NCERT Exemplar Solutions Class 11 Physics Motion in a Straight Line — Very Short, Short and Long Answer Questions Q12–26

NCERT Exemplar Problems, Class 11 Physics — Motion in a Straight Line, Very Short Answer Questions 12 to 16, Short Answer Questions 17 to 22 and Long Answer Questions 23 to 26. All 15 questions solved. The book numbers them 3.12 to 3.26.

Three ideas carry this page. Graphs are calculus you can see: the slope of $x$-$t$ is $v$, the slope of $v$-$t$ is $a$, and the area under $v$-$t$ is displacement. Relative motion removes a common acceleration: two bodies falling freely have no acceleration relative to each other, so their separation changes at a steady rate. Real stopping takes reaction time: the vehicle keeps its full speed for that time before the brakes act, which is often what decides whether there is a collision.

Key insight. Before solving any two-body question, ask what the relative acceleration is. In Questions 21, 24 and 26 it is zero, or zero for a while, and the problem collapses to uniform motion.

Very Short Answer Questions

Question 12

Refer to the graphs in Fig 3.1. Match the following.

Graph	Characteristic
(a)	(i) has $v > 0$ and $a < 0$ throughout.
(b)	(ii) has $x > 0$ throughout and has a point with $v = 0$ and a point with $a = 0$.
(c)	(iii) has a point with zero displacement for $t > 0$.
(d)	(iv) has $v < 0$ and $a > 0$.

Solution. Read each graph by its sign, slope and curvature.

• (a) starts below the axis and rises through $x = 0$, so there is a moment after $t = 0$ with $x = 0$: that is (iii).
• (b) stays above the axis, has a maximum (slope zero, so $v = 0$) and, on the way down, changes from concave down to concave up (a point of inflexion, so $a = 0$): that is (ii).
• (c) falls all the time ($v < 0$) and is concave up, flattening out ($a > 0$): that is (iv).
• (d) rises all the time ($v > 0$) and is concave down, flattening out ($a < 0$): that is (i).

Question 13

A uniformly moving cricket ball is turned back by hitting it with a bat for a very short time interval. Show the variation of its acceleration with time. (Take acceleration in the backward direction as positive).

Solution. Before the hit the ball moves uniformly, so $a = 0$. During the brief contact with the bat its velocity reverses, a large change in a very short time, so the acceleration is large and points backward, which is positive by the stated convention. After the hit it again moves uniformly, so $a = 0$ again. The graph is therefore zero everywhere except for a single tall, narrow positive pulse during the contact time. The area under the pulse equals the change in velocity, $v_{\text{after}} + v_{\text{before}}$ (both measured as speeds), which is why a shorter contact needs a taller pulse.

Question 14

Give examples of a one-dimensional motion where

(a) the particle moving along positive $x$-direction comes to rest periodically and moves forward.

(b) the particle moving along positive $x$-direction comes to rest periodically and moves backward.

Solution.

(a) Take $x(t) = t – \sin t$. Its velocity $v = 1 – \cos t$ is never negative and is zero at $t = 0, 2\pi, 4\pi, \dots$. So the particle keeps moving forward, stopping momentarily once every $2\pi$ s and then carrying on. A point on the rim of a wheel rolling along a road behaves like this horizontally.

(b) Take $x(t) = \sin t$. Its velocity $v = \cos t$ is zero at $t = \pi/2, 3\pi/2, \dots$, and the particle reverses each time it stops. A mass oscillating on a spring is an example.

Question 15

Give example of a motion where $x > 0$, $v < 0$, $a > 0$ at a particular instant.

Solution. We need a particle on the positive side of the origin, moving towards it, and slowing down (since $a$ is opposite to $v$). A lift coming down from a high floor and slowing to stop at the 4th floor, with the ground as origin and up as positive, is one example. A formula that does it at every instant is

$$x(t) = A + Be^{-\gamma t}, \qquad A, B, \gamma > 0,$$

for which $x > 0$, $v = -B\gamma e^{-\gamma t} < 0$ and $a = B\gamma^2e^{-\gamma t} > 0$.

Question 16

An object falling through a fluid is observed to have acceleration given by $a = g – bv$ where $g$ = gravitational acceleration and $b$ is constant. After a long time of release, it is observed to fall with constant speed. What must be the value of constant speed?

Solution. Falling at constant speed means the acceleration is zero. Setting $a = 0$:

$$g – bv = 0 \implies v = \frac{g}{b}.$$

As the speed grows, the drag term $bv$ grows until it cancels $g$, and from then on the speed stops changing. This is the terminal speed.

Short Answer Questions

Question 17

A ball is dropped and its displacement vs time graph is as shown Fig. 3.4 (displacement $x$ is from ground and all quantities are +ve upwards).

(a) Plot qualitatively velocity vs time graph.

(b) Plot qualitatively acceleration vs time graph.

The figure shows $x$ starting at the release height, falling along a parabola to $x = 0$, then a series of parabolic arches, each lower and shorter than the last.

Solution. Between bounces the ball is in free fall, so $a = -g$ and the velocity decreases at a steady rate.

(a) The velocity starts at $0$ and falls linearly to a negative value as the ball reaches the ground. At the bounce it jumps almost instantly to a positive value (a little smaller in size, because each arch is lower), then falls linearly again with the same slope $-g$, through $0$ at the top of the arch, to the negative of that value just before the next bounce. The graph is a set of parallel downward-sloping straight segments, joined by vertical jumps, each shorter than the last.

(b) The acceleration is $-g$ throughout, except during each very short contact with the ground, when a large upward force reverses the velocity and $a$ is a tall, narrow positive spike.

Question 18

A particle executes the motion described by $x(t) = x_0(1 – e^{-\gamma t})$; $t \ge 0$, $x_0 > 0$.

(a) Where does the particle start and with what velocity?

(b) Find maximum and minimum values of $x(t)$, $v(t)$, $a(t)$. Show that $x(t)$ and $a(t)$ increase with time and $v(t)$ decreases with time.

Solution. Differentiating,

$$v(t) = x_0\gamma e^{-\gamma t}, \qquad a(t) = -x_0\gamma^2 e^{-\gamma t}.$$

(a) At $t = 0$: $x = x_0(1 – 1) = 0$ and $v = x_0\gamma$. The particle starts at the origin with velocity $\gamma x_0$.

(b) The factor $e^{-\gamma t}$ falls steadily from $1$ at $t = 0$ towards $0$ as $t \to \infty$. So:

• $x = x_0(1 – e^{-\gamma t})$ increases from a minimum of $0$ towards a maximum of $x_0$ (approached as $t \to \infty$).
• $v = x_0\gamma e^{-\gamma t}$ decreases from a maximum of $\gamma x_0$ towards a minimum of $0$.
• $a = -x_0\gamma^2 e^{-\gamma t}$ is negative and its size shrinks, so it increases from a minimum of $-\gamma^2x_0$ towards a maximum of $0$.

The particle slows steadily and creeps up on $x = x_0$ without ever reaching it.

Question 19

A bird is tossing (flying to and fro) between two cars moving towards each other on a straight road. One car has a speed of 18 m/h while the other has the speed of 27 km/h. The bird starts moving from first car towards the other and is moving with the speed of 36 km/h and when the two cars were separated by 36 km. What is the total distance covered by the bird? What is the total displacement of the bird?

Solution. The “18 m/h” is a misprint for $18$ km/h, which is how the key reads it. Do not follow the bird trip by trip; ask how long it flies. It flies until the cars meet. The cars close the gap at $18 + 27 = 45$ km/h, so

$$t = \frac{36\ \text{km}}{45\ \text{km/h}} = 0.8\ \text{h}.$$

The bird flies at a steady $36$ km/h for all that time, so

$$\text{distance} = 36 \times 0.8 = 28.8\ \text{km}.$$

The bird ends up where the cars meet. It started on the first car, which in $0.8$ h moves $18 \times 0.8 = 14.4$ km towards the second car. So the bird’s displacement is $14.4$ km in the direction from the first car towards the second.

Question 20

A man runs across the roof-top of a tall building and jumps horizontally with the hope of landing on the roof of the next building which is of a lower height than the first. If his speed is $9$ m/s, the (horizontal) distance between the two buildings is $10$ m and the height difference is $9$ m, will he be able to land on the next building? (take $g = 10\ \text{m/s}^2$)

Solution. He leaves horizontally, so his vertical motion starts from rest and is independent of his horizontal motion. The time to fall $9$ m is

$$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 9}{10}} = \sqrt{1.8} \approx 1.34\ \text{s}.$$

In that time he moves horizontally at a steady $9$ m/s, covering

$$9 \times 1.34 \approx 12.1\ \text{m}.$$

That is more than the $10$ m gap, so he is still over the second roof when he has dropped the $9$ m.

Question 21

A ball is dropped from a building of height $45$ m. Simultaneously another ball is thrown up with a speed $40$ m/s. Calculate the relative speed of the balls as a function of time.

Solution. Take up as positive and $g = 10\ \text{m/s}^2$. While both balls are in the air,

$$v_1 = -10t, \qquad v_2 = 40 – 10t, \qquad v_2 – v_1 = 40\ \text{m/s}.$$

Both have the same acceleration $-g$, so neither accelerates relative to the other and the relative speed stays at $40$ m/s.

That lasts only while both are falling freely. The dropped ball hits the ground when $45 = 5t^2$, at $t = 3$ s. After that it is at rest, and the relative speed is simply the speed of the second ball, $|40 – 10t|$, until the second ball lands when $40t – 5t^2 = 0$, at $t = 8$ s. (It reaches its highest point, $80$ m, at $t = 4$ s, so it is still well above the ground at $t = 3$ s.)

The key gives only the constant $40$ m/s, which is the in-flight part of this answer.

Question 22

The velocity-displacement graph of a particle is shown in Fig. 3.5.

(a) Write the relation between $v$ and $x$.

(b) Obtain the relation between acceleration and displacement and plot it.

The figure is a straight line from $v = v_0$ at $x = 0$ down to $v = 0$ at $x = x_0$.

Solution.

(a) A straight line with intercept $v_0$ and slope $-v_0/x_0$:

$$v = v_0 – \frac{v_0}{x_0}x.$$

(b) With $v$ given as a function of $x$, use $a = \dfrac{dv}{dt} = \dfrac{dv}{dx}\dfrac{dx}{dt} = v\dfrac{dv}{dx}$:

$$a = \left(v_0 – \frac{v_0}{x_0}x\right)\left(-\frac{v_0}{x_0}\right) = \frac{v_0^2}{x_0^2}\,x – \frac{v_0^2}{x_0}.$$

This is a straight line in $x$ with positive slope $v_0^2/x_0^2$ and negative intercept $-v_0^2/x_0$. It reaches $0$ at $x = x_0$, where the particle stops. The acceleration is negative throughout, since the particle is slowing down all the way.

Long Answer Questions

Question 23

It is a common observation that rain clouds can be at about a kilometre altitude above the ground.

(a) If a rain drop falls from such a height freely under gravity, what will be its speed? Also calculate in km/h. ($g = 10\ \text{m/s}^2$)

(b) A typical rain drop is about 4 mm diameter. Momentum is mass $\times$ speed in magnitude. Estimate its momentum when it hits ground.

(c) Estimate the time required to flatten the drop.

(d) Rate of change of momentum is force. Estimate how much force such a drop would exert on you.

(e) Estimate the order of magnitude force on umbrella. Typical lateral separation between two rain drops is 5 cm. (Assume that umbrella is circular and has a diameter of 1 m and cloth is not pierced through!!)

Solution.

(a) Falling freely from rest through $h = 1000$ m,

$$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 1000} \approx 141\ \text{m/s} \approx 510\ \text{km/h}.$$

(b) The drop is a sphere of radius $2$ mm and density $10^3\ \text{kg/m}^3$:

$$m = \tfrac43\pi r^3\rho = \tfrac43\pi(2\times10^{-3})^3(10^3) \approx 3.4\times10^{-5}\ \text{kg},$$ $$p = mv \approx 3.4\times10^{-5}\times141 \approx 4.7\times10^{-3}\ \text{kg m/s}.$$

(c) The drop flattens in about the time it takes to travel its own diameter:

$$\Delta t \approx \frac{d}{v} = \frac{4\times10^{-3}}{141} \approx 2.8\times10^{-5}\ \text{s} \approx 30\ \mu\text{s}.$$

(d) The drop loses all its momentum in that time, so

$$F \approx \frac{\Delta p}{\Delta t} = \frac{4.7\times10^{-3}}{2.8\times10^{-5}} \approx 1.7\times10^2\ \text{N}.$$

(e) The umbrella’s area is $\pi d^2/4 = \pi(1)^2/4 \approx 0.8\ \text{m}^2$. With drops $5$ cm apart, each drop has about $(5\times10^{-2})^2 = 2.5\times10^{-3}\ \text{m}^2$ to itself, so about

$$N \approx \frac{0.8}{2.5\times10^{-3}} \approx 320$$

drops strike at almost the same time. The total force is about $320 \times 170 \approx 5\times10^4$ N, of order $10^4$–$10^5$ N.

These numbers are absurdly large, and that is the point: real rain drops do not fall freely for a kilometre. Air drag brings them to a terminal speed of a few metres per second, which is why rain does not hurt.

Question 24

A motor car moving at a speed of $72$ km/h can not come to a stop in less than $3.0$ s while for a truck this time interval is $5.0$ s. On a highway the car is behind the truck both moving at $72$ km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does not bump onto (collide with) the truck. Human response time is $0.5$ s.

(Comment: This is to illustrate why vehicles carry the message on the rear side “Keep safe Distance”.)

Solution. $72$ km/h $= 20$ m/s. The fastest stops give the decelerations

$$a_{\text{truck}} = \frac{20}{5} = 4\ \text{m/s}^2, \qquad a_{\text{car}} = \frac{20}{3}\ \text{m/s}^2.$$

Let the truck brake at $t = 0$. The car driver reacts and brakes at $t = 0.5$ s. The gap between them shrinks as long as the car is faster than the truck, and starts to grow once the car is slower, so the danger moment is when their speeds become equal:

$$20 – \frac{20}{3}(t – 0.5) = 20 – 4t \implies \frac{8}{3}t = \frac{10}{3} \implies t = \frac54\ \text{s}.$$

Distances covered by then:

$$s_{\text{truck}} = 20\left(\tfrac54\right) – \tfrac12(4)\left(\tfrac54\right)^2 = 21.875\ \text{m},$$ $$s_{\text{car}} = 20(0.5) + 20\left(\tfrac54 – 0.5\right) – \tfrac12\cdot\tfrac{20}{3}\left(\tfrac54 – 0.5\right)^2 = 23.125\ \text{m}.$$

The car gains $23.125 – 21.875 = 1.25$ m on the truck. After $t = 5/4$ s the car is always slower than the truck (and stops sooner), so the gap never closes further. The car must therefore be more than $1.25$ m behind the truck.

The small number hides how much the reaction time matters: in the first $0.5$ s alone, the car gains $0.5$ m on the braking truck.

Question 25

A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time $t$ is given by $v(t) = 2t(3 – t)$; $0 < t < 3$ and $v(t) = -(t – 3)(6 – t)$ for $3 < t < 6$ s in m/s. It repeats this cycle till it reaches the height of 20 m.

(a) At what time is its velocity maximum?

(b) At what time is its average velocity maximum?

(c) At what time is its acceleration maximum in magnitude?

(d) How many cycles (counting fractions) are required to reach the top?

Solution.

(a) On the climb, $v = 6t – 2t^2$ and $dv/dt = 6 – 4t = 0$ at $t = 3/2$ s, where $v = 4.5$ m/s. During the slip $v$ is negative, so this is the maximum velocity in each cycle: at $t = 1.5$ s (and $1.5$ s into every later cycle).

(b) The average velocity from the start up to time $t$ during the first climb is

$$\bar v(t) = \frac{1}{t}\int_0^t (6t’ – 2t’^2)\,dt’ = \frac{3t^2 – \tfrac23t^3}{t} = 3t – \tfrac23t^2.$$

$d\bar v/dt = 3 – \tfrac43t = 0$ at $t = 9/4$ s, giving $\bar v = 3.375$ m/s. After $t = 3$ s the monkey slips and the average only falls, so the maximum is at $t = 9/4$ s.

(c) On the climb $a = 6 – 4t$, which is $+6$ at $t = 0$ and $-6$ at $t = 3$ s. On the slip $v = t^2 – 9t + 18$ and $a = 2t – 9$, which runs from $-3$ to $+3$. So the largest magnitude, $6\ \text{m/s}^2$, occurs at $t = 0$ and $t = 3$ s (the end of the climb).

(d) In each cycle the monkey climbs

$$\int_0^3 2t(3 – t)\,dt = 9\ \text{m}$$

and slips back

$$\int_3^6 (t – 3)(6 – t)\,dt = 4.5\ \text{m},$$

a net gain of $4.5$ m. Track the height cycle by cycle:

Cycle	Starts at	Top of climb	Ends at
1	$0$ m	$9$ m	$4.5$ m
2	$4.5$ m	$13.5$ m	$9$ m
3	$9$ m	$18$ m	$13.5$ m
4	$13.5$ m	reaches $20$ m	—

It never reaches $20$ m in the first three cycles. In the fourth it needs $6.5$ m of climb: $3t^2 – \tfrac23t^3 = 6.5$ gives $t \approx 1.96$ s into the cycle. So it reaches the top after $3 + 1.96/6 \approx 3.33$ cycles, at a total time of about $19.96$ s.

Printed key error. The key prints “6 cycles” for (d). Six full cycles would take the monkey to a net $27$ m (with a climb peak of $31.5$ m), well past the top. Its own data give a climb of $9$ m and a slip of $4.5$ m per cycle, and the height table above shows the monkey reaches $20$ m early in the fourth cycle. The correct answer is about $3\tfrac13$ cycles.

Question 26

A man is standing on top of a building $100$ m high. He throws two balls vertically, one at $t = 0$ and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is $+15$ m at $t = 2$ s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.

Solution. Let the first ball be thrown up at $u$ at $t = 0$, and the second at $u/2$ at $t = \tau$. Take $g = 10\ \text{m/s}^2$. A constant gap means the two balls have the same velocity at every moment:

$$u – gt = \frac u2 – g(t – \tau) \implies \frac u2 = g\tau \implies u = 20\tau.$$

Now use the size of the gap at $t = 2$ s. The heights above the roof are

$$y_1 = ut – 5t^2, \qquad y_2 = \frac u2(t – \tau) – 5(t – \tau)^2.$$

With $u = 20\tau$,

$$y_1 – y_2 = 15\tau^2,$$

the same at every $t$, as it should be. Setting $15\tau^2 = 15$ gives $\tau = 1$ s, so $u = 20$ m/s and $u/2 = 10$ m/s.

Common mistakes

• Following the bird trip by trip in Question 19. The bird flies at constant speed for as long as the cars take to meet; one division gives the distance.
• Saying the relative speed is $40$ m/s for ever in Question 21. It is constant only while both balls are in free fall; once the first lands it is just the speed of the second.
• Forgetting the reaction time in Question 24. The car covers $10$ m at full speed before its brakes act, and that is where it gains on the truck.
• Dividing $20$ m by the net $4.5$ m per cycle in Question 25. That ignores that the monkey reaches the top partway up a climb, before the slip; follow the height cycle by cycle.
• Using $a = dv/dx$ in Question 22. The acceleration is $v\,dv/dx$, not $dv/dx$, when $v$ is given as a function of position.
• Drawing a smooth acceleration graph in Question 17. The bounces are impulsive, so $a$ has spikes; between them it is exactly $-g$.

Practise next

• Problem. A ball is dropped from the top of a tower $60$ m high. At the same instant a second ball is thrown straight up from the foot of the tower at $30$ m/s. Taking $g = 10\ \text{m/s}^2$, when and at what height do they meet?

Show answer

After $2$ s, at $40$ m above the ground. As in Question 21, both have the same acceleration, so they close the $60$ m gap at a steady $30$ m/s and meet at $t = 60/30 = 2$ s. The dropped ball has then fallen $\tfrac12(10)(2)^2 = 20$ m, so it is at $40$ m; the thrown ball is at $30(2) – 20 = 40$ m too.

• Problem. A car moving at $72$ km/h can brake at $5\ \text{m/s}^2$. The driver’s reaction time is $0.5$ s. What is the shortest distance in which the car can stop after the driver sees a hazard?

Show answer

$50$ m. At $20$ m/s the car covers $20 \times 0.5 = 10$ m before braking begins, then $v^2/(2a) = 400/10 = 40$ m while braking. The reaction distance is a fifth of the total, which is the point of Question 24.

• Exemplar Motion in a Straight Line: Multiple Choice Questions — sign conventions, graph reading and average speed in one-line form.
• NCERT textbook Chapter 2 — Motion in a Straight Line for the theory of instantaneous velocity, acceleration, the kinematic equations and relative velocity.
• Exemplar Units and Measurement — the previous unit, on units, dimensions and errors.
• Exemplar Motion in a Plane — the next unit, which takes projectile motion like Question 20 further.