Chapter 4
Describing Motion
Around Us
Every question solved — worked Examples, Activities 4.1–4.5, the 5 Pause & Ponder, and all 16 “Revise, Reflect, Refine” exercises — with step-wise numericals, kinematic equations and original graphs.
In-text Questions
Examples · Activities · Pause & Ponder
Two postmen start towards each other from 210 yojanas apart. One covers 9 yojanas/day, the other 5 yojanas/day. In how many days do they meet?
- Combined distance per day = 9 + 5 = 14 yojanas/day
- Days to cover 210 together = 210 ÷ 14 = 15 days
A ball is thrown up from O to top B (120 cm) and falls back to O. Fill Table 4.1, then choose what is true for displacement.
| Position | Total distance from O | Displacement from O |
|---|---|---|
| O (start) | 0 cm | 0 cm |
| A (40, going up) | 40 cm | 40 cm up |
| B (120, top) | 120 cm | 120 cm up |
| C (80, coming down) | 120 + 40 = 160 cm | 80 cm up |
| O (back) | 120 + 120 = 240 cm | 0 cm |
When is the athlete’s displacement zero (Fig. 4.4)? What is the total distance travelled then?
Displacement is zero whenever the athlete returns to the starting point O (start and end positions are the same). In that case the total distance is not zero — it equals the whole path covered. For example, if she runs O → A (100 m) and back to O, the total distance is 100 + 100 = 200 m while the displacement is 0.
Fuel used in a vehicle depends on total distance travelled or displacement? Justify.
The engine burns fuel for every metre the vehicle actually moves, no matter the direction. So fuel depends on the total path length, not on displacement. A vehicle could return to its start (displacement = 0) yet still have burned plenty of fuel.
A ball rolls down an inclined track (Fig. 4.6). Is it straight-line motion? Can O→D be shown on a horizontal line? Are distance and displacement magnitudes equal at A, B, C, D?
Yes, it is straight-line motion — the ball moves along one straight path (the incline), even though that path is tilted.
Yes, the motion from O to D can be drawn on a horizontal number line, because positions along a single straight line can always be marked on one axis.
Since the ball moves in one direction without turning back, the total distance and the magnitude of displacement are equal at every point (10, 20, 30, 40 cm at A, B, C, D).
Sarang swims one length and back in a 25 m pool in 50 s. Find his average speed and average velocity.
Total distance = 25 + 25 = 50 m; displacement = 0 (back to start).
Road trip: 200 km north in 3 h, then 200 km south in 2 h. Find the average speed and average velocity for the whole trip.
- Total distance = 200 + 200 = 400 km
- Total time = 3 + 2 = 5 h
- Average speed = 400 ÷ 5 = 80 km h⁻¹
- Displacement = 200 N − 200 S = 0 km (back to start)
- Average velocity = 0 ÷ 5 = 0 km h⁻¹
Under what conditions is (i) the magnitude of average velocity equal to average speed? (ii) the magnitude of average velocity zero while average speed is not zero?
(i) When the object moves in a straight line in one direction without turning back — then distance = magnitude of displacement, so the two are equal.
(ii) When the object returns to its starting point — displacement (and hence average velocity) is zero, but it has still travelled a real distance, so its average speed is not zero.
A bus speeds up from 36 → 54 km h⁻¹ in 10 s, then later brakes from 54 → 0 km h⁻¹ in 5 s. Find the average acceleration in each interval.
First convert: 36 km h⁻¹ = 10 m s⁻¹, 54 km h⁻¹ = 15 m s⁻¹.
- Speeding up (u=10, v=15, t=10): a = (15 − 10)/10 = +0.5 m s⁻² — along the motion.
- Braking (u=15, v=0, t=5): a = (0 − 15)/5 = −3 m s⁻² — the minus sign means it acts opposite to the motion.
A car’s acceleration is given as the time to go from 0 to 100 km h⁻¹. How do you calculate the average acceleration?
Use a = change in velocity ÷ time. First convert 100 km h⁻¹ to m s⁻¹: 100 × 1000/3600 ≈ 27.8 m s⁻¹.
Example: if a car reaches 100 km h⁻¹ in 8 s: a = 27.8 ÷ 8 ≈ 3.5 m s⁻². A “faster” car (smaller time) has a larger acceleration.
A dropped object’s speed is 9.8, 19.6, 29.4, 39.2 m s⁻¹ at t = 1, 2, 3, 4 s. Find the average acceleration in each second. Is it constant? Direction?
Each successive second the velocity rises by 9.8 m s⁻¹:
So the acceleration is constant = 9.8 m s⁻², directed downward (along the motion). This is the acceleration due to gravity, g.
From the position-time graph (Fig. 4.11c / 4.14), calculate the velocity from the slope of line AB.
Velocity = slope = (change in position) ÷ (change in time). Reading A (2 s, 40 m) and B (4 s, 80 m):
What does the shape of a velocity-time graph tell us about the motion?
- Horizontal line → velocity constant → acceleration = 0.
- Straight line sloping up → velocity increasing → constant acceleration along the motion.
- Straight line sloping down → velocity decreasing → constant acceleration opposite to the motion.
Also: slope of a v-t graph = acceleration, and area under a v-t graph = displacement.
Brakes give a = −4 m s⁻². Find the stopping distance if the car was moving at (i) 54 km h⁻¹, (ii) 108 km h⁻¹.
Rearranged: s = u² ÷ 8 (since 2 × 4 = 8).
- (i) u = 54 km h⁻¹ = 15 m s⁻¹ → s = 15² ÷ 8 = 225/8 ≈ 28.1 m
- (ii) u = 108 km h⁻¹ = 30 m s⁻¹ → s = 30² ÷ 8 = 900/8 = 112.5 m
A marble circles inside a ring; the ring is lifted. What happens, and why? What is the direction of velocity in circular motion?
When the ring is removed, the marble flies off in a straight line — along the tangent to the circle at the point where it was released. It keeps moving in the direction it had at that instant.
So in circular motion the velocity at any point is directed along the tangent, in the direction of motion. Even at constant speed, the direction of velocity keeps changing, so uniform circular motion is accelerated motion.
The kinematic equations for motion in a straight line with constant acceleration.
s = ut + ½at² (4.4b)
v² = u² + 2as (4.4c)
Here u = initial velocity, v = final velocity, a = acceleration, t = time, s = displacement. Two more useful forms: s = ½(u+v)t and s = vt − ½at². For circular motion, average speed for one revolution = 2πR ÷ T.
Revise, Reflect, Refine
End-of-chapter exercise — all 16 questions
Father: home → shop (250 m), back home (forgot bag), shop again, then home. Total distance? Displacement?
He covers the 250 m stretch four times: home→shop→home→shop→home.
He finishes back at home, so displacement = 0 m.
A student goes ground → 4th floor → down to 2nd floor (each floor 3 m). (i) Total vertical distance, (ii) displacement.
- Up: ground → 4th = 4 × 3 = 12 m
- Down: 4th → 2nd = 2 × 3 = 6 m
- (i) Total distance = 12 + 6 = 18 m
- (ii) Displacement = ground → 2nd = 2 × 3 = 6 m upward
A scooter’s speedometer reading is constant. Can it still be accelerating? How?
Yes. The speedometer shows only the speed. If the scooter is turning (changing direction), its velocity is changing even though the speed stays the same — and a change in velocity means acceleration. So on a curved or circular path at constant speed, the scooter is accelerating (the acceleration comes from the change in direction).
A car starts from rest and reaches 24 m s⁻¹ in 6 s. Find the average acceleration and the distance travelled.
- Acceleration (u=0, v=24, t=6): a = (24 − 0)/6 = 4 m s⁻²
- Distance s = ½(u+v)t: s = ½ × 24 × 6 = 72 m
A motorbike at 28 m s⁻¹ stops after 98 m with constant acceleration. Find the acceleration and the time to stop.
- Use v² = u² + 2as (v=0, u=28, s=98): 0 = 784 + 2a(98)
- Solve: a = −784 / 196 = −4 m s⁻² (deceleration)
- Use v = u + at (v=0): 0 = 28 + (−4)t → t = 7 s
Position-time graphs of A and B are straight lines that cross (Fig. 4.27). Do A and B ever have equal velocity? Justify.
For a straight-line position-time graph, the velocity equals the slope, which is constant for each object. Since the two lines have different slopes (that is why they cross), their constant velocities are different. At the crossing point (t = 5 s) they only share the same position — not the same velocity.
A and B (Fig. 4.28) move in a straight line, 0–10 s, with the same initial and final positions. Choose the correct option(s).
- (i) Average velocity of both is equal (same initial & final positions).
- (ii) Average speeds of both are equal (equal distance in equal time).
- (iii) A’s average speed is lower than B’s (shorter distance).
- (iv) A’s average speed is greater than B’s.
Both objects move forward (positions only increase) from the same start to the same end in 10 s. So their displacement is equal → equal average velocity (i). Because the motion is in one direction, distance = displacement, so both cover the same distance → equal average speed (ii). Options (iii) and (iv) are therefore wrong.
A truck slows from 54 km h⁻¹ to 36 km h⁻¹ in 36 s (constant acceleration). What distance did it cover during this time?
- Convert: u = 54 km h⁻¹ = 15 m s⁻¹, v = 36 km h⁻¹ = 10 m s⁻¹
- Use s = ½(u+v)t: s = ½ (15 + 10) × 36
- s = ½ × 25 × 36 = 450 m
A car: rest → 20 m s⁻¹ in 5 s, then 20 m s⁻¹ for 10 s, then brakes to stop in 6 s. Find total distance.
- Accelerating (0→20, 5 s): s₁ = ½ × 20 × 5 = 50 m
- Constant 20 m s⁻¹ for 10 s: s₂ = 20 × 10 = 200 m
- Braking (20→0, 6 s): s₃ = ½ × 20 × 6 = 60 m
- Total: 50 + 200 + 60 = 310 m
A bus at 36 km h⁻¹ sees an obstacle 30 m ahead. Reaction time 0.5 s, then braking at 2.5 m s⁻². Will it stop in time?
- Convert: u = 36 km h⁻¹ = 10 m s⁻¹
- Reaction distance (moving 0.5 s before braking): 10 × 0.5 = 5 m
- Braking distance (v²=u²−2as, v=0): s = 10² ÷ (2 × 2.5) = 100/5 = 20 m
- Total stopping distance = 5 + 20 = 25 m
“The Earth moves around the Sun.” Can an object kept on the Earth be considered at rest?
Rest and motion are relative — they depend on the chosen reference point. With respect to the Earth (e.g. the ground beside it), the object can be at rest, since its position relative to the Earth doesn’t change. But with respect to the Sun, the same object is moving, because it travels along with the Earth around the Sun. So the object is at rest in one frame and in motion in another — both statements are correct for their own reference point.
For the cyclist’s v-t graph (Fig. 4.30, 0–120 s): shade the constant-velocity and decreasing-velocity parts, and find the displacement and average acceleration over 120 s.
- 0–20 s (triangle): ½ × 20 × 3 = 30 m
- 20–100 s (rectangle): 80 × 3 = 240 m
- 100–120 s (trapezium, 3→2): ½ (3 + 2) × 20 = 50 m
- Total displacement ≈ 30 + 240 + 50 = 320 m
Average acceleration over 120 s = (final − initial velocity)/time = (2 − 0)/120 ≈ 0.017 m s⁻².
From the marathon runner’s v-t graph (Fig. 4.31, velocity in km h⁻¹ over ~6 h), estimate the distance she ran.
Distance = area under the v-t graph. The velocity stays close to about 7 km h⁻¹ for roughly 6 hours, so:
So she ran about 42 km — close to a full marathon. (Estimate the area by counting squares under the curve for a more precise value.)
A car moves at 6 m s⁻¹ for 2 min, then accelerates at 1 m s⁻² for 6 s. Find the displacement in 2 min 6 s using a v-t graph.
- Phase 1 (6 m s⁻¹ for 120 s): rectangle area 6 × 120 = 720 m
- Phase 2 (u=6, a=1, t=6): final v = 6 + 1×6 = 12 m s⁻¹; area = trapezium ½(6 + 12) × 6 = 54 m
- Total displacement = 720 + 54 = 774 m
Cars A and B start from rest with constant acceleration. A reaches 5 m s⁻¹ in 5 s; B reaches 3 m s⁻¹ in 10 s. Plot v-t and find each displacement.
- Accelerations: A → 5/5 = 1 m s⁻²; B → 3/10 = 0.3 m s⁻²
- Displacement A (5 s) = ½ × 5 × 5 = 12.5 m
- Displacement B (10 s) = ½ × 3 × 10 = 15 m
From 6:00 to 7:30 PM, for the tip of a 7 cm minute hand, find its (i) distance, (ii) displacement, (iii) speed, (iv) velocity.
Time = 90 min = 1.5 revolutions. Radius r = 7 cm, so circumference = 2πr = 2 × (22/7) × 7 = 44 cm.
- (i) Distance = 1.5 × circumference = 1.5 × 44 = 66 cm
- (ii) Displacement: after 1.5 turns the hand points opposite to start (12 → 6), so it equals the diameter = 2 × 7 = 14 cm
- (iii) Speed = distance/time = 66 cm ÷ 5400 s ≈ 0.012 cm s⁻¹
- (iv) Velocity = displacement/time = 14 cm ÷ 5400 s ≈ 0.0026 cm s⁻¹, directed from 12 toward 6
Derive the two extra kinematic equations: s = ½(u + v)t and s = vt − ½at².
On a v-t graph the area under the line is a trapezium with parallel sides u and v and height t. Area = ½ (sum of parallel sides) × height, so s = ½(u + v)t.
s = vt − ½at²From v = u + at, we get u = v − at. Substitute into s = ut + ½at²: