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

Concept. A body can be treated as a point object when its size is much smaller than the distance it travels (or than the other lengths in the problem) and when its internal motion (rotation, spin, deformation) does not matter for what we want to describe. Then all its parts move almost identically and a single position coordinate describes the whole body.

1. Compare the size of the object with the distance it moves (or the radius of its path).
2. Ask whether rotation, spin or deformation affects the result.
3. If the size is negligible and all parts move alike, use the point-object approximation; otherwise treat the body as extended (Chapter 6).

Examples beyond the question. The Earth is a point object when we study its revolution round the Sun (radius 6.4 × 10⁶ m ≪ orbital radius 1.5 × 10¹¹ m, ratio about 4 × 10⁻⁵); the same Earth cannot be a point object when we study its rotation about its axis or the motion of a cyclone on its surface.

Extra depth. The point-object model is the foundation of kinematics in Chapters 2 and 3: it lets us describe motion with just one position, velocity and acceleration. Even a large body can be a point object if only its centre of mass motion is considered (Chapter 6); for a thrown cricket bat, the centre of mass follows a smooth parabola while the bat tumbles around it.

Conclusion: (a) and (b) can be treated as point objects; (c) and (d) cannot because spin or tumbling matters.

Concept. On a position-time graph the height of a point gives the distance from the school O, the slope of the line gives the velocity (speed here), and the time at which a line begins on the t-axis gives the starting time. Two lines meeting at a point mean that the two children are at the same place at the same instant (one overtakes the other).

1. Speed of A = OP / t_A, speed of B = OQ / t_B; since the slope of B is bigger, B’s speed is greater although it starts later.
2. At the crossing point (the red dot) the two have the same x at the same t: this is the overtaking. Because B is faster, it can overtake only once and stays ahead afterwards; a second crossing would need A to speed up.
3. Both children travel from the school towards their homes, so x increases with t for both (positive velocity).

Check by numbers. Suppose OP = 5 km, OQ = 7.5 km; A walks at 0.5 km per unit time starting at 0, B walks at 1.5 km per unit time starting at 3: A is at 0.5 t, B at 1.5(t − 3); they meet when 0.5t = 1.5t − 4.5, i.e. t = 4.5 at x = 2.25 (inside the road): exactly one meeting.

Extra depth. The slope of a straight x-t line is a constant velocity; a steeper line means a larger speed. The area between the lines has no meaning. A horizontal x-t line would mean rest, and a line with a negative slope would mean motion back towards the school. Remember this reading skill: height = position, slope = velocity, intersection = meeting. In part (d) the answer depends on the end-points of the plotted lines; always read what the figure in front of you shows.

Conclusion: A lives closer than B; A starts earlier; B walks faster; B overtakes A once (the arrival times differ in the figure as printed, B arriving earlier).

Concept. An x-t graph is built from the three stages of motion; each stage with constant velocity is a straight line segment whose slope is the velocity, and a stay at rest is a horizontal line. We first find the time taken in each stage using time = distance/speed.

1. Stage 1 (walking to office): distance 2.5 km at 5 km h⁻¹ takes t₁ = 2.5/5 = 0.5 h = 30 min. She leaves at 9.00 am and reaches the office at 9.30 am (x = 2.5 km). Slope = 5 km h⁻¹.
2. Stage 2 (at the office): she stays from 9.30 am up to 5.00 pm (7.5 h). x = 2.5 km remains constant: a horizontal line, slope zero.
3. Stage 3 (returning by auto): distance 2.5 km at 25 km h⁻¹ takes t₃ = 2.5/25 = 0.1 h = 6 min. She leaves at 5.00 pm and reaches home at 5.06 pm. The slope is −25 km h⁻¹ (x decreases from 2.5 km to 0).

Scales chosen. Time axis: 1 cm = 1 hour from 9 am to 5.06 pm (8.1 h in all); position axis: 1 cm = 0.5 km. The first and last segments will look steep because the times (30 min and 6 min) are short compared with the 7.5 h stay. Average speed over the whole trip = total path 5 km / total time 8.1 h = 0.62 km h⁻¹, but average velocity is zero because she returns to the starting point (displacement 0).

Extra depth. Notice the three kinds of straight-line segments: positive slope (moving away), zero slope (rest) and negative slope (returning). The magnitudes of the slopes (5 and 25 km h⁻¹) show that the return journey is five times faster: the descending line is five times steeper. If the auto had taken the same time as walking, the descending line would be a mirror image of the ascending line.

Conclusion: The graph is the polygon (9.00 am, 0) → (9.30 am, 2.5 km) → (5.00 pm, 2.5 km) → (5.06 pm, 0).

Concept. Each step takes 1 s and covers 1 m, so one complete cycle (5 forward + 3 backward = 8 steps) takes 8 s and gives a net displacement of 5 − 3 = +2 m. The motion is periodic with a slowly advancing mean position; the pit will be reached during a forward run, as soon as his position first equals 13 m.

1. Graphically: draw the zig-zag with slope +1 m s⁻¹ for 5 s and −1 m s⁻¹ for 3 s repeatedly; the line first reaches x = 13 m at t = 37 s.
2. By calculation: after n complete cycles the position is 2n metres at the time 8n s. In the next forward run the maximum position is 2n + 5. He reaches 13 m during that run if 2n + 5 ≥ 13, i.e. n ≥ 4. So n = 4: position at 32 s is 8 m; he needs 13 − 8 = 5 more forward steps = 5 s.
3. Total time = 4 × 8 s + 5 s = 37 s.
4. Check that he does not reach 13 m earlier: in cycle 4 the maximum is 2 × 3 + 5 = 11 m < 13 m.

Common error: dividing 13 m by the net 2 m per cycle gives 6.5 cycles = 52 s, which ignores that he can reach the pit during a forward run, before he steps back. The answer 37 s counts only the 5 forward steps of the last cycle.

Extra depth. The motion illustrates the difference between average velocity and speed: his average speed is 1 m s⁻¹ (every step is 1 m in 1 s) but his average velocity over each 8 s cycle is only 2/8 = 0.25 m s⁻¹. Over the first 37 s the total path length is 37 m, but the displacement is 13 m (average velocity 13/37 = 0.35 m s⁻¹).

Conclusion: The drunkard falls into the pit at t = 37 s (4 complete cycles of 8 s plus 5 forward steps).

Concept. This is uniformly accelerated (here uniformly decelerated) motion in a straight line, so the three kinematic equations apply. Given: initial velocity v₀, final velocity v = 0, distance x; find a and t. Remember to convert km h⁻¹ into m s⁻¹ before using SI units.

1. v₀ = 126 km h⁻¹ = 126 × 5/18 m s⁻¹ = 35 m s⁻¹; v = 0; x = 200 m.
2. Use v² = v₀² + 2ax: 0 = (35)² + 2a(200) ⇒ a = −1225/400 = −3.06 m s⁻². The negative sign means the acceleration is opposite to the velocity; the retardation is 3.06 m s⁻².
3. Use v = v₀ + at: 0 = 35 + (−3.06)t ⇒ t = 35/3.0625 = 11.4 s.
4. Check with x = (v₀ + v)t/2 = (35 + 0)(11.43)/2 = 200 m ✓.

Meaning. A deceleration of 3.06 m s⁻² is about 0.31 g, a normal braking of a car. A car at 126 km h⁻¹ needs 11.4 s and 200 m to stop even before taking into account the driver’s reaction time, which is why speed limits and safe distances are important (see the stopping distance formula d_s = v₀²/2|a|: twice the speed means four times the distance).

Extra depth. Alternate direct formula for time: since the average velocity is (35 + 0)/2 = 17.5 m s⁻¹ and the distance is 200 m, t = 200/17.5 = 11.4 s. This avoids calculating a first. Express the retardation also in terms of g: a/g = 3.06/9.8 = 0.31. If the driver’s reaction time is 0.7 s (see Example 2.7), the car travels another 35 × 0.7 = 24.5 m before the brakes are applied, so the total stopping distance is about 225 m.

More to know — stopping distance and speed. From d_s = v₀²/(2|a|), at the same braking deceleration 3.06 m s⁻²: 60 km h⁻¹ (16.7 m s⁻¹) needs 45 m; 90 km h⁻¹ (25 m s⁻¹) needs 102 m; 126 km h⁻¹ needs 200 m. The reaction distance adds v₀ × t_r. Road-safety rules (speed limits in school zones, minimum following distance) are based on this square law. In the NCERT text (Example 2.6) braking distances of 10, 20, 34 and 50 m are quoted for 11, 15, 20 and 25 m s⁻¹, which correspond to deceleration of about 6 m s⁻² on a dry road — twice as much as in this problem.

Conclusion: The retardation is 3.06 m s⁻² and the car stops in 11.4 s.

Concept. Once the ball leaves the hand, the only force acting on it (neglecting air resistance) is gravity. Hence throughout the flight — going up, at the top and coming down — the ball has the same constant acceleration g = 9.8 m s⁻² directed vertically downward. The velocity changes (up → zero → down) but the acceleration does not.

(a) Direction of acceleration during the upward motion

Vertically downward, with magnitude g = 9.8 m s⁻². The acceleration is opposite to the velocity during ascent, so the speed decreases by 9.8 m s⁻¹ every second: 29.4, 19.6, 9.8, 0 at t = 0, 1, 2, 3 s.

(b) At the highest point

Velocity = 0 (momentarily at rest); acceleration = 9.8 m s⁻² downward (not zero). Zero velocity does not mean zero acceleration: the ball is about to reverse its motion precisely because gravity is still acting.

(c) Signs with origin at the top, downward positive, t = 0 at the top

(d) Maximum height and total time of flight

1. Upward: v = 0, v₀ = 29.4 m s⁻¹, a = −9.8 m s⁻² (upward positive). v² = v₀² + 2ay: 0 = (29.4)² − 2(9.8)H ⇒ H = 864.36/19.6 = 44.1 m.
2. Time to reach the top: v = v₀ + at ⇒ 0 = 29.4 − 9.8 t₁ ⇒ t₁ = 3.0 s.
3. By symmetry (no air resistance) the time of descent is also 3.0 s; the ball returns to the hand with speed 29.4 m s⁻¹ after t = 6.0 s.
4. Check: y = v₀ t − ½ g t² = 29.4(6) − 4.9(36) = 176.4 − 176.4 = 0 ✓.

Extra depth. Why is the acceleration positive in both parts of (c)? The sign depends only on the chosen axis (downward positive), not on whether the speed increases or decreases: gravity points down, i.e. along the positive axis, so a = +9.8 m s⁻² always. During ascent v is negative and a is positive (opposite): speed decreases. During descent both are positive: speed increases. This is the point stressed in ‘Points to ponder’ of the chapter. The time of ascent equals the time of descent, and the speed on return equals the speed of projection, because the motion is symmetric when air resistance is neglected.

Conclusion: (a) downward; (b) v = 0, a = 9.8 m s⁻² downward; (c) up: x +, v −, a +; down: x +, v +, a +; (d) 44.1 m and 6 s.

Concept. Use the definitions: speed is the magnitude of velocity (always ≥ 0), velocity is v = dx/dt, acceleration a = dv/dt. In one-dimensional motion velocity has only a sign (direction) and a magnitude. Test each statement by recalling a simple example or counter-example.

1. (a) example in numbers: v = 29.4 − 9.8t: at t = 3 s v = 0 while a = −9.8 m s⁻² ≠ 0.
2. (d) example in numbers: a car moving left (v = −10 m s⁻¹) with a = +2 m s⁻² (to the right): after 1 s v = −8 m s⁻¹, speed falls from 10 to 8 m s⁻¹: slowing down although a is positive.

Rule: if a and v have the same sign the particle speeds up; if opposite signs it slows down. The sign of a alone says nothing about speeding up or slowing down because it depends on the choice of the positive direction.

Extra depth. The statements are exactly the points to ponder of the chapter: (3) the sign of acceleration does not tell us whether speed is increasing or decreasing; (4) zero velocity does not imply zero acceleration. Statement (c) in two dimensions: uniform circular motion has constant speed v but acceleration v²/r directed to the centre, because the direction of velocity changes.

Conclusion: (a) True; (b) False; (c) True (in one dimension); (d) False.

Concept. The ball falls freely with a = g, then rebounds with 90% of its impact speed and moves up decelerating at g, stops, falls again, and so on. The speed-time graph (speed = |v|, never negative) is made of straight lines of slope ±g, with a sudden drop in speed at every collision. Take g = 9.8 m s⁻².

1. First fall. h = ½ g t² ⇒ t₁ = √(2h/g) = √(180/9.8) = 4.29 s. Speed on reaching the floor v₁ = g t₁ = 9.8 × 4.29 = 42.0 m s⁻¹.
2. After the first collision. Speed = 0.9 × 42.0 = 37.8 m s⁻¹ (upward). The ball decelerates at 9.8 m s⁻² and stops after t = 37.8/9.8 = 3.86 s, i.e. at 4.29 + 3.86 = 8.14 s, rising to 0.9² × 90 = 72.9 m.
3. Second descent. It falls again for 3.86 s and strikes the floor with the same speed 37.8 m s⁻¹ at t = 8.14 + 3.86 = 12.0 s.
4. The graph for 0 ≤ t ≤ 12 s: a straight line from (0, 0) to (4.29 s, 42.0 m s⁻¹); a vertical drop to 37.8 m s⁻¹; a straight line down to (8.14 s, 0); a straight line up to (12.0 s, 37.8 m s⁻¹).

Check of the rebound height. From v² = 2 g H, H = (37.8)²/(2 × 9.8) = 72.9 m = 0.81 × 90 m: since speed is cut to 0.9, the height is cut to 0.9² = 0.81. The slopes of all the straight lines have the same magnitude, 9.8 m s⁻².

Extra depth. The area under a speed-time graph is the total path length: in the first 4.29 s it is ½ × 4.29 × 42.0 = 90 m (the drop height). If the question asked for a velocity-time graph instead, the descending parts would be drawn with negative velocity and the ascending parts with positive velocity (downward negative), giving a zig-zag with the same shape but crossing the time axis. With g = 10 m s⁻² the numbers become t₁ = 4.24 s, v₁ = 42.4 m s⁻¹, and the same shape.

Conclusion: The speed-time graph is a line rising to 42.0 m s⁻¹ at 4.29 s, dropping to 37.8 m s⁻¹, falling linearly to zero at 8.14 s and rising again to 37.8 m s⁻¹ at 12.0 s.

Concept. Displacement is the change of position, Δx = x_final − x_initial; it depends only on the end points. Path length is the actual length covered along the way and depends on the whole route. Displacement is a vector (sign), path length a non-negative scalar.

(a) Magnitude of displacement and total path length

Since we cannot cover a gap between two points in less than the straight distance, path length ≥ |displacement|. Equality holds only if the particle moves along a straight line without reversing its direction. Example: walking 3 m east then 4 m west gives displacement 1 m west but path 7 m.

(b) Magnitude of average velocity and average speed

Dividing the inequality of (a) by the same positive time interval: s/Δt ≥ |Δx|/Δt, i.e. average speed ≥ magnitude of average velocity. Equality holds when the particle moves in a straight line in one direction only (no turning back). If the particle returns to the starting point, the average velocity is zero while the average speed is not.

Extra depth. In one dimension, path length s = ∫|v|dt and displacement Δx = ∫v dt, and |∫v dt| ≤ ∫|v| dt (the modulus of an integral does not exceed the integral of the modulus), which proves both inequalities. A runner who completes a lap of a 400 m track in 50 s has average speed 8 m s⁻¹ but average velocity zero. This is why speed is defined with path length and not as the magnitude of average velocity (Ex 2.10).

Conclusion: Displacement depends on end points only while path length depends on the route; |displacement| ≤ path length and |average velocity| ≤ average speed, with equality only for motion in a straight line without reversal.

Concept. First find the times of the two parts of the journey from time = distance/speed, then locate the man at the end of each interval and use: average velocity = displacement/time and average speed = total path length/time. Convert minutes to hours: 30 min = 0.5 h.

1. Going: t₁ = 2.5 km / 5 km h⁻¹ = 0.5 h = 30 min (the market is reached at t = 30 min).
2. Returning: t₂ = 2.5 km / 7.5 km h⁻¹ = 1/3 h = 20 min; he is back home at t = 50 min.
3. (i) 0 to 30 min: displacement = 2.5 km (home to market), path = 2.5 km, time = 0.5 h.
4. (ii) 0 to 50 min: displacement = 0 (back home), path = 5 km, time = 50/60 = 5/6 h.
5. (iii) 0 to 40 min: first 30 min 2.5 km out; in the next 10 min = 1/6 h he walks back 7.5 × (1/6) = 1.25 km. Position = 2.5 − 1.25 = 1.25 km from home (displacement 1.25 km); path = 2.5 + 1.25 = 3.75 km; time = 2/3 h.

Arithmetic check for (iii): |v̄| = 1.25/(2/3) = 1.875 km h⁻¹; average speed = 3.75/(2/3) = 5.625 km h⁻¹. In (ii) the average speed 5/(5/6) = 6 km h⁻¹ lies between 5 and 7.5 km h⁻¹ (weighted by time), as expected.

Extra depth. The NCERT note says that you would not like to tell the tired man that his average speed on returning home was zero: this is why average speed is defined as path length divided by time and not as the magnitude of average velocity. Note also that average speed (6 km h⁻¹) is not the arithmetic mean of 5 and 7.5 (= 6.25) because the two speeds are maintained for different times; the equal-distance case gives the harmonic mean 2 × 5 × 7.5/(5 + 7.5) = 6 km h⁻¹, which is the correct average speed over the whole 5 km.

Conclusion: (i) 5 km h⁻¹ and 5 km h⁻¹; (ii) 0 and 6 km h⁻¹; (iii) 1.875 km h⁻¹ and 5.625 km h⁻¹.

Concept. Instantaneous velocity is v = lim_Δt→0 Δx/Δt = dx/dt and instantaneous speed is the limit of (path length)/(time) as the time interval shrinks to zero, v_speed = lim_Δt→0 Δs/Δt = ds/dt. The question is why ds/dt = |dx/dt|.

1. For a finite interval the path length can exceed the magnitude of displacement because the particle may turn back in between (Ex 2.9).
2. As Δt is made smaller and smaller, the interval becomes so short that the particle has no time to reverse its direction within it; velocity is a continuous function of time and cannot change sign abruptly. So within an infinitesimal interval the motion is in one direction only.
3. In a one-way motion the path length equals the magnitude of the displacement: Δs = |Δx| in the limit, i.e. ds = |dx|.
4. Hence speed = ds/dt = |dx|/dt = |dx/dt| = |v|. Instantaneous speed = magnitude of instantaneous velocity at every instant.

Example. A ball thrown up returns to the hand after 6 s: over the 6 s the average velocity is zero but the average speed is 2H/6 = 14.7 m s⁻¹. At any single instant, say t = 1 s, v = +19.6 m s⁻¹ and the speed is 19.6 m s⁻¹: the speedometer of a car reads |v|, not the average velocity.

Extra depth. A car’s speedometer reads the instantaneous speed, which is the magnitude of the instantaneous velocity; there is no meter for velocity direction. The argument fails only if the velocity changes abruptly (a kink), which would need an infinite acceleration; in real motion functions are smooth (chapter text, Section 2.3). In two or three dimensions the same result holds: speed is the magnitude of the velocity vector, |dr/dt| = ds/dt, because for a tiny interval the arc and the chord coincide.

Conclusion: Over an infinitesimal interval the path length equals the magnitude of displacement (no time to turn back), so ds/dt = |dx/dt|: instantaneous speed always equals the magnitude of instantaneous velocity.

Concept. A graph can represent a real one-dimensional motion only if it obeys three basic requirements: (1) at any instant the particle has only one position and only one velocity (a function of time must be single-valued); (2) speed cannot be negative; (3) the total path length can never decrease with time (it is an accumulated distance). A graph violating any of these cannot describe the motion.

• Graph (a): a vertical line at a given t cuts the curve more than once ⇒ the position is not unique.
• Graph (b): similarly, a vertical line at some t cuts the circle in two points.
• Graph (c): the portions below the axis would require negative speed.
• Graph (d): the descending segments would mean that the particle ‘un-travels’ some of its path, which is impossible; a real particle that returns toward the start still adds to its path length.

Conclusion. None of the four graphs (a), (b), (c) or (d) can represent one-dimensional motion of a particle.

Extra depth. A correct x-t graph can have any slope sign (the particle moves forward or backward) and can double back in x, but it cannot double back in t: time flows only forward. The vertical-line test is a quick check: a vertical line at any t must cut the graph exactly once. A real v-t graph can go below the axis (negative velocity means motion in the negative direction), but a speed-time graph cannot. Path length versus time has non-negative slope everywhere (its slope is the speed).

Conclusion: All four graphs (a)–(d) are impossible: (a) two positions at one time, (b) two velocities at one time, (c) negative speed, (d) decreasing path length.

Concept. An x-t graph is not the path of the particle. It plots the position coordinate x (along a straight line) against time. The shape of the graph tells us how the position changes with time, not the shape of the trajectory. Here the problem says the motion is one-dimensional, so the particle always moves along a straight line.

1. For t < 0: the graph is along the t-axis, i.e. x = 0 for all these times. The particle is at rest at the origin; it is not moving along a straight line ‘with a path’; it simply stays at x = 0.
2. For t > 0: the graph is a parabola x = ½ a t² (a constant). Its slope v = dx/dt = a t increases linearly with time, and its second derivative a is constant. This describes uniformly accelerated straight-line motion starting from rest at the origin at t = 0.
3. Therefore it is not correct to say that the particle moves along a parabolic path for t > 0: the path is still the straight line of the x-axis; only the x-t graph is a parabola.

Suitable physical context. A ball (or any object) kept at rest at the origin until t = 0 and then released: x measured downward from the release point. For t > 0 it falls freely with a = g and x = ½ g t². Another example: a car standing at a traffic signal that starts at t = 0 with constant acceleration.

Extra depth. The same graph for the velocity would be v = 0 for t < 0 and a straight line v = at for t > 0 passing through the origin; and the acceleration graph would be zero for t < 0 and a constant a for t > 0. A kink at t = 0 in an x-t graph is not a problem: x and its first derivative (v) are continuous there, only the acceleration jumps (a step from 0 to a), which is the idealised start of the motion.

Conclusion: No: the x-t graph is not the trajectory; the particle is at rest at x = 0 for t < 0 and moves in a straight line with uniform acceleration for t > 0, e.g. a ball dropped from rest at t = 0.

Concept. The muzzle speed of the bullet (150 m s⁻¹) is measured relative to the gun, i.e. relative to the police van. The damage to the thief’s car depends on the speed of the bullet relative to the car. So we need the velocity of the bullet relative to the ground, then the relative velocity with respect to the thief’s car: v_BC = v_B − v_C.

1. Convert to m s⁻¹: police van v_V = 30 km h⁻¹ = 30 × 5/18 = 8.33 m s⁻¹; thief’s car v_C = 192 km h⁻¹ = 192 × 5/18 = 53.33 m s⁻¹.
2. Velocity of the bullet relative to the ground = muzzle speed + van speed (fired in the direction of motion): v_B = 150 + 8.33 = 158.33 m s⁻¹.
3. Velocity of the bullet relative to the car: v_BC = v_B − v_C = 158.33 − 53.33 = 105 m s⁻¹.
4. The bullet hits the thief’s car with a speed of 105 m s⁻¹ relative to the car.

Why not 150 or 158.33? What matters for the damage is the closing speed: the car moves away at 53.33 m s⁻¹ in the same direction, so the bullet closes in on it at only 158.33 − 53.33 = 105 m s⁻¹. If the bullet and car moved in opposite directions the relative speed would be the sum of the speeds.

Extra depth. Relative velocity: v_AB = v_A − v_B (all velocities with respect to the same frame). The direction is the same for all the three velocities, so the relation is algebraic. The impact speed of 105 m s⁻¹ is smaller than the muzzle speed because the target is running away. A faster car (above 158 m s⁻¹ = 570 km h⁻¹, impossible for a car) would never be hit.

Conclusion: The bullet hits the thief’s car with a relative speed of 105 m s⁻¹.

Concept. To suggest a situation we must read three things from every graph: the type of quantity plotted (x, v or a), the slope/shape of the segments (constant velocity, constant acceleration, impulse), and the sign changes. Then a real example must match all of them.

(a) x-t graph: at rest at the origin, moves away to A, then back through the origin to B, and stops

1. Flat part at x = 0: the object is at rest at the origin.
2. Straight line rising to the peak A: it moves with constant velocity in the positive direction up to a farthest point A.
3. At A the slope changes sign: it reverses and moves back with constant (negative) velocity, passing through x = 0 and reaching a point B on the negative side, where it stops (flat again).

Situation: a ball kicked from the origin towards a wall (or a person walking away from home) which at A hits an obstacle or decides to turn back and then walks through the starting point to a shop on the other side, B, where it stops. The motion is assumed to be at constant speed between turns.

(b) v-t graph: segments of equal negative slope with shrinking size

1. The slope of each segment is constant and negative: a constant acceleration a = −g.
2. Each segment starts at a positive velocity, falls to a negative velocity, so the object moves up, stops, comes back down.
3. At the end of each segment the velocity jumps suddenly to a smaller positive value (the object rebounds with reduced speed) and the cycle repeats with decreasing amplitude.

Situation: a ball bouncing on the floor with energy loss at every bounce (compare Ex 2.8). Taking upward as positive, the acceleration is −g all the time; at each impact the velocity reverses sign and its magnitude decreases, so the successive bounces become shorter until the ball comes to rest.

(c) a-t graph: zero except for a short interval

Situation: a cricket ball moving with constant velocity hit by a bat for a very short time: the acceleration is zero before and after and has a sharp pulse during the contact. Another: a billiard ball struck by a cue; a car whose accelerator is pressed briefly.

Extra depth. Interpretation tools: slope of x-t = v; slope of v-t = a; area under a-t = change in velocity (here the area of the bump = the velocity given to the ball by the bat). In (c) the velocity before the bump is constant, and after the bump it is constant at a different value: the graph describes an impulsive change of velocity. The three cases (a), (b), (c) each correspond to an everyday event: turning back, bouncing and hitting.

Conclusion: (a) A body at rest at the origin that moves away, reverses, passes the origin and stops; (b) a ball bouncing on the floor with loss of speed at each bounce; (c) a ball struck by a bat for a short time.

Concept. In simple harmonic motion (SHM) the particle oscillates about the origin; the displacement x, velocity v = dx/dt (slope of the x-t graph) and acceleration a = d²x/dt² = −ω² x are related. Two facts decide all the signs: (1) velocity = slope of the tangent to the x-t curve, and (2) acceleration has the opposite sign to the position (it always points towards the mean position, a = −ω² x).

1. Reading the graph: the curve crosses the axis at t = 0, ±1, ±2, … (period 2 s), is negative between 0 and 1 s, positive between 1 and 2 s and negative between −2 and −1 s, with the trough at t = 0.5 s and crest at 1.5 s.
2. At t = 0.3 s the curve is in the first (below-axis) loop, going down (the trough is at 0.5 s): x < 0, slope < 0, so v < 0, a = −ω² x > 0.
3. At t = 1.2 s the curve is above the axis and still rising towards the crest at 1.5 s: x > 0, v > 0, a < 0.
4. At t = −1.2 s the curve is below the axis (between −2 and −1) and rising towards the next zero at −1 s: x < 0, v > 0, a > 0.

Check with the equation. The curve is x = −A sin(πt): at t = 0.3, x = −0.81A (negative); v = −Aπ cos(0.3π) < 0; a = +Aπ² sin(0.3π) > 0 ✓. At t = 1.2: x = −A sin(1.2π) = +0.59A, v = −Aπ cos(1.2π) = +0.81Aπ > 0, a < 0 ✓. At t = −1.2: x = +A sin(1.2π) = −0.59A, v = −Aπ cos(−1.2π) = +0.81Aπ > 0, a = −ω² x > 0 ✓.

Extra depth. This exercise combines the three graphical rules: height gives position, slope gives velocity, curvature gives acceleration (concave up means a > 0 and concave down means a < 0). Where the graph is below the axis it is concave up (U-shaped), so a is positive; above the axis it is concave down (∩-shaped), so a is negative: consistent with a = −ω² x. Note that at t = −1.2 s time is negative: this just means the graph extended back before the instant we call t = 0.

Conclusion: t = 0.3 s: x −, v −, a +; t = 1.2 s: x +, v +, a −; t = −1.2 s: x −, v +, a +.

Concept. In a one-dimensional x-t plot, over an interval the average velocity is Δx/Δt, the slope of the chord joining the end points; with no reversal of direction inside an interval its magnitude equals the average speed. With equal intervals Δt, the interval with the largest |Δx| has the largest average speed.

1. Average speed (interval) ≈ |Δx|/Δt; for equal Δt compare the steepness of the curve.
2. Interval 3 has the steepest slope, so the average speed is greatest in interval 3, and the direction of motion is negative (x decreasing).
3. Interval 2 lies near the maximum of the curve where the slope is nearly zero, so the average speed is least in interval 2.
4. Signs of the average velocity: interval 1: positive; interval 2: positive (very small); interval 3: negative.

Reminder. The sign of velocity tells the direction (along or opposite to the positive x-axis); the speed is its magnitude. A negative velocity of large magnitude (interval 3) represents a fast motion in the negative direction.

Extra depth. The tangent at the crest (just after interval 2) has zero slope: at the instant of the maximum the velocity is zero and changes from positive to negative. Hence just after interval 2 the particle turns back. If the question had asked for speed over the whole period 1 to 3 including the turning point, one would need the total path length rather than the net displacement.

Conclusion: The average speed is greatest in interval 3 and least in interval 2; the average velocity is positive in intervals 1 and 2 and negative in interval 3.

Concept. For motion in a constant direction, speed = |v| = v (positive direction chosen along the motion). On a speed-time graph the slope of the chord over an interval is the average acceleration and the height is the speed. The slope of the tangent at a point is the instantaneous acceleration.

1. Greatest average acceleration (magnitude): interval 2, where the speed drops most steeply (steepest chord slope).
2. Greatest average speed: interval 3, where the curve is highest (it lies around the highest peak D).
3. Signs: the particle moves in the chosen positive direction, so v is positive in all three intervals. a is positive when the speed increases (intervals 1 and 3) and negative when it decreases (interval 2).
4. Accelerations at A, B, C, D: A is the starting point of a flat part, B a maximum, C a minimum and D a maximum of the speed curve; the tangent is horizontal at each, so a = dv/dt = 0 at A, B, C and D.

Reminder. The acceleration at maximum or minimum points of a speed-time graph is zero even though the speed is greatest or least there; the acceleration is largest in magnitude where the graph is steepest, not at the peaks.

Extra depth. Speed-time graphs cannot go below the axis, but a speed graph does not carry information about direction. Here the direction is constant, so we can treat the speed as the (positive) velocity and the slope as acceleration. The area under the graph is the distance travelled, so the interval where the curve is highest also contributes the largest distance in equal time.

Conclusion: Greatest |a|: interval 2; greatest average speed: interval 3; v is + in all, a is +, −, + in intervals 1, 2, 3; a = 0 at A, B, C and D.