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

These Class 11 Physics Chapter 2 solutions cover Motion in a Straight Line from the NCERT textbook (session 2026–27). You get every end-of-chapter Exercise question reproduced verbatim with a complete, step-by-step solution — numericals worked out with correct units and cross-checked final answers, and conceptual questions explained in clear, exam-ready language. The chapter builds the foundations of kinematics: instantaneous velocity and speed, acceleration, the three kinematic equations of uniformly accelerated motion, and relative velocity.

Class: 11 Subject: Physics Chapter: 2 Name: Motion in a Straight Line Exercises: 2.1 – 2.18 Session: 2026–27
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Class 11 Physics Chapter 2 – Overview

Chapter 2, Motion in a Straight Line, studies rectilinear motion — the motion of an object along a single straight line, treating the object as a point object. It is part of kinematics, where we describe how things move without asking what causes the motion. The chapter sharpens the idea of average velocity into instantaneous velocity (the derivative dx/dt, the slope of the tangent to the position–time graph), introduces acceleration as the rate of change of velocity with time, and derives the three kinematic equations for constant acceleration. Special cases such as free fall (a = −g), stopping distance and relative velocity are explored. A recurring skill is reading and interpreting x–t, v–t and a–t graphs, where slope gives velocity/acceleration and the area under a v–t graph gives displacement.

Key Concepts & Definitions

Point object: an object treated as a point because its size is much smaller than the distance it moves — a valid approximation in many real-life problems.

Distance vs displacement: distance is the total path length (a scalar, always positive); displacement is the change in position (a vector, can be positive, negative or zero). |displacement| ≤ distance.

Average velocity = displacement ÷ time interval. Average speed = total path length ÷ time interval; average speed ≥ |average velocity|.

Instantaneous velocity (v): the limit of average velocity as Δt → 0, i.e. v = dx/dt. It equals the slope of the tangent to the x–t graph at that instant.

Instantaneous speed: the magnitude of instantaneous velocity. Unlike averages, instantaneous speed always equals the magnitude of instantaneous velocity.

Acceleration (a): the rate of change of velocity with time. Average acceleration = Δv/Δt; instantaneous acceleration a = dv/dt = slope of the v–t graph. It can be positive, negative (retardation) or zero.

Free fall: motion of a body under gravity alone (air resistance neglected), a uniformly accelerated motion with a = g ≈ 9.8 m s−2 directed downward.

Relative velocity of A with respect to B: vAB = vA − vB.

Graphs: slope of x–t graph = velocity; slope of v–t graph = acceleration; area under v–t graph = displacement.

Important Formulas

Instantaneous velocity: v = dx/dt   |   Instantaneous acceleration: a = dv/dt = d2x/dt2

Kinematic equations (constant acceleration a):

1. v = v0 + at

2. x = v0t + ½ at2

3. v2 = v02 + 2ax

(If the object is at x0 at t = 0, replace x by (x − x0).)

Free fall (taking up as positive): a = −g; for a body dropped from rest, v = −gt, y = −½gt2.

Stopping distance: ds = v02/(2a)   |   Relative velocity: vAB = vA − vB

Useful conversion: 1 km h−1 = 5/18 m s−1 ≈ 0.2778 m s−1.

NCERT Exercises 2.1–2.18 — Solutions

Questions are reproduced verbatim from the NCERT textbook. Numerical answers are worked step by step with units and cross-checked against the textbook answer key. Figure/graph questions are answered in words from the given data.

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.

ANSWER (a) and (b) can be treated as point objects. (a) The distance between the two stations is very large compared with the size of the railway carriage, so the carriage can be treated as a point object. (b) The size of the monkey is very small compared with the size of the circular track, so it too can be treated as a point object. (c) The sharp turn of a spinning ball is caused by, and is comparable to, its own size and spin, so its size cannot be neglected — not a point object. (d) A tumbling beaker rotates as it falls, and this tumbling depends on its size and shape, so it is not a point object.

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. 2.9. 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).

ANSWER (a) A lives closer to the school than B (P is nearer to O than Q on the position axis). (b) A starts from the school earlier than B (A’s line begins at the origin t = 0, B starts later). (c) B walks faster than A (B’s line is steeper, i.e. greater slope = greater speed). (d) A and B reach home at the same time (both lines end at the same value of t). (e) B overtakes A on the road once (the two lines cross exactly once).

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

ANSWER Going to office (walking): time = distance/speed = 2.5 km ÷ 5 km h−1 = 0.5 h = 30 min. She leaves at 9:00 am and reaches the office (x = 2.5 km) at 9:30 am. At office: from 9:30 am to 5:00 pm she stays put, so x = 2.5 km stays constant (a horizontal line) for 7.5 hours. Returning home (auto): time = 2.5 km ÷ 25 km h−1 = 0.1 h = 6 min. She leaves at 5:00 pm and reaches home (x = 0) at 5:06 pm. Graph: Take the x-axis from 0 to 2.5 km and the t-axis in hours from 9 am. Plot three line segments — (i) a straight rising line from (9:00, 0) to (9:30, 2.5), (ii) a horizontal line from (9:30, 2.5) to (17:00, 2.5), and (iii) a steeper falling line from (17:00, 2.5) to (17:06, 0). The return line is much steeper because the auto is five times faster.

2.4 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 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

ANSWER One cycle: 5 steps forward then 3 steps backward = net displacement of (5 − 3) = 2 m, taking (5 + 3) = 8 s. So after each 8-second cycle his net advance is 2 m. Net progress: after 1 cycle → 2 m (8 s); after 2 cycles → 4 m (16 s); after 3 cycles → 6 m (24 s); after 4 cycles → 8 m (32 s). Reaching the pit: at the start of the 5th cycle he is at 8 m at t = 32 s. In this cycle he steps forward 5 m, reaching 8 + 5 = 13 m. This forward stretch takes 5 s, so he reaches the pit at t = 32 + 5 = 37 s and falls in. Verification: matches the textbook answer of 37 s. The x-t graph is a zig-zag: each cycle rises 5 m over 5 s, then drops 3 m over 3 s, the saw-tooth drifting upward until it touches x = 13 m at t = 37 s.

2.5 A car moving along a straight highway with speed of 126 km h–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 ?

ANSWER Given: v0 = 126 km h−1 = 126 × (5/18) = 35 m s−1; final v = 0; distance x = 200 m. Retardation: using v2 = v02 + 2ax → 0 = (35)2 + 2 × a × 200 a = −1225 / 400 = −3.06 m s−2 (the negative sign shows retardation of 3.06 m s−2). Time to stop: using v = v0 + at → 0 = 35 + (−3.06)t → t = 35 / 3.06 = 11.4 s. Verification: matches the textbook answers (3.06 m s−2; 11.4 s).

2.6 A player throws a ball upwards with an initial speed of 29.4 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 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 m s–2 and neglect air resistance).

ANSWER (a) During upward motion the acceleration is due to gravity and is directed vertically downwards (g acts downward at every instant, whether the ball goes up or down). (b) At the highest point the ball is momentarily at rest, so velocity = 0; the acceleration = 9.8 m s−2 directed vertically downwards (it does not become zero). (c) With the origin at the highest point and the downward direction taken as positive: position x > 0 in both the upward and downward journeys (the ball is always below or at the chosen origin while in hand-to-top region… here the ball is below the top, so x > 0). During the upward motion the ball moves opposite to the positive (downward) axis, so velocity v < 0; during the downward motion v > 0. Acceleration is downward (the positive direction) throughout, so a > 0 in both phases. (d) Height risen: using v2 = v02 − 2gh with v = 0, v0 = 29.4 m s−1: 0 = (29.4)2 − 2(9.8)h → h = 864.36 / 19.6 = 44.1 m. Time of flight: time to rise = v0/g = 29.4 / 9.8 = 3 s. By symmetry the time to fall back is also 3 s, so total time to return to the hands = 3 + 3 = 6 s. Verification: matches the textbook answers (44.1 m; 6 s).

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.

ANSWER (a) True. A ball thrown vertically up has zero speed at its highest point, yet its acceleration there is g = 9.8 m s−2 downward. (b) False. Speed is the magnitude of velocity; if speed is zero, velocity must also be zero. They cannot differ in magnitude. (c) True (for one-dimensional/straight-line motion). If the speed is constant and the direction is fixed, velocity does not change, so acceleration is zero. (In two dimensions, e.g. uniform circular motion, constant speed can still have acceleration because direction changes.) (d) False. Whether the particle speeds up depends on the sign of acceleration relative to the velocity, not on its own sign. A ball thrown up with the upward direction chosen positive has a downward (negative) acceleration while moving up. Equivalently, if velocity is negative and acceleration positive, the particle slows down. So positive acceleration does not always mean speeding up — it is true only when the chosen positive direction is along the direction of motion.

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

ANSWER First fall: dropped from rest, h = 90 m. Time to hit floor: h = ½gt2 → 90 = ½(10)t2 → t2 = 18 → t ≈ 4.24 s (taking g = 10 m s−2). Speed just before impact: v = gt = 10 × 4.24 ≈ 42.4 m s−1 (or v = √(2gh) = √1800 ≈ 42.4 m s−1). After the first bounce: the ball loses one-tenth of its speed, so it rebounds with 0.9 × 42.4 ≈ 38.2 m s−1. It rises, momentarily stops, then falls back and strikes the floor again at 38.2 m s−1; the next rebound speed is 0.9 × 38.2 ≈ 34.3 m s−1, and so on. Speed–time graph: From t = 0, speed increases linearly (slope = g = 10 m s−2) from 0 to about 42.4 m s−1 at t ≈ 4.24 s — this is a straight rising line. At the collision the speed drops abruptly (a vertical drop) from 42.4 to 38.2 m s−1. Then the speed decreases linearly to 0 as the ball rises, and increases linearly back to 38.2 m s−1 as it falls (forming a downward triangle/V-shape), before dropping abruptly to 34.3 m s−1 at the next bounce. The pattern of saw-tooth peaks of decreasing height repeats up to t = 12 s.

2.9 Explain clearly, with examples, the distinction between : (a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval; (b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only].

ANSWER (a) The magnitude of displacement is the straight-line distance between the start and end points (shortest distance), while the total path length is the actual length of the path travelled. Example: a particle goes 5 m east then 3 m west. Magnitude of displacement = 5 − 3 = 2 m, but total path length = 5 + 3 = 8 m. Since the path can never be shorter than the straight-line distance, total path length ≥ magnitude of displacement. (b) Dividing both quantities in (a) by the same time interval: magnitude of average velocity = |displacement|/time and average speed = total path length/time. Because path length ≥ |displacement|, average speed ≥ |average velocity|. When equality holds: in both (a) and (b) the equality sign is true when the particle moves along a straight line without reversing its direction (motion entirely in one direction). Then the path length equals the magnitude of displacement, and the two quantities become equal.

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

ANSWER Set-up: Home to market (2.5 km at 5 km h−1) takes 2.5/5 = 0.5 h = 30 min. Market to home (2.5 km at 7.5 km h−1) takes 2.5/7.5 = 1/3 h = 20 min. So he reaches the market at 30 min and is back home at 50 min. (i) 0 to 30 min: he has just reached the market (2.5 km away, no return). Displacement = 2.5 km, path length = 2.5 km, time = 0.5 h. Magnitude of average velocity = 2.5/0.5 = 5 km h−1; average speed = 2.5/0.5 = 5 km h−1. (ii) 0 to 50 min: he is back home, so displacement = 0. Magnitude of average velocity = 0/… = 0. Total path length = 2.5 + 2.5 = 5 km, time = 50 min = 5/6 h. Average speed = 5 ÷ (5/6) = 6 km h−1. (iii) 0 to 40 min: first 30 min he reaches the market (2.5 km); in the next 10 min (1/6 h) he walks back at 7.5 km h−1, covering 7.5 × 1/6 = 1.25 km. So his position is 2.5 − 1.25 = 1.25 km from home. Time = 40 min = 2/3 h. Magnitude of average velocity = 1.25 ÷ (2/3) = 15/8 = 1.875 km h−1. Total path length = 2.5 + 1.25 = 3.75 km; average speed = 3.75 ÷ (2/3) = 45/8 = 5.625 km h−1. Verification: matches the textbook answers (a) 5, 0, 15/8 km h−1; (b) 5, 6, 45/8 km h−1. This shows why average speed (not |average velocity|) is the meaningful measure of how tired the man is.

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?

ANSWER Instantaneous quantities are defined over an infinitesimally small time interval (Δt → 0). In such a vanishingly small interval the particle cannot reverse its direction, so the path it traces is an infinitesimally short straight segment. For this tiny straight segment, the length of the path equals the magnitude of the displacement. Therefore the instantaneous speed (path length ÷ Δt) equals the magnitude of the instantaneous velocity (|displacement| ÷ Δt). The difference between speed and velocity that appears over a finite interval simply disappears in the limit.

2.12 Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

ANSWER None of the four graphs can represent one-dimensional motion of a particle. (a) The x-t graph shows two positions at the same instant of time — impossible, because a particle can be at only one position at any given instant. (b) The v-t graph shows the particle with velocity in two opposite directions at the same instant — impossible, a particle has a single velocity at any instant. (c) The speed-time graph shows negative speed — impossible, because speed is the magnitude of velocity and can never be negative. (d) The graph shows the total path length decreasing with time — impossible, since the total path length covered can only stay the same or increase, never decrease.

2.13 Figure 2.11 shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t >0 ? If not, suggest a suitable physical context for this graph.

ANSWER No, it is not correct. An x-t graph is not the actual path (trajectory) of the particle in space — it only shows how its position along one straight line varies with time. The motion is one-dimensional (along a single line) throughout; the graph being a straight line or a parabola does not mean the particle physically travels in a straight or parabolic path. Suitable context: A ball at rest is held at the origin (x = 0) for all t < 0, then is dropped (released from rest) at t = 0 from the top of a tower. After release it falls under gravity, so x increases as ½gt2, giving the parabolic x-t curve for t > 0.

2.14 A police van moving on a highway with a speed of 30 km h–1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h–1. If the muzzle speed of the bullet is 150 m s–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).

ANSWER Convert to m s−1: speed of van = 30 × (5/18) = 8.33 m s−1; speed of thief’s car = 192 × (5/18) = 53.33 m s−1. Both move in the same (positive) direction. Speed of bullet in ground frame: the muzzle speed (150 m s−1) is relative to the van, so the bullet’s speed over the ground = 150 + 8.33 = 158.33 m s−1. Relevant speed = speed relative to the thief’s car: vbullet, car = 158.33 − 53.33 = 105 m s−1. This is the speed with which the bullet actually strikes (and can damage) the car. Verification: matches the textbook answer of 105 m s−1.

2.15 Suggest a suitable physical situation for each of the following graphs (Fig 2.12):

ANSWER (a) The x-t graph rises, then briefly becomes a sharp horizontal jump (sudden change), then continues. Situation: a ball at rest on a smooth floor is kicked; it rebounds from a wall with reduced speed and then moves to the opposite wall, which stops it. (b) The v-t graph shows the sign of velocity reversing abruptly with the magnitude decreasing each time. Situation: a ball thrown up with some initial velocity, rebounding from the floor with reduced speed after each hit. (c) The a-t graph shows a short, sharp pulse of acceleration. Situation: a uniformly moving cricket ball is turned back by hitting it with a bat for a very short time interval (a large force acts briefly).

2.16 Figure 2.13 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 13). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.

ANSWER For simple harmonic motion x = A cos(ωt), velocity v = dx/dt ∝ −sin(ωt), and acceleration a = −ω2x always points opposite to x. Reading the standard SHM curve: At t = 0.3 s: the particle is on the negative side, moving towards the mean position. So x < 0, v < 0, a > 0 (acceleration is opposite to x). At t = 1.2 s: the particle is on the positive side. So x > 0, v > 0, a < 0. At t = −1.2 s: the particle is on the negative side but moving towards the positive side. So x < 0, v > 0, a > 0. Verification: matches the textbook answers (x < 0, v < 0, a > 0; x > 0, v > 0, a < 0; x < 0, v > 0, a > 0).

2.17 Figure 2.14 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least ? Give the sign of average velocity for each interval.

ANSWER Average speed over an interval = (magnitude of slope of the x-t graph) and is greatest where the curve is steepest. Average speed is greatest in interval 3 and least in interval 2 (interval 3 has the steepest curve, interval 2 the flattest). Sign of average velocity: it follows the direction of the slope — positive in intervals 1 and 2 (position increasing) and negative in interval 3 (position decreasing). Verification: matches the textbook answer (greatest in 3, least in 2; v > 0 in 1 and 2, v < 0 in 3).

2.18 Figure 2.15 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 ?

ANSWER Average acceleration = magnitude of slope of the speed-time graph; it is greatest where the speed changes most steeply — this is in interval 2. Average speed is greatest where the curve is highest overall — this is in interval 3. Signs (positive direction = direction of motion): the particle always moves in the positive direction, so v > 0 in all three intervals (1, 2 and 3). Acceleration = slope of the speed curve: speed rising in interval 1 → a > 0; speed falling in interval 2 → a < 0; speed rising again in interval 3 → a > 0. At points A, B, C and D the curve has its peaks/troughs (turning points), where the tangent is horizontal, so the slope — and hence the acceleration — is zero at A, B, C and D. Verification: matches the textbook answer (acceleration magnitude greatest in 2; speed greatest in 3; v > 0 in 1, 2, 3; a > 0 in 1 and 3, a < 0 in 2; a = 0 at A, B, C, D).

Extra Practice Questions

Short Answer Type Questions

Q1. Define instantaneous velocity. How is it obtained from a position–time graph?

ANSWERInstantaneous velocity is the limit of average velocity as the time interval tends to zero, v = dx/dt. On a position–time graph it equals the slope of the tangent drawn to the curve at that instant.

Q2. A car accelerates uniformly from rest at 4 m s−2. Find its velocity after 5 s and the distance covered in this time.

ANSWERv = v0 + at = 0 + 4 × 5 = 20 m s−1. Distance x = v0t + ½at2 = 0 + ½ × 4 × 52 = ½ × 4 × 25 = 50 m.

Q3. Why can the acceleration due to gravity be taken as negative for a body thrown upward?

ANSWERIf we take the upward direction as positive, then g acts downward (opposite to the chosen positive direction), so a = −g. This negative acceleration decelerates the body on the way up and accelerates it on the way down.

Q4. State two differences between distance and displacement.

ANSWER(i) Distance is a scalar (only magnitude) while displacement is a vector (magnitude and direction). (ii) Distance is always positive and equals the total path length, whereas displacement is the shortest straight-line change in position and can be positive, negative or zero.

Q5. A body covers half its total path in the last second of its free fall from rest. Without finding the height, state which kinematic relation links distance and time here.

ANSWERThe relation is y = ½gt2 (starting from rest, v0 = 0). The distance in the last second is y(t) − y(t−1), and setting this equal to ½y(t) gives an equation in t that can be solved for the total time of fall.

Long Answer Type Questions

Q1. Derive the three kinematic equations of motion for a uniformly accelerated body using the velocity–time graph.

ANSWERConsider a body with initial velocity v0 and constant acceleration a; its v–t graph is a straight line. First equation: the slope of the line gives a = (v − v0)/t, so v = v0 + at. Second equation: displacement = area under the v–t graph = area of rectangle (v0t) + area of triangle (½ × t × (v − v0)) = v0t + ½t(at) = x = v0t + ½at2. Third equation: the area can also be written as the trapezium ½(v + v0)t; substituting t = (v − v0)/a from the first equation gives x = (v + v0)(v − v0)/(2a) = (v2 − v02)/(2a), i.e. v2 = v02 + 2ax.

Q2. A stone is dropped from the top of a tower 80 m high. Taking g = 10 m s−2, find (a) the time to reach the ground and (b) its speed on hitting the ground.

ANSWER(a) Dropped from rest, v0 = 0, height h = 80 m. Using h = ½gt2: 80 = ½ × 10 × t2 → t2 = 16 → t = 4 s. (b) v = v0 + gt = 0 + 10 × 4 = 40 m s−1. (Check: v2 = 2gh = 2 × 10 × 80 = 1600 → v = 40 m s−1.)

Q3. Explain, with the help of velocity–time graphs, how the area under the graph gives displacement and the slope gives acceleration. Why is this useful?

ANSWEROn a velocity–time graph, the slope of the line at any point equals dv/dt, which is the acceleration; a steeper line means greater acceleration, a horizontal line means zero acceleration. The area enclosed between the graph and the time axis equals Σ(v × Δt), which is the displacement over that interval (for uniform velocity u over time T the area is the rectangle u × T = displacement). These graphical results are powerful because they let us read off displacement and acceleration directly, handle motion even when acceleration is not constant (by measuring the area/slope), and quickly compare different motions without solving equations.

Multiple Choice Questions (MCQs)

1. The slope of the tangent to a position–time graph at any instant gives the:

(a) distance    (b) instantaneous velocity    (c) acceleration    (d) average speed

2. The area under a velocity–time graph represents the:

(a) acceleration    (b) speed    (c) displacement    (d) jerk

3. A speed of 90 km h−1 is equal to:

(a) 18 m s−1    (b) 25 m s−1    (c) 30 m s−1    (d) 45 m s−1

4. At the highest point of its path, a ball thrown vertically upward has:

(a) zero velocity and zero acceleration    (b) zero velocity and acceleration g downward    (c) maximum velocity    (d) zero acceleration

5. For a body in free fall from rest, the distance fallen in time t is:

(a) gt    (b) ½gt    (c) ½gt2    (d) 2gt2

6. Which kinematic equation does NOT contain time t?

(a) v = v0 + at    (b) x = v0t + ½at2    (c) v2 = v02 + 2ax    (d) x = vt

7. The stopping distance of a vehicle is proportional to:

(a) the initial velocity    (b) the square of the initial velocity    (c) the deceleration    (d) the time only

8. For an object moving with constant velocity, the position–time graph is a:

(a) parabola    (b) straight line inclined to the time axis    (c) horizontal straight line    (d) circle

9. Two trains move in the same direction at 60 km h−1 and 40 km h−1. The velocity of the first relative to the second is:

(a) 100 km h−1    (b) 20 km h−1    (c) 0    (d) 50 km h−1

10. The average speed of a particle is always:

(a) less than the magnitude of average velocity    (b) equal to the magnitude of average velocity    (c) greater than or equal to the magnitude of average velocity    (d) zero

Answer key: 1-(b), 2-(c), 3-(b), 4-(b), 5-(c), 6-(c), 7-(b), 8-(b), 9-(b), 10-(c).

Assertion–Reason Questions

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: A body can have zero velocity and still have non-zero acceleration.

Reason: At the highest point of vertical motion, a ball has zero velocity but acceleration equal to g.

A-R 2. Assertion: The average speed of a moving particle can never be less than the magnitude of its average velocity.

Reason: The total path length is always greater than or equal to the magnitude of the displacement.

A-R 3. Assertion: Instantaneous speed is always equal to the magnitude of instantaneous velocity.

Reason: Over an infinitesimally small time interval the path length equals the magnitude of the displacement.

A-R 4. Assertion: A positive value of acceleration always means the particle is speeding up.

Reason: Acceleration and velocity are both vector quantities.

A-R 5. Assertion: The position–time graph of a uniformly accelerated body is a parabola.

Reason: For uniform acceleration, position varies as the square of time (x = v0t + ½at2).

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).

Common Mistakes to Avoid

Watch out for these

  • Confusing distance with displacement (and average speed with magnitude of average velocity) — remember speed/distance count the whole path, velocity/displacement count only net change.
  • Thinking acceleration is zero at the highest point of a thrown ball — the velocity is zero there, but acceleration stays g = 9.8 m s−2 downward.
  • Forgetting to convert km h−1 to m s−1 (multiply by 5/18) before using SI-unit formulas.
  • Mixing up signs: fix one positive direction first, then assign signs to v0, v, a and x consistently (e.g. up positive → g is negative).
  • Assuming a positive acceleration means speeding up — it depends on the direction of acceleration relative to velocity.
  • Treating an x-t graph as the actual path of the particle — it only shows position versus time, not the trajectory in space.

Exam Tips

How to score full marks in this chapter

Always write the given data with units, choose a sign convention, and state the formula before substituting. For free-fall problems, decide your positive direction first and keep g consistent (use g = 9.8 or 10 m s−2 exactly as the question says). Pick the kinematic equation that involves only the known quantities plus the unknown — use v2 = v02 + 2ax when time is neither given nor asked. For graph questions, remember the golden rule: slope of x-t = velocity, slope of v-t = acceleration, area under v-t = displacement. Round the final answer to the correct significant figures and clearly state the unit.

Frequently Asked Questions

What is Class 11 Physics Chapter 2 Motion in a Straight Line about?

Chapter 2 deals with rectilinear (straight-line) motion in kinematics — instantaneous velocity and speed, acceleration, the three kinematic equations for uniformly accelerated motion, special cases such as free fall and stopping distance, relative velocity, and the interpretation of x-t, v-t and a-t graphs.

How many exercise questions are there in Class 11 Physics Chapter 2?

The NCERT textbook (session 2026-27) has Exercises numbered 2.1 to 2.18 at the end of the chapter. All eighteen are reproduced here with complete step-by-step solutions, and the numericals are cross-checked against the official answer key.

What are the three kinematic equations of motion?

For constant acceleration a: (1) v = v0 + at, (2) x = v0t + half a t squared, and (3) v squared = v0 squared + 2ax. If the body is at position x0 at t = 0, x is replaced by (x – x0).

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