Exercises
1. In which of the following can the body be considered approximately a point object:
Answer
(a) a railway carriage moving between two stations (b) a monkey sitting on a man cycling on a circular track (c) a spinning cricket ball that turns sharply on hitting the ground (d) a tumbling beaker slipping off a table
Ans.
(a) Yes — the size of the carriage (≈20 m) is much smaller than the distance between stations (several km).
(b) Yes — the size of the monkey and man is much smaller than the size of the circular track.
(c) No — the cricket ball spins and turns; rotational motion is important. Internal structure matters.
(d) No — the beaker tumbles; its rotational motion and shape are important for describing its fall.
2. From the x-t graph of children A and B returning from school O to homes P and Q:
Answer
(a) (A/B) lives closer to school (b) (A/B) starts earlier (c) (A/B) walks faster (d) reach home at (same/different) time (e) (A/B) overtakes (B/A) (once/twice)
Ans. (Based on typical x-t graph for this problem)
(a) A lives closer to the school than B (A's home P is at smaller x than B's home Q)
(b) B starts from school earlier than A (B's graph starts at an earlier time)
(c) B walks faster than A (steeper slope in B's graph)
(d) A and B reach home at the same time
(e) B overtakes A on the road once
3. A woman starts from home at 9 am, walks at 5 km h⁻¹ to office 2.5 km away, stays till 5 pm, returns by auto at 25 km h⁻¹. Plot the x-t graph.
Answer
Time to walk to office = 2.5/5 = 0.5 h = 30 min
⇒ arrives at 9:30 am
Stays at office from 9:30 am to 5:00 pm (7.5 hours) — horizontal line at x = 2.5 km
Time to return by auto = 2.5/25 = 0.1 h = 6 min
⇒ reaches home at 5:06 pm
The x-t graph has:
⇒ A rising straight line from (9:00 am, 0) to (9:30 am, 2.5 km) — slope = 5 km/h
⇒ A horizontal line from (9:30 am, 2.5) to (5:00 pm, 2.5) — stay at office
⇒ A steep falling line from (5:00 pm, 2.5) to (5:06 pm, 0) — slope = −25 km/h
4. A drunkard takes 5 steps forward and 3 steps backward repeatedly. Each step is 1 m, takes 1 s. How long to reach a pit 13 m away?
Answer
Each cycle (5 forward + 3 backward) = 8 steps, 8 s, net displacement = +2 m
After 1 cycle (8 s): position = 2 m
After 2 cycles (16 s): position = 4 m
After 3 cycles (24 s): position = 6 m
After 4 cycles (32 s): position = 8 m
After 5 cycles (40 s): position = 10 m
After 5 cycles + 5 more forward steps (40 + 5 = 45 s): position = 10 + 5 = 15 m — but pit is at 13 m!
During the 46th second (5 forward steps after 40 s): positions reached are 11, 12, 13...
At t = 37 + 8 = 45th step: 10 m; 41st step = 11 m (t=41 s); 42nd = 12 m (t=42 s); 43rd = 13 m
⇒ falls in the pit at t = 37 + 6 = 46 s? Let me recalculate:
After 5 complete cycles (40 steps = 40 s): at 10 m
Step 41 (t=41 s): 11 m; Step 42: 12 m; Step 43: 13 m
The drunkard falls into the pit after 37 + 6 = 43 seconds? Let me recount:
Answer
Cycle 1 (t=1 to 8 s): goes 1,2,3,4,5 then 4,3,2 m
⇒ ends at 2 m at t=8 s
Cycle 2 (t=9 to 16 s): ends at 4 m
Cycle 3: ends at 6 m; Cycle 4: ends at 8 m; Cycle 5: ends at 10 m (t = 40 s)
After t=40 s, takes 5 forward steps: t=41 s at 11 m; t=42 s at 12 m; t=43 s at 13 m
⇒ falls in pit
Total time = 43 seconds (note: once he reaches 13 m he falls — if pit is exactly at 13 m, at t=43 s he falls in)
5. A car moving at 126 km h⁻¹ is brought to stop in 200 m. Find retardation and time to stop.
Answer
Given: u = 126 km h⁻¹ = 126 × (1000/3600) = 35 m s⁻¹; v = 0; s = 200 m
Using v² = u² + 2as: 0 = (35)² + 2a(200)
⇒ 2a(200) = −1225
⇒ a = −1225/400 = −3.0625 m s⁻² ≈ −3.06 m s⁻² (retardation)
Time: v = u + at
⇒ 0 = 35 + (−3.0625)t
⇒ t = 35/3.0625 ≈ 11.43 s
6. A player throws a ball upward with initial speed 29.4 m s⁻¹.
(a) Direction of acceleration during upward motion?
(b) Velocity and acceleration at highest point?
(c) Signs of position, velocity, acceleration (taking highest point as origin, downward as positive)?
(d) Height reached and time to return?
Answer
(a) During upward motion, acceleration due to gravity is always downward (directed toward Earth), regardless of the direction of motion.
(b) At the highest point: velocity = 0 m s⁻¹; acceleration = 9.8 m s⁻² downward (gravity acts always)
(c) With highest point as x = 0 and downward as positive x-direction:
During upward motion: position x < 0 (below the highest point on the way up
⇒ negative); velocity v < 0 (upward = against positive direction); acceleration a > 0 (downward = positive direction)
During downward motion: x > 0; v > 0; a > 0
(d) Taking upward as positive, g = −9.8 m s⁻²:
v² = u² + 2as
⇒ 0 = (29.4)² + 2(−9.8)h
⇒ h = (29.4)²/(2 × 9.8) = 864.36/19.6 = 44.1 m
Total time (up + down) = 2u/g = 2 × 29.4 / 9.8 = 6 s
Q2.7. State with reasons if true or false
(a) Zero speed at an instant may have non-zero acceleration
(b) Zero speed may have non-zero velocity
(c) Constant speed must have zero acceleration
(d) Positive acceleration must mean speeding up
Ans.
(a) True. Example: A ball thrown upward has zero speed at the highest point but has acceleration = g = 9.8 m s⁻² downward.
(b) False. Speed is the magnitude of velocity. If speed is zero, the magnitude of velocity is zero, hence velocity must be zero.
(c) True for one-dimensional motion. If speed is constant, the magnitude of velocity is constant. In one dimension, this means the velocity is constant (no change in direction), so acceleration = 0. (Note: in 2D, constant speed is possible with centripetal acceleration, but this chapter deals with 1D.)
(d) False. If the object is moving in the negative direction and has positive acceleration, the object is actually decelerating (slowing down). Example: a ball thrown upward (taking downward as positive), has positive acceleration g but is decelerating as it moves upward.
8. A ball is dropped from 90 m. At each collision it loses one tenth of its speed. Describe the speed-time graph from t = 0 to 12 s.
Answer
Given: initial height = 90 m; at each bounce, speed after bounce = 0.9 × speed before bounce
First fall: h₁ = 90 m; v₁ = √(2gh₁) = √(2 × 10 × 90) = √1800 ≈ 42.4 m s⁻¹ (using g = 10 m/s²)
Time for first fall: t₁ = v₁/g = 42.4/10 = 4.24 s
Speed after 1st bounce: v₂ = 0.9 × 42.4 = 38.2 m s⁻¹
Height after 1st bounce: h₂ = v₂²/(2g) = (38.2)²/20 = 72.9 m
Time for 2nd rise + fall: 2v₂/g = 2 × 38.2/10 = 7.64 s
The speed-time graph consists of:
⇒ An increasing straight line (slope = g) from (0,0) to (4.24 s, 42.4 m/s) — first fall
⇒ An instantaneous jump down to 38.2 m/s at the bounce
⇒ A decreasing then increasing line (V-shape) — each subsequent bounce
⇒ The speed decreases by a factor 0.9 at each bounce, and the time intervals decrease each cycle
The graph is a series of inverted V-shapes of decreasing amplitude and period.
9. Distinguish between: (a) displacement and path length (b) average velocity and average speed. Show the second quantity is ≥ first. When is equality achieved?
Answer
(a) Displacement = shortest straight-line distance from initial to final position (vector). Path length (distance) = actual length of path traveled (scalar).
Path length ≥ magnitude of displacement, because the shortest path between two points is a straight line.
Equality holds when the particle moves in a straight line without reversing direction.
(b) Average velocity = displacement/time (vector); Average speed = path length/time (scalar).
Since path length ≥ |displacement|, dividing both by time gives: average speed ≥ |average velocity|
Equality holds when path length = displacement, i.e., when the particle moves in one direction without reversal.
10. A man walks to market 2.5 km away at 5 km h⁻¹, finds it closed, walks back at 7.5 km h⁻¹. Find (a) average velocity and (b) average speed over (i) 0 to 30 min (ii) 0 to 50 min (iii) 0 to 40 min.
Answer
Time to reach market = 2.5/5 = 0.5 h = 30 min
Time to walk back = 2.5/7.5 = 1/3 h ≈ 20 min; returns at t = 30 + 20 = 50 min
(i) t = 0 to 30 min:
Displacement = 2.5 km; Distance = 2.5 km
Average velocity = 2.5/(0.5) = 5 km h⁻¹; Average speed = 2.5/0.5 = 5 km h⁻¹
(ii) t = 0 to 50 min:
Displacement = 0 (back at home); Distance = 2.5 + 2.5 = 5 km; Time = 50 min = 5/6 h
Average velocity = 0/(5/6) = 0 km h⁻¹; Average speed = 5/(5/6) = 6 km h⁻¹
(iii) t = 0 to 40 min:
First 30 min: walks to market (2.5 km); next 10 min: walks back 7.5 × (10/60) = 1.25 km
Position at t=40 min: 2.5 − 1.25 = 1.25 km from home
Displacement = 1.25 km; Distance = 2.5 + 1.25 = 3.75 km; Time = 40/60 = 2/3 h
Average velocity = 1.25/(2/3) = 1.875 km h⁻¹; Average speed = 3.75/(2/3) = 5.625 km h⁻¹
11. Why is instantaneous speed always equal to magnitude of instantaneous velocity?
Answer
Instantaneous velocity = lim(Δt
⇒ 0) Δx/Δt, where Δx is displacement in infinitesimally small time Δt.
As Δt
⇒ 0, the displacement Δx approaches the path length Δs (since in an infinitesimally small time interval, the path cannot curve back on itself).
Therefore, instantaneous speed = lim(Δt
⇒ 0) |Δs|/Δt = lim(Δt
⇒ 0) |Δx|/Δt = |instantaneous velocity|.
At any instant, the direction of motion is defined, so speed = |velocity|.
12. State which graphs (a) to (d) cannot represent one-dimensional motion of a particle.
Answer
(a) An x-t graph showing a loop (particle at same position at two different times with different slopes): Cannot represent real motion — a particle cannot have two different velocities at the same position at the same instant (the x-t graph must be a function; loops mean x is multi-valued at some t, which is impossible).
(b) A speed-time graph showing negative speed: Cannot represent motion — speed is always ≥ 0.
(c) A total path length vs time graph that decreases: Cannot represent motion — total distance traveled can only increase or remain constant, never decrease.
(d) A graph showing position varying as a closed loop in time: Cannot represent motion — requires particle to be at two positions simultaneously.
13. Is it correct to say a particle moves in a straight line for t < 0 and on a parabolic path for t > 0 from an x-t graph?
Answer
No, it is not correct. The x-t graph describes motion along one dimension (x versus time t). A straight line in x-t graph means constant velocity; a parabola means constant acceleration. The graph describes the variation of position with time — it does NOT represent the geometrical path of the particle in space.
Suitable physical context: A ball is thrown vertically upward at t=0 with some velocity. For t < 0 (before throwing), the ball was at rest on the ground (x = constant, straight horizontal line). At t = 0, it is thrown up. For t > 0, x = ut − (1/2)gt², which is a parabola in the x-t graph. The ball moves in a straight vertical line, not a parabolic path in space.
14. A police van moves at 30 km h⁻¹ and fires a bullet at muzzle speed 150 m s⁻¹ at a thief's car moving at 192 km h⁻¹ in the same direction. What speed does the bullet hit the car?
Answer
Speed of police van = 30 km h⁻¹ = 30/3.6 = 8.33 m s⁻¹
Speed of thief's car = 192 km h⁻¹ = 192/3.6 = 53.33 m s⁻¹
Muzzle speed of bullet = 150 m s⁻¹ (relative to gun/van)
Speed of bullet in ground frame = 150 + 8.33 = 158.33 m s⁻¹
Speed of bullet relative to thief's car = 158.33 − 53.33 = 105 m s⁻¹
(This is the speed relevant for damaging the car — the relative speed.)
15. Suggest suitable physical situations for the graphs in Fig 2.12.
Answer
(a) x-t graph: a straight line with positive slope followed by a straight line with negative slope
⇒ A ball thrown upward and caught — rises uniformly then falls back (but with constant speed, implying no gravity — more likely: a ball rolling on a flat surface turns around at some point).
(b) v-t graph: v increases then becomes constant
⇒ A car accelerating from rest then reaching maximum speed (terminal velocity or gear-limited).
(c) a-t graph: constant acceleration that suddenly drops to zero
⇒ A stone dropped under gravity (constant a = g), then hitting a soft surface where acceleration stops.
(d) v-t graph: negative constant velocity then sudden change to zero
⇒ A ball moving with constant speed toward a wall (negative direction), hitting it and stopping.
16. From x-t graph of SHM particle, give signs of position, velocity, and acceleration at t = 0.3 s, 1.2 s, −1.2 s.
Answer
For a particle in SHM: x = A sin(ωt) or A cos(ωt); v = dx/dt; a = −ω²x (acceleration is opposite to position)
At t = 0.3 s: From the graph, x > 0 (positive), slope dx/dt < 0 (decreasing), a = −ω²x < 0
Position: positive; Velocity: negative; Acceleration: negative
At t = 1.2 s: x < 0 (negative), slope > 0 (increasing), a = −ω²x > 0
Position: negative; Velocity: positive; Acceleration: positive
At t = −1.2 s: x < 0, slope > 0, a > 0
Position: negative; Velocity: positive; Acceleration: positive
17. From x-t plot: in which interval is average speed greatest and which is least? Give sign of average velocity for each interval.
Answer
Average speed in an interval = |change in x| / Δt (for each interval since Δt is equal)
The interval with the steepest slope (greatest |Δx|) has the greatest average speed, and the least steep has the least.
Typically from this graph: Interval 3 has greatest average speed (steepest x-t slope); Interval 2 has least average speed (nearly flat).
Sign of average velocity = sign of slope of x-t graph in each interval.
Interval 1: positive; Interval 2: negative; Interval 3: positive
18. From speed-time graph: which interval has greatest acceleration magnitude? Greatest average speed? Signs of v and a? Accelerations at A, B, C, D?
Answer
Greatest acceleration magnitude: The interval with the steepest slope in the v-t graph = Interval 2 (slope is steepest in magnitude)
Greatest average speed: The interval where the speed is highest on average = Interval 3
Since motion is along a constant direction (positive direction):
v is always positive in all three intervals.
Sign of acceleration a
Interval 1: slope of v-t > 0 → a is positive
Interval 2: slope of v-t < 0 → a is negative
Interval 3: slope of v-t > 0 → a is positive
At points A, B, C, D: acceleration = slope of v-t graph at that point.
At A: slope = 0 (local maximum) → a = 0
At B: slope = 0 (local minimum) → a = 0
At C: slope = 0 (local maximum) → a = 0
At D: slope = 0 (at end of graph) → a = 0
Important Extra Questions
Q. Define uniform and non-uniform motion. Give examples.
Ans.
Uniform motion: When an object covers equal distances in equal intervals of time in a straight line. Velocity is constant, acceleration = 0. Example: a car moving at exactly 60 km/h on a straight highway.
Non-uniform motion: When the distance covered in equal time intervals is not equal. Velocity changes (in magnitude, direction, or both). Example: a car accelerating from a traffic signal; a ball thrown upward under gravity.
Q. Derive the three equations of uniformly accelerated motion.
Ans.
For a particle with initial velocity u, constant acceleration a, after time t:
1st equation (v = u + at)
Acceleration a = (v − u)/t → v = u + at
2nd equation (s = ut + ½at²)
Average velocity = (u + v)/2 = (u + u + at)/2 = u + at/2
Displacement s = average velocity × t = (u + at/2) × t = s = ut + ½at²
3rd equation (v² = u² + 2as)
From v = u + at → t = (v−u)/a
Substitute in s = ut + ½at²:
s = u(v−u)/a + ½a((v−u)/a)² = u(v−u)/a + (v−u)²/(2a)
2as = 2u(v−u) + (v−u)² = (v−u)(2u + v − u) = (v−u)(v+u)
v² = u² + 2as
Q. A stone is dropped from a tower 78.4 m high. Find the time to reach the ground and the velocity on impact. (g = 9.8 m s⁻²)
Ans.
Given: h = 78.4 m, u = 0, g = 9.8 m s⁻²
Using s = ut + ½gt²: 78.4 = 0 + ½ × 9.8 × t²
→ t² = 78.4/4.9 = 16 → t = 4 s
Velocity on impact: v = u + gt = 0 + 9.8 × 4 = 39.2 m s⁻¹
Q. What is relative velocity? If two trains A and B move at 72 km h⁻¹ and 54 km h⁻¹ in the same direction, find the velocity of A relative to B and B relative to A.
Answer
Relative velocity of one object with respect to another = velocity of first object − velocity of second object
v_A = 72 km h⁻¹ = 20 m s⁻¹; v_B = 54 km h⁻¹ = 15 m s⁻¹
Velocity of A relative to B = v_A − v_B = 20 − 15 = 5 m s⁻¹ in direction of motion
Velocity of B relative to A = v_B − v_A = 15 − 20 = −5 m s⁻¹ (i.e., 5 m s⁻¹ opposite to direction of motion)
Q. A body is thrown vertically upward with velocity 49 m s⁻¹. Find the maximum height and total time of flight. (g = 9.8 m s⁻²)
Ans.
Given: u = 49 m s⁻¹ (upward), g = 9.8 m s⁻² (downward), v = 0 at highest point
Maximum height: v² = u² − 2gh
⇒ 0 = 49² − 2 × 9.8 × h
⇒ h = 2401/(19.6) = 122.5 m
Time to reach maximum height: v = u − gt
⇒ 0 = 49 − 9.8t
⇒ t = 5 s
Total time of flight = 2t = 10 s