Motion in a Straight Line Class 11 Physics NCERT Solutions

Step-by-step solution

1. 1A body is treated as a point object when its size is negligible compared with the distance it moves or the size of its path.
2. 2(a) The size of the carriage is very small compared with the distance between two stations, so it can be a point object.
3. 3(b) The size of the monkey is very small compared with the size of the circular track, so it too can be a point object.
4. 4(c) The size of the spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground — it cannot be a point object.
5. 5(d) The size of the tumbling beaker is comparable to the height of the table from which it slips — it cannot be a point object.

Step-by-step solution

1. 1(a) From the graph, the distance OP is less than OQ, so A lives closer to the school than B.
2. 2(b) At x = 0, t = 0 for A, but for B at x = 0, t has some finite positive value — so A starts from the school earlier than B.
3. 3(c) The slope of the x-t graph gives speed; the slope of B’s graph is greater than A’s, so B walks faster than A.
4. 4(d) Both graphs end at the same time on the t-axis, so A and B reach home at the same time.
5. 5(e) B, who starts later but walks faster, meets A exactly once — so B overtakes A once on the road.

Step-by-step solution

1. 1Time for the outward walk: t = distance/speed = 2.5/5 = 0.5 h = 30 min. She reaches the office at 9.30 am.
2. 2Return trip by auto: t = 2.5/25 = 0.1 h = 6 min. She leaves the office at 5.00 pm and reaches home at 5.06 pm.
3. 3On the x-t graph: a straight line of slope 5 km h⁻¹ from t = 0 to t = 30 min (position rising to 2.5 km), a horizontal line at x = 2.5 km from 30 min to 480 min (9.30 am to 5.00 pm), and a steeper falling straight line from x = 2.5 km to x = 0 between 480 min and 486 min, slope 25 km h⁻¹.
4. 4Suitable scales: 1 cm = 30 min on the time axis and 1 cm = 0.5 km on the position axis make the three legs of the graph clearly visible.

Step-by-step solution

1. 1Each step is 1 m long and takes 1 s, so 5 steps forward take 5 s and 3 steps backward take 3 s.
2. 2Net distance per cycle = 5 − 3 = 2 m, covered in 5 + 3 = 8 s.
3. 3After one cycle (8 s) he is at 2 m; after two (16 s) at 4 m; after three (24 s) at 6 m; after four (32 s) at 8 m.
4. 4From 8 m, he takes 5 more forward steps covering 5 m in 5 s, reaching 8 + 5 = 13 m at t = 32 + 5 = 37 s, and falls into the pit.
5. 5The x-t graph is a staircase: rising 1 m per second for 5 s, falling 1 m per second for 3 s, repeating until the pit is reached at 13 m.

Step-by-step solution

1. 1Initial speed u = 126 km h⁻¹ = 126 × (5/18) = 35 m s⁻¹; final speed v = 0; stopping distance s = 200 m.
2. 2Use v² − u² = 2as: 0 − (35)² = 2 × a × 200.
3. 3
   $$a=-\frac{35\times 35}{2\times 200}=-3.06{\text{m s}}^{-2}$$
4. 4So the retardation (magnitude of the deceleration) is 3.06 m s⁻².
5. 5Use v = u + at: t = (v − u)/a = (0 − 35)/(−3.06) ≈ 11.44 s.

Step-by-step solution

1. 1(a) Irrespective of direction of motion, acceleration (due to gravity) always acts downward, towards the centre of the Earth.
2. 2(b) At the highest point the velocity is zero, while the acceleration is still g = 9.8 m s⁻² acting downward.
3. 3(c) With x = 0 at the highest point and downward positive: position x > 0 for both motions; velocity v < 0 during upward motion and v > 0 during downward motion; acceleration a > 0 throughout.
4. 4(d) For the rise: v² − u² = 2(−g)(s), with v = 0, u = 29.4 m s⁻¹: s = (0² − 29.4²)/(2 × (−9.8)) = 44.1 m.
5. 5Time of ascent using v = u + at: t = (0 − 29.4)/(−9.8) = 3 s.
6. 6Time of ascent equals time of descent, so total time = 3 + 3 = 6 s.

Step-by-step solution

1. 1(a) True. When an object is thrown vertically up, its speed becomes zero at the maximum height, yet its acceleration is g = 9.8 m s⁻² downward at that instant.
2. 2(b) False. Speed is the magnitude of velocity; if speed is zero, the magnitude of velocity is zero, so the velocity itself is zero.
3. 3(c) True. A car moving on a straight highway with constant speed has constant velocity, and acceleration — the rate of change of velocity — is zero.
4. 4(d) False in general. If acceleration is positive while velocity is negative (e.g., a particle projected upwards after the origin is chosen), the particle slows down until velocity becomes zero. The statement is true only when both velocity and acceleration are positive (e.g., falling vertically downward).

Step-by-step solution

1. 1First fall: s = ut + ½at² with u = 0, a = g: 90 = ½ × 9.8 × t², so t = √18.38 ≈ 4.29 s.
2. 2Impact speed: v = u + at = 0 + 9.8 × 4.29 ≈ 42.04 m s⁻¹.
3. 3After the collision the ball keeps 9/10 of its speed: rebound speed uᵣ = (9/10)(42.04) ≈ 37.84 m s⁻¹.
4. 4Time to reach maximum height on rebound: 0 = uᵣ + (−g)t′, so t′ = 37.84/9.8 ≈ 3.86 s; total time so far = 4.29 + 3.86 = 8.15 s.
5. 5Time of ascent = time of descent, so the second impact is at t = 8.15 + 3.86 ≈ 12.01 s, with rebound speed (9/10)(37.84) ≈ 34.05 m s⁻¹.
6. 6Speed-time graph: for each fall the speed rises linearly from 0 to the impact value; at each bounce it drops discontinuously to 9/10 of that value and falls again linearly.

Step-by-step solution

1. 1(a) The magnitude of displacement is the shortest distance (a straight line) between the initial and final positions; the path length is the actual distance traversed. E.g., moving from A to B and back to C: magnitude of displacement = AC, while path length = AB + BC.
2. 2The path length can equal the displacement only when there is no reversal of direction — AB + BC ≥ AC, so path length ≥ magnitude of displacement.
3. 3(b) Magnitude of average velocity = |displacement| / Δt; average speed = total path length / Δt.
4. 4In the example above, average speed = (AB + BC)/t while |average velocity| = AC/t; since AB + BC > AC, average speed is greater than magnitude of average velocity.
5. 5Equality holds exactly when the motion never reverses direction, i.e., the particle moves along a straight line without turning back.

Step-by-step solution

1. 1Time outward = 2.5/5 = 0.5 h = 30 min; time return = 2.5/7.5 = 1/3 h = 20 min; total = 50 min.
2. 2(i) 0 to 30 min: displacement = 2.5 km, time = 0.5 h. Average velocity magnitude = 2.5/0.5 = 5 km h⁻¹; average speed = 5 km h⁻¹ too (no reversal yet).
3. 3(ii) 0 to 50 min: net displacement = 0, time = 5/6 h. Average velocity = 0; total distance = 5 km, so average speed = 5/(5/6) = 6 km h⁻¹.
4. 4(iii) 0 to 40 min: he has been returning for 10 min, covering 7.5 × (10/60) = 1.25 km. Net displacement = 2.5 − 1.25 = 1.25 km; total distance = 2.5 + 1.25 = 3.75 km.
5. 5Average velocity magnitude = 1.25/(40/60) = 1.875 ≈ 1.9 km h⁻¹; average speed = 3.75/(40/60) = 5.625 ≈ 5.6 km h⁻¹.

Step-by-step solution

1. 1Instantaneous velocity is the first derivative of position with respect to time.
2. 2In the infinitesimal interval dt, the displacement is so small that the particle does not change the direction of its motion within it.
3. 3Because there is no reversal of direction in that interval, the path length travelled equals the magnitude of displacement in dt.
4. 4Hence instantaneous speed (ds/dt) equals the magnitude of instantaneous velocity (|dx/dt|) — the distinction vanishes at the instantaneous limit.

Step-by-step solution

1. 1Graph (a) is an x-t graph that is not one-valued in position: it cannot represent one-dimensional motion because a particle cannot have two positions at the same instant of time.
2. 2Graph (b) is a v-t graph that is not one-valued in velocity: a particle can never have two values of velocity at the same instant of time.
3. 3Graph (c) is a v-t graph taking negative speed values: speed, being a scalar magnitude, can never be negative.
4. 4Graph (d) shows the x-t curve turning back so that the position decreases: the total path length travelled can never decrease with time.

Step-by-step solution

1. 1No — the x-t graph does not give the shape of the particle’s path; t and x are both coordinates, so the ‘path’ picture is a misreading of the graph.
2. 2The claim that the particle ‘moves in a straight line for t < 0 and on a parabolic path for t > 0’ is therefore not correct.
3. 3A suitable physical situation matching the plot: a freely falling body that is held at a height for some time (so x stays constant), then released from rest at t = 0 and falls under gravity, giving the parabolic x-t section beyond t = 0.

Step-by-step solution

1. 1Convert speeds: police van vₚ = 30 km h⁻¹ = 8.33 m s⁻¹; thief’s car vₜ = 192 km h⁻¹ = 53.33 m s⁻¹; muzzle speed of bullet in the van’s frame v_b = 150 m s⁻¹.
2. 2Since the bullet is fired from the moving van, its resultant speed in the ground frame = 150 + 8.33 = 158.33 m s⁻¹.
3. 3Both vehicles move in the same direction, so the speed of the bullet relative to the thief’s car = 158.33 − 53.33 = 105 m s⁻¹.

Step-by-step solution

1. 1Graph (a), an x-t graph: the body is initially at rest, its speed builds up to a constant value, drops to zero, then increases in the opposite direction and levels off. A suitable situation: a football (initially at rest) kicked towards a rigid wall, rebounding with reduced speed, passing the kicker, and finally coming to rest.
2. 2Graph (b), a v-t graph: the sign of velocity changes and its magnitude decreases with time. A suitable situation: a ball dropped on a hard floor from a height — it strikes with some velocity, rebounds with reduced velocity, and this repeats until the ball comes to rest.
3. 3Graph (c), an a-t graph: the body moves with a uniform velocity, its acceleration rises for a short interval and drops back to zero, and the uniform velocity resumes. A suitable situation: a hammer moving with uniform velocity striking a nail.

Step-by-step solution

1. 1For SHM, acceleration a = −ω²x, so the acceleration always points opposite to the displacement; the velocity sign comes from the slope of the x-t graph.
2. 2At t = 0.3 s: x is negative, and the slope of the x-t graph is negative, so v is negative; from a = −ω²x, a is positive. Answer: (−, −, +).
3. 3At t = 1.2 s: x is positive and the slope is positive; hence v is positive and a is negative. Answer: (+, +, −).
4. 4At t = −1.2 s: x is negative, and with t negative the slope gives a positive velocity; a is positive. Answer: (−, +, +).

Step-by-step solution

1. 1Average speed over an interval is given by the magnitude of the slope of the x-t graph in that interval.
2. 2From the graph, the slope is maximum in interval 3 and minimum in interval 2: average speed is greatest in interval 3 and least in interval 2.
3. 3The sign of the average velocity follows the sign of the slope: positive in intervals 1 and 2 (slope positive), negative in interval 3 (slope negative).

Step-by-step solution

1. 1Acceleration is the slope of the speed-time graph; the slope is greatest (in magnitude) in interval 2, so the average acceleration is greatest in interval 2.
2. 2The height of the curve above the time-axis gives the speed; it is greatest in interval 3, so the average speed is greatest in interval 3.
3. 3Interval 1: slope positive, so a > 0; speed positive, so v > 0. Interval 2: slope negative, so a < 0; speed (a scalar) still positive. Interval 3: slope zero, so a = 0; speed is uniform and positive.
4. 4At the points A, B, C and D the v-t curve runs parallel to the time-axis, making the slope — and hence the acceleration — zero at each point.