Equations used: v = u + at | s = ut + ½at2 | v2 = u2 + 2as | 1 km h−1 = 5/18 m s−1
1My father went to a shop 250 m from home on a straight road. He came home to take a cloth bag, went to the shop again, bought provisions and came back home. How much was the total distance travelled by him? What was his displacement from home?
He covered the 250 m path four times (home → shop → home → shop → home).
Distance = 4 × 250 m = 1000 m Displacement = 0 m (he ends at the starting point)
2A student runs from the ground floor to the fourth floor to collect a book and then comes down to the classroom on the second floor. If the height of each floor is 3 m, find (i) the total vertical distance travelled, and (ii) the displacement from the starting point.
Up: ground to 4th floor = 4 × 3 = 12 m
Down: 4th to 2nd floor = 2 × 3 = 6 m
(i) Distance = 12 + 6 = 18 m
(ii) Displacement = 2 × 3 = 6 m in the upward direction
3A girl is riding her scooter and finds that its speedometer reading is constant. Is it possible for her scooter to be accelerating and if so, how?
Yes. A speedometer shows only speed, not direction. Velocity has both speed and direction. If she takes a turn or moves on a curved road, her direction changes. So her velocity changes even though the speed is constant. A change in velocity means she is accelerating.
4A car starts from rest and its velocity reaches 24 m s−1 in 6 s. Find the average acceleration and the distance travelled in these 6 s.
u = 0, v = 24 m s−1, t = 6 s
a = (v − u)/t = 24/6 = 4 m s−2 (in the direction of velocity)
s = ut + ½at2 = 0 + ½ × 4 × 36 = 72 m
5A motorbike moving with initial velocity 28 m s−1 and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.
u = 28 m s−1, v = 0, s = 98 m
v2 = u2 + 2as ⇒ 0 = 784 + 2 × a × 98
a = −784/196 = −4 m s−2 (opposite to the velocity)
t = (v − u)/a = (0 − 28)/(−4) = 7 s
6Fig. 4.27 shows a position-time graph of two objects A and B moving along parallel tracks in the same direction. Do objects A and B ever have equal velocity? Justify your answer.
No. Both graphs are straight lines, so each object moves with a constant velocity. The slope of A is greater than the slope of B. So A always moves faster than B. At t = 5 s they are at the same position, but their velocities are still different.
7A graph in Fig. 4.28 shows the change in position with time for two objects A and B moving in a straight line from 0 to 10 seconds. Choose the correct option(s).
Answer: (i) and (ii)
Both start and end at the same positions in 10 s, so their average velocities are equal. Both keep moving forward, so the distance covered by each is the same. Hence their average speeds are also equal. B is slower in some parts and faster in others, but the totals are the same.
8A truck driver driving at 54 km h−1 notices a speed limit sign of 40 km h−1. He slows down to 36 km h−1 in 36 s. What was the distance travelled during this time? (Assume constant acceleration.)
u = 54 km h−1 = 15 m s−1, v = 36 km h−1 = 10 m s−1, t = 36 s
s = ½(u + v) × t = ½ × (15 + 10) × 36 = 450 m
9A car starts from rest and accelerates uniformly to 20 m s−1 in 5 s. It then travels at 20 m s−1 for 10 s and finally stops in 6 s with uniform acceleration. Find the total distance travelled.
Part 1: s1 = ½ × (0 + 20) × 5 = 50 m
Part 2: s2 = 20 × 10 = 200 m
Part 3: s3 = ½ × (20 + 0) × 6 = 60 m
Total distance = 50 + 200 + 60 = 310 m
10A bus is travelling at 36 km h−1 when the driver sees an obstacle 30 m ahead. He takes 0.5 s to react. Then the bus slows down with a constant acceleration of 2.5 m s−2. Will the bus be able to stop before reaching the obstacle?
u = 36 km h−1 = 10 m s−1
Distance during reaction time = 10 × 0.5 = 5 m
Braking distance: 0 = 102 − 2 × 2.5 × s ⇒ s = 100/5 = 20 m
Total stopping distance = 5 + 20 = 25 m
25 m is less than 30 m. So yes, the bus stops 5 m before the obstacle.
11A student said, “The Earth moves around the Sun”. In this context, discuss whether an object kept on the Earth can be considered to be at rest.
Rest and motion are relative. They depend on the reference point.
An object kept on a table does not change its position with respect to the Earth. So it is at rest relative to the Earth.
But the object moves along with the Earth around the Sun. So it is in motion relative to the Sun.
Thus, the object can be considered at rest only when the Earth is taken as the reference point.
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12The velocity-time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas representing the displacement (i) while the cyclist is moving with constant velocity, (ii) when the velocity is decreasing. Also, calculate the displacement and average acceleration in the 120 s time interval.
Blue area (I): constant velocity from 20 s to 100 s. Orange area (II): velocity decreasing from 100 s to 120 s.
0 – 20 s (triangle): ½ × 20 × 3 = 30 m
20 – 100 s (rectangle, area I): 80 × 3 = 240 m
100 – 120 s (trapezium, area II): ½ × (3 + 2) × 20 = 50 m
Total displacement = 30 + 240 + 50 = 320 m
Average acceleration = (2 − 0)/120 = 1/60 m s−2 ≈ 0.017 m s−2
13A girl is preparing for her first marathon. The graph (Fig. 4.31) depicts her velocity versus time. Estimate the distance she ran based on the graph.
Distance = area under the velocity-time graph. Reading the graph in parts:
0 – 0.6 h: 7 × 0.6 = 4.2 km
0.6 – 1.6 h: ½(7 + 7.5) × 1 = 7.25 km
1.6 – 3.0 h: 7.5 × 1.4 = 10.5 km
3.0 – 4.6 h: ½(7.5 + 7) × 1.6 = 11.6 km
4.6 – 5.6 h: ½(7 + 6.5) × 1 = 6.75 km
5.6 – 6.6 h: 6.5 × 1 = 6.5 km
Total ≈ 47 km
Quick check: her average velocity is about 7 km h−1 for about 6.6 h. 7 × 6.6 ≈ 46 km. So she ran roughly 45 – 47 km.
14A car moves with a constant velocity of 6 m s−1 for 2 minutes and then accelerates at 1 m s−2 for 6 seconds. Find the displacement in the 2 min 6 s interval by drawing a velocity-time graph.
Velocity-time graph of the car. Displacement = shaded area.
Velocity after 6 s of acceleration = 6 + 1 × 6 = 12 m s−1
0 – 120 s (rectangle): 6 × 120 = 720 m
120 – 126 s (trapezium): ½(6 + 12) × 6 = 54 m
Total displacement = 720 + 54 = 774 m
15Car A attains 5 m s−1 in 5 s from rest. Car B attains 3 m s−1 in 10 s from rest. Plot the velocity-time graphs for both cars on the same graph and calculate the displacement in the two time intervals.
aA = 5/5 = 1 m s−2 → velocities at 0, 1, 2, 3, 4, 5 s: 0, 1, 2, 3, 4, 5 m s−1
aB = 3/10 = 0.3 m s−2 → velocities at 0, 2, 4, 6, 8, 10 s: 0, 0.6, 1.2, 1.8, 2.4, 3 m s−1
Velocity-time graphs of Car A and Car B.
Car A (0 – 5 s): area = ½ × 5 × 5 = 12.5 m
Car B (0 – 10 s): area = ½ × 10 × 3 = 15 m
16Rohan studies from 6 PM to 7:30 PM. Consider the tip of the minute hand (length 7 cm) of the wall clock. During this time, find its (i) distance travelled, (ii) displacement, (iii) speed, and (iv) velocity.
In 1 hour 30 minutes, the minute hand makes 1½ rounds. At 6 PM it points at 12, and at 7:30 PM it points at 6. Time = 90 min = 5400 s.
(i) Distance = 1.5 × 2πr = 1.5 × 2 × (22/7) × 7 = 66 cm
(ii) Displacement = diameter = 2 × 7 = 14 cm (from 12 to 6, downwards)
(iii) Average speed = 66/5400 = 11/900 cm s−1 ≈ 0.012 cm s−1
(iv) Average velocity = 14/5400 = 7/2700 cm s−1 ≈ 0.0026 cm s−1 (from 12 towards 6)
