NCERT Exemplar Solutions Class 11 Physics Motion in a Straight Line — Multiple Choice Questions Q1–11

NCERT Exemplar Problems, Class 11 Physics — Motion in a Straight Line, MCQ I and MCQ II. All 11 questions solved.

The book numbers these questions 3.1 to 3.11 (in the exemplar this is Unit 3; in the current textbook it is Chapter 2); here they are Questions 1 to 11. Questions 1 to 6 are MCQ I (exactly one option is correct) and Questions 7 to 11 are MCQ II (one or more options may be correct). Three ideas carry the whole page:

• Signs mean directions. Once an origin and a positive direction are fixed, $x$, $v$ and $a$ each carry a sign of their own, and the three signs are independent of one another.
• Graphs are read by slope and curvature. On a position-time graph the slope is the velocity and the curvature is the acceleration: concave up means $a > 0$, concave down means $a < 0$.
• Distance is not displacement. Average velocity uses the net change in position; average speed uses the total path length, which is never smaller.

Key insight. Speeding up means $v$ and $a$ have the same sign; slowing down means they have opposite signs. Questions 2, 8 and 10 turn on that one rule.

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 1

Among the four graphs (Fig. 3.1), there is only one graph for which average velocity over the time interval $(0, T)$ can vanish for a suitably chosen $T$. Which one is it?

Solution. Average velocity over $(0, T)$ is $[x(T) – x(0)]/T$, so it vanishes only if the particle comes back to the position it started from: the graph must cross the horizontal line through its own starting value. In (b) the particle starts at a positive $x$, goes up to a maximum and then comes down below its starting value, so somewhere on the way down $x(T) = x(0)$. In (a) $x$ rises from its starting value and never returns to it (it only crosses $x = 0$, which is not the starting value). In (c) $x$ only ever decreases and in (d) it only ever increases, so neither can return.

Question 2

A lift is coming from 8th floor and is just about to reach 4th floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct?

Solution. Take the three signs one at a time.

• Position. The 4th floor is above the ground floor, which is the origin, so $x > 0$.
• Velocity. The lift is going down, from the 8th floor to the 4th, so $v < 0$.
• Acceleration. A lift about to reach a floor is slowing down so that it can stop there. Slowing down means $a$ is opposite to $v$, so $a > 0$.

That is option (c).

Printed key error. The key prints (a). Option (a) has $x < 0$, which would put the lift below the ground floor, in a basement; nothing in the question allows that. The velocity and acceleration signs in (a) agree with (c), so the slip is in the position sign only. The correct answer is (c).

Question 3

In one dimensional motion, instantaneous speed $v$ satisfies $0 \le v < v_0$.

Solution. Speed is the magnitude of velocity, so the condition says only that $|v| < v_0$; the velocity itself can be positive or negative. In time $T$ the particle can travel less than $v_0T$ in either direction, so its displacement lies strictly between $-v_0T$ and $v_0T$: (b) is true. Because the particle may move backwards, the displacement can be negative, so (a) is false. Nothing about the speed bound fixes the sign of the acceleration, so (c) is false. And the particle may stop and turn back (speed $0$ is allowed), so it can have turning points: (d) is false.

Question 4

A vehicle travels half the distance $L$ with speed $V_1$ and the other half with speed $V_2$, then its average speed is

Solution. Average speed is total distance divided by total time. The two halves take times $\dfrac{L/2}{V_1}$ and $\dfrac{L/2}{V_2}$, so

$$\bar v = \frac{L}{\dfrac{L}{2V_1} + \dfrac{L}{2V_2}} = \frac{2V_1V_2}{V_1 + V_2}.$$

This is the harmonic mean, not the arithmetic mean in (a), because the vehicle spends more time on the slower half, which pulls the average towards $V_{\text{slow}}$. Option (b) is not even a speed (it has no dimensions of velocity), and (d) has the dimensions of time.

Question 5

The displacement of a particle is given by $x = (t – 2)^2$ where $x$ is in metres and $t$ in seconds. The distance covered by the particle in first 4 seconds is

Solution. The velocity is $v = 2(t – 2)$, which is negative for $t < 2$ and positive for $t > 2$, so the particle turns round at $t = 2$ s. Distance must be added up separately on each side of the turning point:

• $t = 0$: $x = 4$ m; $t = 2$: $x = 0$. Distance $4$ m.
• $t = 2$: $x = 0$; $t = 4$: $x = 4$ m. Distance $4$ m.

Total distance $= 8$ m. The displacement over the same 4 s is $x(4) – x(0) = 0$, which is why simply subtracting end positions gives the wrong answer here.

Question 6

At a metro station, a girl walks up a stationary escalator in time $t_1$. If she remains stationary on the escalator, then the escalator take her up in time $t_2$. The time taken by her to walk up on the moving escalator will be

Solution. Let the length of the escalator be $L$. Her walking speed relative to the steps is $L/t_1$ and the escalator’s speed relative to the ground is $L/t_2$. Walking up a moving escalator, the two velocities add, so

$$t = \frac{L}{L/t_1 + L/t_2} = \frac{t_1t_2}{t_1 + t_2}.$$

As a check, the answer is shorter than both $t_1$ and $t_2$, as it must be when both help her up. Option (a) is longer than the shorter of the two, and (d) can be negative.

Question 7

The variation of quantity A with quantity B, plotted in Fig. 3.2 describes the motion of a particle in a straight line.

The figure is a straight line with positive slope and a positive intercept on the A axis (A on the vertical axis, B on the horizontal axis).

Solution. The graph says $A = A_0 + kB$ with $A_0 > 0$ and $k > 0$. Take B as time, which is allowed, so (a) is true.

• In uniform motion $x = x_0 + vt$: displacement is a straight line in time with slope $v$ and intercept $x_0$, exactly this shape. So (c) is true.
• In uniform motion the velocity is constant, so a velocity-time graph is a horizontal line, not a sloping one. So (b) is false.
• In uniformly accelerated motion $v = u + at$: velocity is a straight line in time with slope $a$ and intercept $u$. So (d) is true.

Question 8

A graph of $x$ versus $t$ is shown in Fig. 3.3. Choose correct alternatives from below.

Solution. Read velocity as the slope and acceleration as the curvature.

• (a) At A, at $t = 0$, the graph is flat, so the slope (velocity) is zero there. The particle starts from rest. True.
• (b) B is a maximum of $x$. A maximum is where the graph is concave down, so $a < 0$ at B. False.
• (c) At C the graph flattens into a shoulder and then carries on falling: it is a point of inflexion with a horizontal tangent. The slope is zero there, so $v = 0$, and the curvature changes sign there, so $a = 0$. True.
• (d) The graph is lower at D than at A, so the displacement from A to D is negative and so is the average velocity. False.
• (e) At D the graph is at its steepest, plunging through the axis; at E it is rising but much less steeply. The magnitude of the slope is larger at D, so the speed at D is greater. True.

Question 9

For the one-dimensional motion, described by $x = t – \sin t$

Solution. Differentiate twice:

$$v = 1 – \cos t, \qquad a = \sin t.$$

• (a) $\sin t < t$ for every $t > 0$, so $x = t – \sin t > 0$. True.
• (b) $v = 1 – \cos t$ is never negative, but it equals $0$ at $t = 2\pi, 4\pi, \dots$, so it is not strictly positive for all $t > 0$. False.
• (c) $a = \sin t$ is negative for $\pi < t < 2\pi$. False.
• (d) Since $-1 \le \cos t \le 1$, $v = 1 – \cos t$ lies between $0$ and $2$. True.

Question 10

A spring with one end attached to a mass and the other to a rigid support is stretched and released.

Solution. The spring force is proportional to the stretch from the equilibrium position, $F = -kx$, so $|a| = kx/m$ is largest where the stretch is largest.

• (a) At the instant of release the stretch is the full amount the spring was pulled, so $|a|$ is a maximum. True.
• (b) At equilibrium the stretch, and so the force and acceleration, are zero. False.
• (c) The mass speeds up all the way in to the equilibrium position (force and velocity in the same direction) and slows down beyond it, so its speed is greatest there. True.
• (d) The speed is minimum (zero) at both ends of the swing. In this chapter displacement means change of position from where the motion started, which is the release point. At the release end the displacement is then zero, its least value, even though the speed is minimum there. So “always maximum” fails. False.

Read (d) with displacement measured from the equilibrium position instead and it would be true, since both ends of the swing are then at the largest distance from equilibrium. The question asks about the motion from release, and the key gives (a), (c), so the page follows that reading.

Question 11

A ball is bouncing elastically with a speed $1$ m/s between walls of a railway compartment of size $10$ m in a direction perpendicular to walls. The train is moving at a constant velocity of $10$ m/s parallel to the direction of motion of the ball. As seen from the ground,

Solution. In the train the ball crosses the 10 m compartment in 10 s at $1$ m/s, then turns back. Relative to the ground its velocity is the train’s velocity plus its own:

• moving forward in the train: $10 + 1 = 11$ m/s;
• moving backward in the train: $10 – 1 = 9$ m/s.

Both are forward, so as seen from the ground the ball never reverses: (a) is false. Its speed switches between 11 m/s and 9 m/s every 10 s: (b) is true. Any 20 s interval contains exactly 10 s at each speed, so the ground distance is $110 + 90 = 200$ m whatever the starting moment and the average speed is always $10$ m/s: (c) is true. The train moves at constant velocity, so it is an inertial frame and accelerations (the brief jolts at the walls) are the same in both frames: (d) is true.

Common mistakes

• Picking (a) in Question 2. The lift is above the ground floor, so $x > 0$; only the velocity and acceleration are opposite in sign, because it is slowing down.
• Averaging the speeds in Question 4. Average speed is total distance over total time; equal distances give the harmonic mean.
• Subtracting end positions in Question 5. That gives displacement, $0$; distance must be added up on each side of the turning point at $t = 2$ s.
• Reading a sloping $v$-$t$ line as uniform motion in Question 7. Uniform motion has a flat velocity graph; a sloping one means constant acceleration.
• Calling a maximum of $x$ a point of positive acceleration in Question 8. A peak on an $x$-$t$ graph is concave down, so $a < 0$.
• Forgetting that $v = 1 – \cos t$ touches zero in Question 9. Non-negative is not the same as positive.
• Thinking the ball reverses as seen from the ground in Question 11. Its $1$ m/s backward motion is swamped by the train’s $10$ m/s.

Practise next

• Problem. A car covers the first third of a journey at $20$ km/h and the remaining two thirds at $60$ km/h. Find its average speed.

• Problem. A particle moves along a line with $x = (t – 3)^2$ metres. Find the distance it covers in the first $5$ s and its displacement over the same interval.

• Exemplar Motion in a Straight Line: Very Short, Short and Long Answer Questions — where the same ideas are applied to graphs, relative motion, free fall and stopping distances.
• 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 the same ideas into two dimensions.