Motion
NCERT Solutions • Class 9 Science • Chapter 7Chapter Exercises
1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Given: Diameter ($D$) = 200 m $\Rightarrow$ Radius ($r$) = 100 m. Time for 1 round ($t$) = 40 s.
Total time = 2 min 20 s = $120 + 20 = 140 \text{ s}$.
Number of rounds: $\frac{\text{Total Time}}{\text{Time per round}} = \frac{140}{40} = 3.5 \text{ rounds}$.
Distance: $3.5 \times \text{Circumference} = 3.5 \times 2\pi r = 3.5 \times 2 \times \frac{22}{7} \times 100 = 2200 \text{ m}$.
Displacement: After 3.5 rounds, the athlete is at the opposite end of the diameter.
Displacement = Diameter = 200 m.
Total time = 2 min 20 s = $120 + 20 = 140 \text{ s}$.
Number of rounds: $\frac{\text{Total Time}}{\text{Time per round}} = \frac{140}{40} = 3.5 \text{ rounds}$.
Distance: $3.5 \times \text{Circumference} = 3.5 \times 2\pi r = 3.5 \times 2 \times \frac{22}{7} \times 100 = 2200 \text{ m}$.
Displacement: After 3.5 rounds, the athlete is at the opposite end of the diameter.
Displacement = Diameter = 200 m.
2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
(a) A to B
Distance = 300 m. Time = 2 min 30 s = 150 s. Displacement = 300 m.
Avg Speed = $\frac{300}{150} = 2 \text{ m/s}$.
Avg Velocity = $\frac{300}{150} =$ $2 \text{ m/s}$.
Distance = 300 m. Time = 2 min 30 s = 150 s. Displacement = 300 m.
Avg Speed = $\frac{300}{150} = 2 \text{ m/s}$.
Avg Velocity = $\frac{300}{150} =$ $2 \text{ m/s}$.
(b) A to C
Total Distance = $300 + 100 = 400 \text{ m}$.
Total Time = $150 + 60 = 210 \text{ s}$.
Displacement (A to C) = $300 – 100 = 200 \text{ m}$.
Avg Speed = $\frac{400}{210} \approx$ $1.90 \text{ m/s}$.
Avg Velocity = $\frac{200}{210} \approx$ $0.95 \text{ m/s}$.
Total Distance = $300 + 100 = 400 \text{ m}$.
Total Time = $150 + 60 = 210 \text{ s}$.
Displacement (A to C) = $300 – 100 = 200 \text{ m}$.
Avg Speed = $\frac{400}{210} \approx$ $1.90 \text{ m/s}$.
Avg Velocity = $\frac{200}{210} \approx$ $0.95 \text{ m/s}$.
3. Abdul, while driving to school, computes the average speed for his trip to be $20 \text{ km h}^{-1}$. On his return trip along the same route, there is less traffic and the average speed is $30 \text{ km h}^{-1}$. What is the average speed for Abdul’s trip?
Let the distance to school be $x$ km.
Time for trip 1 ($t_1$) = $\frac{x}{20}$. Time for trip 2 ($t_2$) = $\frac{x}{30}$.
Total Distance = $x + x = 2x$.
Total Time = $\frac{x}{20} + \frac{x}{30} = \frac{3x + 2x}{60} = \frac{5x}{60} = \frac{x}{12}$.
Average Speed = $\frac{\text{Total Distance}}{\text{Total Time}} = \frac{2x}{x/12} = 2 \times 12 =$ $24 \text{ km h}^{-1}$.
Time for trip 1 ($t_1$) = $\frac{x}{20}$. Time for trip 2 ($t_2$) = $\frac{x}{30}$.
Total Distance = $x + x = 2x$.
Total Time = $\frac{x}{20} + \frac{x}{30} = \frac{3x + 2x}{60} = \frac{5x}{60} = \frac{x}{12}$.
Average Speed = $\frac{\text{Total Distance}}{\text{Total Time}} = \frac{2x}{x/12} = 2 \times 12 =$ $24 \text{ km h}^{-1}$.
4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of $3.0 \text{ m s}^{-2}$ for $8.0 \text{ s}$. How far does the boat travel during this time?
Given: $u = 0$, $a = 3.0 \text{ m s}^{-2}$, $t = 8.0 \text{ s}$.
Formula: $s = ut + \frac{1}{2}at^2$
$s = (0 \times 8) + \frac{1}{2} \times 3 \times (8)^2$
$s = 0 + \frac{1}{2} \times 3 \times 64 = 3 \times 32 =$ $96 \text{ m}$.
Formula: $s = ut + \frac{1}{2}at^2$
$s = (0 \times 8) + \frac{1}{2} \times 3 \times (8)^2$
$s = 0 + \frac{1}{2} \times 3 \times 64 = 3 \times 32 =$ $96 \text{ m}$.
5. A driver of a car travelling at $52 \text{ km h}^{-1}$ applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at $3 \text{ km h}^{-1}$ in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Car 1: $u = 52 \text{ km/h} = 14.44 \text{ m/s}$, $t = 5 \text{ s}$, $v = 0$.
Distance ($s_1$) = Area under graph = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 14.44 = 36.1 \text{ m}$.
Car 2: $u = 3 \text{ km/h} = 0.83 \text{ m/s}$, $t = 10 \text{ s}$, $v = 0$.
Distance ($s_2$) = $\frac{1}{2} \times 10 \times 0.83 = 4.15 \text{ m}$.
Conclusion: The first car travelled farther.
Distance ($s_1$) = Area under graph = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 14.44 = 36.1 \text{ m}$.
Car 2: $u = 3 \text{ km/h} = 0.83 \text{ m/s}$, $t = 10 \text{ s}$, $v = 0$.
Distance ($s_2$) = $\frac{1}{2} \times 10 \times 0.83 = 4.15 \text{ m}$.
Conclusion: The first car travelled farther.
6. Study the distance-time graph of three objects A, B and C and answer the following questions:
(a) Fastest: Object B (It has the steepest slope).
(b) Same Point: No, all three lines never intersect at a single point simultaneously.
(c) C when B passes A: Approx 8 km (Check graph intersection of B and A, then find corresponding distance for C).
(d) B when it passes C: Approx 5.7 km (Check intersection of B and C, read Y-axis).
(b) Same Point: No, all three lines never intersect at a single point simultaneously.
(c) C when B passes A: Approx 8 km (Check graph intersection of B and A, then find corresponding distance for C).
(d) B when it passes C: Approx 5.7 km (Check intersection of B and C, read Y-axis).
7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of $10 \text{ m s}^{-2}$, with what velocity will it strike the ground? After what time will it strike the ground?
Given: $u = 0$, $s = 20 \text{ m}$, $a = 10 \text{ m s}^{-2}$.
Find Velocity ($v$): $v^2 – u^2 = 2as$
$v^2 – 0 = 2 \times 10 \times 20 = 400 \Rightarrow v = \sqrt{400} =$ $20 \text{ m/s}$.
Find Time ($t$): $v = u + at$
$20 = 0 + 10t \Rightarrow t =$ $2 \text{ s}$.
Find Velocity ($v$): $v^2 – u^2 = 2as$
$v^2 – 0 = 2 \times 10 \times 20 = 400 \Rightarrow v = \sqrt{400} =$ $20 \text{ m/s}$.
Find Time ($t$): $v = u + at$
$20 = 0 + 10t \Rightarrow t =$ $2 \text{ s}$.
8. The speed-time graph for a car is shown in Fig. 7.11.
(a) Find how far does the car travel in the first 4 seconds. Shade the area…
(b) Which part of the graph represents uniform motion of the car?
(a) Find how far does the car travel in the first 4 seconds. Shade the area…
(b) Which part of the graph represents uniform motion of the car?
(a) Distance in first 4s:
The graph represents non-uniform acceleration (curve). We approximate the area under the curve (usually treated as a triangle or counting squares).
Area $\approx \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 6 = 12 \text{ m}$.
(b) Uniform Motion:
The part of the graph that is a straight horizontal line (parallel to time axis) represents uniform motion (constant speed). Usually from $t = 6 \text{ s}$ onwards.
The graph represents non-uniform acceleration (curve). We approximate the area under the curve (usually treated as a triangle or counting squares).
Area $\approx \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 6 = 12 \text{ m}$.
(b) Uniform Motion:
The part of the graph that is a straight horizontal line (parallel to time axis) represents uniform motion (constant speed). Usually from $t = 6 \text{ s}$ onwards.
9. State which of the following situations are possible and give an example for each of these:
(a) Constant acceleration, zero velocity: Possible. A ball thrown vertically upwards reaches its maximum height where velocity is zero, but acceleration due to gravity ($g$) is still acting downwards.
(b) Acceleration with uniform speed: Possible. An object in uniform circular motion has constant speed but changing velocity (direction), thus it has centripetal acceleration.
(c) Acceleration perpendicular to direction of motion: Possible. An airplane flying horizontally while gravity acts vertically downwards (projectile motion) or an object in circular motion.
(b) Acceleration with uniform speed: Possible. An object in uniform circular motion has constant speed but changing velocity (direction), thus it has centripetal acceleration.
(c) Acceleration perpendicular to direction of motion: Possible. An airplane flying horizontally while gravity acts vertically downwards (projectile motion) or an object in circular motion.
10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Given: Radius ($r$) = 42250 km. Time ($t$) = 24 h.
Formula: Speed ($v$) = $\frac{\text{Distance}}{\text{Time}} = \frac{2\pi r}{t}$
$v = \frac{2 \times 3.14 \times 42250}{24}$
$v = \frac{265330}{24} \approx$ $11055.4 \text{ km/h}$
(Or approx $3.07 \text{ km/s}$).
Formula: Speed ($v$) = $\frac{\text{Distance}}{\text{Time}} = \frac{2\pi r}{t}$
$v = \frac{2 \times 3.14 \times 42250}{24}$
$v = \frac{265330}{24} \approx$ $11055.4 \text{ km/h}$
(Or approx $3.07 \text{ km/s}$).