Exercises — Motion in a Straight Line

18 questions✓ Free · step-by-step

2.12 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.1
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.
Hint. A body counts as a 'point object' only when its own size is negligible compared to the distances relevant to the motion being described — check whether the body's shape, spin, or orientation actually matters to the question at hand.
Step 1 — (a) Railway carriage between stations. The distance between stations is enormously large compared to the carriage's own length, so its size can be ignored: a point object.

Step 2 — (b) Monkey on a cyclist on a circular track. The radius of the track is large compared to the size of the man-and-monkey system, so for describing their overall path around the track, they can be treated as a point object too.

Step 3 — (c) Spinning cricket ball turning sharply. The sharp turn on hitting the ground depends on the ball's own spin and surface, which are properties tied directly to its finite size, so its dimensions cannot be ignored here — not a point object.

Step 4 — (d) Tumbling beaker. Tumbling is a rotation of the beaker about its own shape and size as it falls, so its dimensions again matter directly to the motion being described — not a point object.

✦ Answer: (a) Yes (b) Yes (c) No (d) No.
Where students slip. Assuming (b) must be 'No' because a monkey and a cyclist are clearly not points in real life — the test isn't whether the body looks small, but whether its size matters to the specific motion being tracked, and here only the overall path around the large circular track is asked about.
2.23 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.2
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).
Hint. Read the graph feature by feature: which line starts at t = 0, which is steeper, where their heights end up, and where the two lines cross.
Step 1 — Starting times and final positions. Line A begins right at t = 0 (leaves school immediately) and ends at the lower marked height P; line B only begins later, where it crosses the time axis to the right of the origin, and ends at the higher marked height Q.

Step 2 — Comparing distances from school. Since P is lower on the x-axis than Q, A's home is closer to school than B's home.

Step 3 — Comparing slopes (speeds). Line B climbs far more steeply than line A over the same kind of time span, since a steeper x-t line means covering more distance per unit time, so B walks faster.

Step 4 — The single crossing point. The two lines cross exactly once. Before the crossing, A is already ahead in position while B hasn't even left school yet; after the crossing, B has caught up and moved ahead, since B moves faster despite starting later — so B overtakes A once.

Step 5 — Arrival times. Line B's endpoint (at Q) sits at an earlier time value than line A's endpoint (at P), so despite starting later, B's much greater speed gets it home first — the two do not arrive at the same time.

✦ Answer: (a) A lives closer to the school than B. (b) A starts from school earlier than B. (c) B walks faster than A. (d) A and B reach home at different times. (e) B overtakes A on the road once.
Where students slip. Assuming that because B starts later, B must also arrive later — B's steeper slope (greater speed) more than makes up for the late start, so it actually reaches home first, at an earlier time than A.
2.33 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.3
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.
Hint. Break the day into three separate motion segments and work out the time each one takes before trying to plot anything.
Step 1 — Walking to the office. Time = distance / speed = 2.5 km / 5 km h⁻¹ = 0.5 h = 30 min, so she reaches the office at 9.30 am, at x = 2.5 km.

Step 2 — Staying at the office. She remains at x = 2.5 km, unmoving, from 9.30 am until 5.00 pm — a flat, horizontal segment on the x-t graph lasting 7.5 h.

Step 3 — Returning home by auto. Time = 2.5 km / 25 km h⁻¹ = 0.1 h = 6 min, so she is back home (x = 0) by 5.06 pm.

✦ Answer: The x-t graph is three straight segments — a steep rise from (9.00 am, 0) to (9.30 am, 2.5 km); a flat line at x = 2.5 km from 9.30 am to 5.00 pm; and a very steep fall from (5.00 pm, 2.5 km) back down to (5.06 pm, 0), since the much higher speed of the auto covers the same 2.5 km in a much shorter time than the walk out.
Where students slip. Drawing the return trip with the same slope (steepness) as the outward trip — the auto travels five times faster than her walking speed, so the return segment must be drawn much steeper (covering the same distance in a fifth of the time), not parallel to the outward line.
2.43 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.4
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.
Hint. Track his net position at the end of each complete forward-plus-backward cycle first, then zoom into the one cycle where he actually crosses 13 m.
Step 1 — Net progress per cycle. Each cycle is 5 steps forward (+5 m) then 3 steps backward (−3 m), a net gain of +2 m over 8 s.

Step 2 — Position at the end of each cycle. After cycle 1 (t = 8 s): x = 2 m. Cycle 2 (t = 16 s): x = 4 m. Cycle 3 (t = 24 s): x = 6 m. Cycle 4 (t = 32 s): x = 8 m — each cycle simply adds another 2 m.

Step 3 — The critical cycle. Cycle 5 begins at x = 8 m (t = 32 s). Since position rises by 1 m per second during the forward steps, it reaches 9 m (t = 33 s), 10 m (34 s), 11 m (35 s), 12 m (36 s), 13 m (37 s) — landing on the pit exactly on the fifth forward step of cycle 5, before he would have started stepping backward again.

✦ Answer: The drunkard falls into the pit 37 seconds after he starts, right at the end of his forward steps in the fifth cycle.
Where students slip. Assuming the net +2 m per cycle means the pit at 13 m is reached partway through a backward stretch — checking cycle-by-cycle shows 13 m actually falls exactly at the peak of the forward stretch in cycle 5, not during any backward stretch.
2.52 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.5
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?
Hint. Convert the speed to m/s first, then use v² = u² − 2as with the final speed v = 0.
Step 1 — Convert the initial speed. 126 km h⁻¹ = 126 × 1000 / 3600 = 35 m s⁻¹.

Step 2 — Find the retardation. Using v² = u² − 2as with v = 0, u = 35 m s⁻¹, s = 200 m: 0 = 35² − 2a(200), so a = 1225 / 400 ≈ 3.06 m s⁻².

Step 3 — Find the time to stop. Using v = u − at: 0 = 35 − (3.06)t, so t = 35 / 3.06 ≈ 11.4 s.

✦ Answer: Retardation ≈ 3.06 m s⁻²; the car takes about 11.4 s to stop.
Where students slip. Plugging 126 km h⁻¹ directly into the equations without converting to m s⁻¹ first — the kinematic equations here need speed and distance in consistent SI units (m and s), not a mix of km/h and metres.
2.64 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.6
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).
Hint. For (c), remember that acceleration due to gravity never switches off or reverses — only the ball's own velocity changes sign, since the origin and positive axis are fixed by the problem's own choice.
Step 1 — (a) Direction of acceleration going up. Gravity always pulls straight downward, regardless of which way the ball itself is moving, so the acceleration during the upward motion is directed vertically downward.

Step 2 — (b) At the highest point. The ball is momentarily at rest, so velocity = 0, but gravity has not switched off — acceleration is still g = 9.8 m s⁻² directed downward, since zero velocity at an instant does not mean zero acceleration at that instant.

Step 3 — (c) Signs with downward as positive, origin at the top. Below the highest point (which is where the ball is during both the upward approach and the downward fall) counts as positive x, since downward is positive. During the upward motion, the ball sits below the top (x > 0) but is moving toward the origin, i.e. moving in the negative (upward) direction, so v < 0; acceleration stays positive throughout, since gravity is always downward. During the downward motion, the ball again sits below the top (x > 0), now moving further downward, i.e. in the positive direction, so v > 0, with a still positive.

Step 4 — (d) Maximum height and total time. Max height h = u² / (2g) = 29.4² / (2 × 9.8) = 864.36 / 19.6 = 44.1 m. Time to the top: t = u/g = 29.4/9.8 = 3 s, and by symmetry the fall back down takes the same 3 s, so the ball returns to the player's hands after 6 s in total.

✦ Answer: (a) downward (b) v = 0, a = 9.8 m s⁻² downward (c) upward motion: x > 0, v < 0, a > 0; downward motion: x > 0, v > 0, a > 0 (d) rises 44.1 m, returns after 6 s.
Where students slip. In (c), assuming acceleration must change sign to match whichever way the ball happens to be moving — acceleration due to gravity keeps the same sign (downward = positive here) throughout the entire flight; only the velocity's sign flips between the upward and downward halves.
2.74 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.7
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.
Hint. For each part, ask whether the claim is forced to be true in every possible case, or whether a single counterexample already breaks it.
Step 1 — (a) Zero speed, non-zero acceleration. True. A ball thrown straight up has zero speed for an instant at its highest point, yet gravity is still acting on it there, giving a non-zero acceleration g.

Step 2 — (b) Zero speed, non-zero velocity. False. Speed is defined as the magnitude of velocity, so if speed is exactly zero, the velocity vector itself must also be zero at that instant — the two cannot disagree.

Step 3 — (c) Constant speed, zero acceleration. False in general, since speed only tracks magnitude, not direction. A classic counterexample is a particle in uniform circular motion: its speed never changes, yet its direction is changing every instant, so its velocity (and hence acceleration) is non-zero throughout — this is exactly why physics treats velocity and acceleration as vectors and speed as only their magnitude.

Step 4 — (d) Positive acceleration must mean speeding up. False, since this depends on which way the particle is currently moving relative to the chosen positive direction. If a particle is moving in the negative direction while its acceleration is positive, the acceleration opposes the motion and slows the particle down, rather than speeding it up.

✦ Answer: (a) True (b) False (c) False (d) False.
Where students slip. In (d), assuming 'positive acceleration' always means 'gaining speed' — whether acceleration speeds a particle up or slows it down depends on whether acceleration and velocity point the same way, not on the acceleration's sign alone.
2.83 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.8
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.
Hint. Work out the fall time and impact speed first using the standard free-fall formulas, then handle each bounce as a fresh, slower free-fall/rise pair.
Step 1 — Time and speed of the first fall. From h = ½gt², t = √(2h/g) = √(2 × 90/9.8) ≈ 4.29 s. Impact speed v = √(2gh) = √(2 × 9.8 × 90) = √1764 = 42 m s⁻¹.

Step 2 — After the first bounce. The ball loses a tenth of its speed, leaving it with 42 × 0.9 = 37.8 m s⁻¹ upward. It decelerates under gravity, reaching zero speed after 37.8/9.8 ≈ 3.86 s, then falls back down over the same 3.86 s, reaching 37.8 m s⁻¹ again just before the second bounce.

Step 3 — Check the total time. 4.29 s (first fall) + 3.86 s (rise) + 3.86 s (fall back) ≈ 12.0 s, since these three legs sum to exactly the requested window, landing right at its edge.

✦ Answer: The graph rises in a straight line from (0, 0) to (4.29 s, 42 m/s), drops instantly to (4.29 s, 37.8 m/s) at the bounce, falls in a straight line to (8.14 s, 0) as the ball rises and slows, then climbs in a straight line back up to about (12.0 s, 37.8 m/s) as it falls again toward the second bounce.
Where students slip. Drawing the speed as negative while the ball moves upward after the bounce — this is a speed-time graph, and speed is always a non-negative quantity by definition, so the upward leg is drawn as a falling straight line toward zero, never dipping below the time axis.
2.93 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.9
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. 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?
Hint. Try a concrete round-trip example — walking away and then partway back — and compute both quantities directly to see why the path length is never smaller.
Step 1 — (a) A concrete example. Suppose a person walks 4 m east then 3 m west. Net displacement magnitude = 4 − 3 = 1 m, but total path length = 4 + 3 = 7 m — the path length counts every metre actually walked, while displacement only cares about the net change in position.

Step 2 — (a) Why path length ≥ displacement magnitude. Any reversal of direction adds extra distance to the path without adding (and sometimes even subtracting from) the net displacement, so the total path length can never fall below the magnitude of the net displacement.

Step 3 — (b) Extending to average velocity and average speed. Dividing both of the quantities in (a) by the same time interval preserves the inequality, since average speed = path length / time and the magnitude of average velocity = |displacement| / time.

Step 4 — When equality holds. The two quantities become equal exactly when the particle never reverses direction during the interval, since only then does the path length exactly equal the magnitude of the displacement.

✦ Answer: Path length ≥ |displacement|, and average speed ≥ |average velocity|, with equality only when the motion is one-way throughout the interval (no change of direction).
Where students slip. Assuming the two quantities are always equal for 'simple' back-and-forth motion — equality specifically requires no reversal at all; even a single change of direction during the interval is enough to make the path length strictly greater than the displacement's magnitude.
2.104 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.10
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?
Hint. Work out how far he's travelled and how far from home he actually is at each of the three checkpoints (30, 50, 40 minutes) before computing anything.
Step 1 — Time for each leg of the trip. Going: 2.5 km / 5 km h⁻¹ = 0.5 h = 30 min. Returning: 2.5 km / 7.5 km h⁻¹ = 1/3 h ≈ 20 min. Total round trip: 50 min.

Step 2 — (i) 0 to 30 min. This is exactly the outward trip: displacement = 2.5 km, path length = 2.5 km, both over 0.5 h. Average velocity magnitude = average speed = 2.5/0.5 = 5 km h⁻¹.

Step 3 — (ii) 0 to 50 min. This is the full round trip: net displacement = 0 (he's back home), so average velocity magnitude = 0. But total path length = 5 km over 50/60 h, giving average speed = 5/(5/6) = 6 km h⁻¹.

Step 4 — (iii) 0 to 40 min. The first 30 min covers the outward 2.5 km; the remaining 10 min = 1/6 h is on the return leg, covering 7.5 × 1/6 = 1.25 km back toward home, leaving him 2.5 − 1.25 = 1.25 km from home. Displacement magnitude = 1.25 km over 40/60 h, giving average velocity magnitude = 1.25/(2/3) = 1.875 km h⁻¹. Path length = 2.5 + 1.25 = 3.75 km over the same time, giving average speed = 3.75/(2/3) = 5.625 km h⁻¹.

✦ Answer: (i) both 5 km h⁻¹ (ii) 0 km h⁻¹ and 6 km h⁻¹ (iii) 1.875 km h⁻¹ and 5.625 km h⁻¹.
Where students slip. In (ii), reporting average speed as 0 as well just because average velocity is 0 — the man still walked the full 5 km even though he ended up back where he started, and average speed only cares about that total path length, not the net displacement.
2.112 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.11
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?
Hint. Think about what happens to the difference between 'path length' and 'displacement magnitude' as the time interval used to measure them shrinks toward zero.
Step 1 — Why the gap exists over a finite interval. Over a finite interval, a particle can change direction partway through, making the actual path length longer than the net (straight-line) displacement.

Step 2 — What happens as the interval shrinks to zero. As the time interval Δt shrinks toward zero, the particle has vanishingly little time to change direction within it, so the tiny path it covers becomes indistinguishable from a straight segment, meaning path length and displacement magnitude converge to the same value in that limit.

✦ Answer: In the limit Δt → 0, the path traversed is effectively a single straight segment with no room for a direction reversal, so the distinction between path length and displacement magnitude vanishes — which is exactly why instantaneous speed always equals the magnitude of the instantaneous velocity.
Where students slip. Treating this as a separate, unrelated fact from Q2.9/2.10 — it's the exact same path-length-vs-displacement gap discussed there, just examined in the special case where the time interval has shrunk all the way to zero.
2.123 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.12
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.
Hint. For a graph to represent real one-dimensional motion, the plotted quantity must be a genuine single-valued function of time, and speed specifically can never be negative or ever decrease as total distance covered.
Step 1 — (a) The x-t graph. This curve loops back on itself, so a single value of t inside the loop corresponds to two different values of x at once — impossible, since a particle can only be in one place at a given instant.

Step 2 — (b) The v-t graph. This is a closed circle, which likewise assigns two different values of v to most values of t (the upper and lower halves of the circle) — impossible for the same reason.

Step 3 — (c) The speed-time graph. The curve dips below the time axis into negative values, but speed is defined as a magnitude and can never be negative — impossible.

Step 4 — (d) The total-path-length-vs-time graph. This curve decreases over some stretches, but total path length only ever accumulates as time passes — it can never shrink — so a decreasing segment is impossible.

✦ Answer: All four graphs (a)-(d) are impossible for one-dimensional motion — (a) and (b) are not single-valued functions of time, (c) shows negative speed, and (d) shows total path length decreasing.
Where students slip. Assuming only one of the four graphs is the 'trick' answer — this question deliberately uses a different, independent reason to disqualify each of the four graphs, and all four fail.
2.133 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.13
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.
Hint. Look very carefully at what the curve actually does for t < 0 — a flat line lying along the time axis is a very specific, limited kind of motion, not just any 'straight line.'
Step 1 — Why the t < 0 description is misleading. For t < 0, the curve does not merely lie along some sloped straight line — it coincides exactly with the time axis itself, meaning x stays at zero throughout. This describes a particle sitting completely at rest at the origin, not a particle moving with some constant velocity along a line.

Step 2 — 'Straight line' language is about the path in space, not the shape of the x-t graph. 'Motion in a straight line' describes the trajectory in physical space, which for one-dimensional motion is trivially always a straight line (the x-axis itself) — so calling out t < 0 as 'a straight line' adds no real information; the graph's flatness there really means zero velocity, not constant velocity.

Step 3 — A suitable physical context. A particle initially at rest at the origin, which starts moving only from t = 0 onward as an ever-increasing force (or field) is switched on — for instance, a charged particle released at rest in a region where the accelerating field grows stronger with time, so its position rises with a continually increasing slope, not necessarily a plain parabola.

✦ Answer: No — for t < 0 the particle is simply at rest at the origin, not moving in a straight line with constant velocity; the t > 0 portion need not be exactly parabolic either. A fitting context is a particle at rest until t = 0, after which it begins accelerating under a force that keeps growing with time.
Where students slip. Reading 'moves in a straight line' at face value for t < 0 — every one-dimensional path is trivially a straight line, so the graph being flat there actually signals rest (zero velocity), which is a much stronger and more specific claim than merely 'straight-line motion.'
2.143 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.14
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?
Hint. Muzzle speed is always quoted relative to the gun itself, so first add the van's own speed to get the bullet's speed relative to the ground, then subtract the thief's car's speed to get the relative approach speed.
Step 1 — Convert all speeds to m s⁻¹. Van: 30 × 1000/3600 ≈ 8.33 m s⁻¹. Thief's car: 192 × 1000/3600 ≈ 53.33 m s⁻¹.

Step 2 — Bullet's speed relative to the ground. Since the muzzle speed (150 m s⁻¹) is measured relative to the moving van, and both point the same way: bullet's ground speed = 8.33 + 150 = 158.33 m s⁻¹.

Step 3 — Bullet's speed relative to the thief's car. This is what actually matters for the impact, since both the bullet and the car are moving: relative speed = 158.33 − 53.33 = 105 m s⁻¹.

✦ Answer: The bullet hits the thief's car at a relative speed of 105 m s⁻¹.
Where students slip. Reporting the bare muzzle speed (150 m s⁻¹) as the final answer — that number is the bullet's speed relative to the van that fired it, not relative to the moving car it actually strikes; both the van's and the car's own speeds must be folded in first.
2.153 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.15
Suggest a suitable physical situation for each of the following graphs (Fig 2.12).
Hint. All three panels share a theme — read them as a single story about a ball being struck, in flight, and landing, rather than three unrelated graphs.
Step 1 — (a) What the position graph shows. x stays at zero for t < 0 (the object is at rest at some reference position), then rises sharply to a peak and falls back down past the starting level to settle at a small negative value. This fits a ball resting on the ground that is suddenly struck upward, rises, falls back down, and embeds itself slightly into a soft or sandy landing surface, ending up just below its original level.

Step 2 — (b) The v-t graph. Repeated straight-line segments, each sloping steadily downward and then jumping back up to a slightly lower peak than before. This matches a ball bouncing repeatedly on a hard floor: each downward-sloping segment is free fall under gravity (constant deceleration/acceleration), and each abrupt jump is the near-instant reversal of velocity at a bounce, with each bounce a little weaker than the last since some speed is lost on impact.

Step 3 — (c) The a-t graph. Acceleration sits at zero, then spikes sharply for a very brief moment before returning to zero. This fits a ball being struck by a bat: before and after the brief contact there is no applied force (acceleration ≈ 0, ignoring gravity), but during the short contact itself the force — and hence acceleration — is very large.

✦ Answer: (a) a ball hit upward from rest that falls back and settles slightly below its start (e.g. into sand) (b) a ball bouncing repeatedly on a hard floor, losing some speed at each bounce (c) a ball struck briefly by a bat, with acceleration near zero except during the short impact.
Where students slip. Treating the three graphs as needing three completely unrelated stories — they are far easier to justify, and match the intended teaching point, as three different views (position, velocity, acceleration) of the same general kind of impact event.
2.163 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.16
Figure 2.13 gives the x-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 s, 1.2 s, - 1.2 s.
Hint. Read the period directly off the graph (the zero-crossings repeat every 1 s), then track, for each given time, whether the curve is above or below the axis and whether it is heading up or down at that instant.
Step 1 — Reading the graph's shape. The curve crosses zero every 1 s and, just after t = 0, immediately dips down toward a trough around t = 0.5 s before rising back up to a peak around t = 1.5 s — consistent with x(t) = −A sin(πt).

Step 2 — At t = 0.3 s. This sits between the zero-crossing at t = 0 and the trough at t = 0.5 s, so the particle is below the axis (x < 0) and still heading toward the trough (v < 0). In SHM, acceleration always points back toward the centre (a = −ω²x), so with x negative, acceleration is positive.

Step 3 — At t = 1.2 s. This sits between the zero-crossing at t = 1 s and the peak at t = 1.5 s, so the particle is above the axis (x > 0) and still rising toward the peak (v > 0), giving a negative acceleration (restoring force pulling it back down from a positive displacement).

Step 4 — At t = −1.2 s. This sits between the trough at t = −1.5 s and the zero-crossing at t = −1 s, so the particle is below the axis (x < 0) but climbing back up out of the trough toward zero (v > 0), with acceleration again positive since x is negative.

✦ Answer: t = 0.3 s: x < 0, v < 0, a > 0. t = 1.2 s: x > 0, v > 0, a < 0. t = -1.2 s: x < 0, v > 0, a > 0.
Where students slip. Assuming acceleration's sign tracks whether the particle is speeding up or slowing down — in SHM, acceleration's sign is fixed entirely by the sign of the displacement (a = −ω²x, always pointing back toward the centre), regardless of which way the particle happens to be moving at that instant.
2.173 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.17
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.
Hint. Average speed over an interval tracks how steep the curve is there — compare the steepness of the curve across the three marked bands rather than their height.
Step 1 — Locating the three intervals. Interval 1 sits on the rising part of the curve before its peak. Interval 2 sits right where the curve reaches its peak, where the tangent is nearly flat. Interval 3 sits well past the peak, on the sharply falling section of the curve.

Step 2 — Comparing steepness. Interval 2, straddling the peak, has the smallest slope of the three (the curve is nearly horizontal there), so it has the least average speed. Interval 3, on the steep falling stretch, is visibly the steepest of the three, giving it the greatest average speed.

Step 3 — Signs of average velocity. In interval 1, x is still rising, so average velocity is positive. In interval 2, x is still edging up toward the peak (or essentially flat), so average velocity is positive but very small. In interval 3, x is falling sharply past zero toward negative values, so average velocity is negative.

✦ Answer: Average speed is greatest in interval 3 and least in interval 2. Signs of average velocity: interval 1 positive, interval 2 positive (small), interval 3 negative.
Where students slip. Assuming the interval with the highest position on the graph (interval 2, at the peak) must have the greatest average speed — average speed depends on the slope of the curve within the interval, not on how high up the curve happens to sit there.
2.184 marksNCERT Cl-11 Physics Part I, Ch2 Exercises, Q2.18
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?
Hint. Since the motion never reverses direction, velocity has the same sign as speed throughout — focus on comparing how steeply and how high the curve runs in each of the three marked bands, and check whether each labelled point sits at a peak, a trough, or neither.
Step 1 — Locating the intervals and points. A is near the start (a nearly flat, low part of the curve), B is a local peak, C is a local trough after B, and D is the highest peak of all, reached by climbing steeply from C. Interval 1 sits on the A-to-B rise, interval 2 on the B-to-C fall, and interval 3 on the steep C-to-D rise.

Step 2 — Comparing the three intervals. The C-to-D climb (interval 3) is visibly the steepest of the three rises/falls shown, so it has the greatest average acceleration magnitude. Since D is also the highest point on the whole graph, the speeds within interval 3 are the largest of the three intervals too, giving it the greatest average speed as well.

Step 3 — Signs of v and a. Because the particle never reverses direction, v is positive throughout all three intervals (speed and velocity coincide here). In interval 1 speed is rising, so a > 0; in interval 2 speed is falling, so a < 0; in interval 3 speed is rising again, so a > 0.

Step 4 — Accelerations at A, B, C, D. Each of these four points sits at a place where the speed-time curve's tangent is momentarily horizontal — A at the flat start, B and D at peaks, C at a trough — and a horizontal tangent means zero slope, so acceleration is zero at all four points.

✦ Answer: Interval 3 has both the greatest average acceleration magnitude and the greatest average speed. v > 0 in all three intervals; a > 0 in interval 1, a < 0 in interval 2, a > 0 in interval 3. Acceleration is zero at A, B, C and D, since the curve's slope is momentarily flat at each of these points.
Where students slip. Assuming acceleration must be non-zero at A, B and D just because they look like 'important' labelled points on a rising curve — B, C and D are turning points (peak, trough, peak) where the slope is momentarily zero, and A is drawn with the same flat starting tangent, so all four in fact have zero acceleration.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Physics Part I textbook, Reprint 2026-27 (keph102.pdf) — one end-of-chapter Exercises set (18 questions, 2.1-2.18); figures 2.9-2.15 were rendered directly from the PDF at high resolution and read visually to verify every graph-based question rather than assumed from memory. Questions are referenced from the NCERT textbook for identification.

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