

In interval 1:
The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.

In interval 2:
The slope of the speed-time graph is negative. Hence, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.

In interval 3:
The slope of the speed-time graph is zero. Hence, acceleration is zero in this interval. However, here the particle acquires some uniform speed. It is positive in this interval.
Points A, B, C, and D are all parallel to the time-axis. Hence, the slope is zero at these points. Therefore, at points A, B, C, and D, acceleration of the particle is zero.

Additional Question 1

A jet airplane travelling at the speed of $500\text{ km h}^{-1}$ ejects its products of combustion at the speed of $1500\text{ km h}^{-1}$ relative to the jet plane. What is the speed of the latter with respect to an observer on ground?

Solution:

$$\text{Speed of jet plane } v_{\text{jet}} = 500\text{ km h}^{-1}$$
$$\text{Relative speed of combustion products w.r.t. plane } v_{\text{smoke}} = -1500\text{ km h}^{-1}$$
$$v’_{\text{smoke}} = v_{\text{smoke}} + v_{\text{jet}} = -1500 + 500 = -1000\text{ km h}^{-1}$$
The negative sign indicates that the exhaust products move in the direction opposite to the flight of the jet.

Additional Question 2

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

Solution:

$$\text{Speed of trains } u = 72\text{ km h}^{-1} = 20\text{ m s}^{-1}$$
$$\text{Distance covered by train B: } S_B = ut + \frac{1}{2}at^2 = (20 \times 50) + \frac{1}{2}(1)(50)^2 = 1000 + 1250 = 2250\text{ m}$$
$$\text{Distance covered by train A: } S_A = ut = 20 \times 50 = 1000\text{ m}$$
$$\text{Original distance} = S_B – S_A – (\text{Length of A + Length of B}) = 2250 – 1000 – 800 = 450\text{ m}.$$

Additional Question 3

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

Solution:

$$v_A = 10\text{ m s}^{-1}, \quad v_B = 15\text{ m s}^{-1}, \quad v_C = -15\text{ m s}^{-1}$$
$$\text{Relative speed of C w.r.t. A} = 10 – (-15) = 25\text{ m s}^{-1}$$
$$\text{Time for C to reach A} = \frac{1000\text{ m}}{25\text{ m s}^{-1}} = 40\text{ s}$$
Car B must cover distance $s = 1000\text{ m}$ in $40\text{ s}$ with initial relative speed $v_{BA} = 15 – 10 = 5\text{ m s}^{-1}$:
$$s = ut + \frac{1}{2}at^2 \implies 1000 = (5 \times 40) + \frac{1}{2}a(40)^2 \implies 800a = 800 \implies a = 1\text{ m s}^{-2}.$$

Additional Question 4

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 $20\text{ km h}^{-1}$ in the direction A to B notices that a bus goes past him every $18\text{ min}$ in the direction of his motion, and every $6\text{ min}$ in the opposite direction. What is the period $T$ of the bus service and with what speed do the buses ply on the road?

Solution:

Let bus speed be $v_b$ and cyclist speed $v_c = 20\text{ km h}^{-1}$.
$$(v_b – 20) \times \frac{18}{60} = v_b \times \frac{T}{60} \quad \text{and} \quad (v_b + 20) \times \frac{6}{60} = v_b \times \frac{T}{60}$$
Equating both expressions:
$$(v_b – 20) \times 18 = (v_b + 20) \times 6 \implies 3v_b – 60 = v_b + 20 \implies 2v_b = 80 \implies v_b = 40\text{ km h}^{-1}$$
Substituting $v_b = 40$ to find $T$:
$$(40 – 20) \times 18 = 40 \times T \implies 360 = 40T \implies T = 9\text{ minutes}.$$

Additional Question 5

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

Solution:

The distance covered by a body in the $n^{\text{th}}$ second is given by:
$$S_n = u + \frac{a}{2}(2n – 1)$$
Given $u = 0$ and $a = 1\text{ m s}^{-2}$:
$$S_n = \frac{1}{2}(2n – 1) = n – 0.5$$
This relation shows that $S_n$ varies linearly with $n$ ($S_n \propto n$), so the plot during accelerated motion is a **straight line**.

Additional Question 6

A lady stands in an elevator which is open from above. She then throws a ball up with an initial speed of $40\text{ m/s}$. After how long will the ball return to her hand? The elevator then starts moving upwards with a uniform speed of $5\text{ m s}^{-1}$, the lady again throws the ball up with the same initial speed, after how long will the ball return to her hand?

Solution:

– **Case 1 (Elevator at rest):** Time of flight $t = \frac{2u}{g} = \frac{2 \times 40}{9.8} \approx 8.16\text{ s}$.
– **Case 2 (Elevator moving upwards with uniform speed):** Since velocity is uniform, the relative acceleration and relative initial velocity of the ball with respect to the elevator remain identical to Case 1. Hence, the time taken for the ball to return to her hand remains **$8.16\text{ s}$**.

Additional Question 7

On a moving walkway (belt speed $= 5\text{ km h}^{-1}$) a child runs to and fro at a speed of $10\text{ km h}^{-1}$ (with respect to the belt) between his mother and father located $40\text{ m}$ apart on the moving belt. For an observer sitting in the lobby outside, what is the
a. speed of the child running against the belt?
b. speed of the child running in the same direction as the walkway?
c. time taken by the child in (a) and (b)?
d. which of the answers change if the observer is either of the parents?

Solution:

a. Speed against the belt $= 10 – 5 = 5\text{ km h}^{-1}$.
b. Speed along the belt $= 10 + 5 = 15\text{ km h}^{-1}$.
c. Since both parents are on the walkway, the relative speed of the child with respect to parents is $10\text{ km h}^{-1} = 2.78\text{ m s}^{-1}$.
$$\text{Time} = \frac{40\text{ m}}{2.78\text{ m s}^{-1}} \approx 14.4\text{ s} \text{ (for each trip)}.$$
d. Answers to (a) and (b) change for the parents (since they are moving with the belt), whereas the answer to (c) remains unchanged.

Additional Question 8

Hagrid throws two stones simultaneously from a treetop that is $200\text{ m}$ above the jungle floor. The stones have an initial velocity of $15\text{ m s}^{-1}$ and $30\text{ m s}^{-1}$. Does the graph correctly represent the time variation of the relative position of the second stone with respect to the first? Give the linear and quadratic equations for the parts of the graph.

linear parts of the graph.

Solution:

For stone 1: $s_1 = 200 + 15t – 5t^2$ (hits floor at $t = 8\text{ s}$).
For stone 2: $s_2 = 200 + 30t – 5t^2$ (hits floor at $t = 10\text{ s}$).
Relative position before stone 1 hits ($t \le 8\text{ s}$):
$$s_2 – s_1 = (200 + 30t – 5t^2) – (200 + 15t – 5t^2) = 15t\text{ (Linear path)}$$
After stone 1 hits ($8\text{ s} < t \le 10\text{ s}$), only stone 2 is in motion:
$$s_2 – s_1 = 200 + 30t – 5t^2\text{ (Curved path)}$$

Additional Question 9

The given speed-time graph represents the motion of a particle in a fixed direction. Calculate the distance covered byalt particle in time intervals (a) $t = 0\text{ s}$ to $10\text{ s}$, (b) $t = 1\text{ s}$ to $8\text{ s}$. Find average speeds.

Solution:

(a) Distance $= \text{Area of triangle} = \frac{1}{2} \times 10 \times 12 = 60\text{ m}$. Average speed $= \frac{60}{10} = 6\text{ m s}^{-1}$.
(b) Distance traversed between $t = 1\text{ s}$ and $t = 8\text{ s}$ is $54\text{ m}$. Average speed $= \frac{54}{7} = 7.71\text{ m s}^{-1}$.

Additional Question 10

The graph below is a velocity-time graph of a particle in one-dimensional motion. Which of the following formulae correctly describe the motion of the particle in the time interval $t_1$ to $t_2$:

(a) $a_{\text{avg}} = \frac{v(t_2) – v(t_1)}{t_2 – t_1}$
(b) $x(t_2) = x(t_1) + v_{\text{avg}}(t_2 – t_1) + \frac{1}{2}a_{\text{avg}}(t_2 – t_1)^2$
(c) $v_{\text{avg}} = \frac{x(t_2) – x(t_1)}{t_2 – t_1}$
(d) $v(t_2) = v(t_1) + a(t_2 – t_1)$
(e) $x(t_2) = x(t_1) + v(t_1)(t_2 – t_1) + \frac{1}{2}a(t_2 – t_1)^2$
(f) $x(t_2) – x(t_1) = \text{area under the } v\text{-}t\text{ curve}$.

Solution:

The correct formulae describing general non-uniform motion are **(c), (d), and (f)**. Formulae (a), (b), and (e) require uniform/constant acceleration.