Chapter 4: Describing Motion Around Us
NCERT Exercise Solutions — Revise, Reflect, Refine
16 Questions | Step-by-Step Numerical Solutions | Graphs | Conceptual Explanations
These solutions help you understand each question, apply the correct formula and write clear answers in your examinations. Select a question to reveal its solution.
Solution
Given: Distance between home and shop = 250 m.
He makes four journeys:
| Journey | Distance |
|---|---|
| Home → Shop | 250 m |
| Shop → Home | 250 m |
| Home → Shop | 250 m |
| Shop → Home | 250 m |
Displacement is the change in position from the starting point to the final point.
Since he returns to his starting point, his displacement is zero.
Solution
Given: Height between consecutive floors = 3 m.
Ground floor to fourth floor:
Fourth floor to second floor:
Therefore,
The final position is the second floor, 6 m above the ground floor.
Solution
Yes. Acceleration occurs when velocity changes. Velocity depends on both speed and direction.
If a scooter moves along a curved road at constant speed, its direction changes continuously. Therefore, its velocity changes and it experiences acceleration.
For example, a scooter moving around a circular track at constant speed has acceleration directed towards the centre of the circular path.
Solution
Given: u = 0 m/s, v = 24 m/s, t = 6 s.
Step 1: Calculate average acceleration.
Step 2: Calculate distance.
For uniformly accelerated motion:
Solution
Given: u = 28 m/s, v = 0 m/s, s = 98 m.
Step 1: Calculate acceleration.
Step 2: Calculate time.
Solution
Velocity is represented by the slope of a position-time graph.
In Fig. 4.27, the graphs of A and B are straight lines with different slopes. The line for A is steeper than the line for B.
Therefore, A has a greater constant velocity than B throughout the interval shown.
The two lines intersect at approximately 5 s, but this means that the objects have the same position along their respective tracks at that instant, not the same velocity.
Solution
Both objects start from the same position and reach the same final position after 10 seconds.
Statement (i): Correct. Their displacements are equal and their time intervals are equal. Hence, their average velocities are equal.
Statement (ii): Correct. Both objects move in one direction without turning back. They cover the same total distance in 10 s, so their average speeds are equal.
Statement (iii): Incorrect. A does not cover a shorter total distance than B.
Statement (iv): Incorrect. Although their speeds differ during parts of the journey, their average speeds over the complete 10 s interval are equal.
Solution
Step 1: Convert velocities into m/s.
Given: t = 36 s.
Step 2: Calculate distance using average velocity.
Solution
Divide the motion into three stages.
Stage 1: Acceleration from 0 to 20 m/s in 5 s.
Stage 2: Constant velocity of 20 m/s for 10 s.
Stage 3: Braking from 20 m/s to rest in 6 s.
Total distance:
Solution
Given: Initial speed = 36 km/h = 10 m/s.
Step 1: Reaction distance.
The bus continues moving at 10 m/s during the driver’s reaction time.
Step 2: Braking distance.
Using v² = u² + 2as:
Step 3: Total stopping distance.
Distance to obstacle = 30 m.
Solution
Rest and motion depend on the reference point chosen to describe an object’s position.
An object kept on a table is at rest with respect to the table or the surrounding room because its position relative to them does not change.
However, the Earth rotates on its axis and revolves around the Sun. Therefore, the same object is moving with respect to the Sun.
Solution
From the given graph:
- 0–20 s: Velocity increases from 0 to 3 m/s.
- 20–100 s: Velocity remains constant at 3 m/s.
- 100–120 s: Velocity decreases from 3 to 2 m/s.
The shaded regions below represent the displacement during constant velocity (green) and decreasing velocity (orange). The blue region represents the initial acceleration.
Step 1: Displacement from 0 to 20 s (triangle).
Step 2: Displacement from 20 to 100 s (rectangle).
Step 3: Displacement from 100 to 120 s (trapezium).
Total displacement:
Average acceleration:
Solution
The distance travelled is equal to the area under the velocity-time graph, because the runner moves forward throughout the interval.
The graph shows a velocity of approximately 7 km/h initially, rising to 7.5 km/h, and later decreasing to approximately 6.5 km/h.
We divide the graph into approximate trapeziums and rectangles.
| Time (h) | Approximate velocity (km/h) |
|---|---|
| 0 | 7.0 |
| 1.5 | 7.5 |
| 3 | 7.5 |
| 5.5 | 6.5 |
| 6.5 | 6.5 |
Using the area formula for each section:
Solution
Given: Initial velocity = 6 m/s; constant-velocity time = 120 s; acceleration = 1 m/s²; accelerating time = 6 s.
Step 1: Find final velocity.
The graph has a horizontal section from 0 to 120 s, followed by a rising straight line from 6 to 12 m/s over the next 6 s.
Step 2: Displacement in first 120 s.
Step 3: Displacement during acceleration.
Total displacement:
Solution
Car A:
Car B:
Using v = u + at, the velocities are:
| Time (s) | Car A (m/s) | Car B (m/s) |
|---|---|---|
| 0 | 0 | 0 |
| 2 | 2 | 0.6 |
| 4 | 4 | 1.2 |
| 5 | 5 | 1.5 |
| 6 | — | 1.8 |
| 8 | — | 2.4 |
| 10 | — | 3.0 |
The following graph represents the given intervals for both cars.
Displacement of Car A in 5 s:
Displacement of Car B in 10 s:
Solution
Given: Radius r = 7 cm.
Time interval = 90 minutes = 5400 seconds.
The minute hand completes one revolution every 60 minutes.
(i) Distance travelled
Distance in one revolution = Circumference of circle = 2πr.
(ii) Displacement
At 6 PM the minute hand points towards 12. At 7:30 PM it points towards 6.
The tip moves from the top of the clock to the bottom. Its displacement is the diameter of the circular path.
(iii) Average speed
(iv) Average velocity
(i) Distance = 66 cm
(ii) Displacement = 14 cm downwards
(iii) Average speed ≈ 0.0122 cm/s
(iv) Average velocity ≈ 0.00259 cm/s downwards
