Detailed Answers to CBSE Textbook Questions

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?

Answer:

• Diameter of the track = 200 m
• Radius of the track = 100 m (half of the diameter)
• Time for one complete round = 40 s

The total distance covered by the athlete in one round is the circumference of the circle, which is given by:

$$\text{Distance for one round} = 2\pi r = 2 \times \pi \times 100 = 628.3 \, \text{m}$$

The athlete runs for 2 minutes 20 seconds, which is equal to:

$$2 \, \text{minutes} + 20 \, \text{seconds} = 140 \, \text{seconds}$$

To find the number of rounds the athlete completes, divide the total time by the time for one round:

$$\text{Number of rounds} = \frac{140 \, \text{seconds}}{40 \, \text{seconds}} = 3.5 \, \text{rounds}$$

• Distance covered = 3.5 rounds × 628.3 m = 2199.05 m

Since the athlete has completed 3 full rounds and is halfway through the 4th round, the displacement will be the straight-line distance from the start to the point directly opposite to the starting point (diameter of the circle).

• Displacement = 200 m (diameter of the circle)

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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?

Answer:

• From A to B:

  • Distance = 300 m
  • Time = 2 minutes 30 seconds = 150 s
  • Average speed = $\frac{\text{Distance}}{\text{Time}} = \frac{300}{150} = 2 \, \text{m/s}$
  • Displacement = 300 m (straight line from A to B)
  • Average velocity = $\frac{\text{Displacement}}{\text{Time}} = \frac{300}{150} = 2 \, \text{m/s}$

• From A to C:

  • Total distance = 300 m + 100 m = 400 m
  • Total time = 150 s + 60 s = 210 s
  • Average speed = $\frac{\text{Total Distance}}{\text{Total Time}} = \frac{400}{210} \approx 1.90 \, \text{m/s}$
  • Displacement = 200 m (since Joseph jogs 100 m back, the displacement is the straight-line distance from A to C)
  • Average velocity = $\frac{\text{Displacement}}{\text{Time}} = \frac{200}{210} \approx 0.95 \, \text{m/s}$

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3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h–1. On his return trip along the same route, there is less traffic and the average speed is 30 km h–1. What is the average speed for Abdul’s trip?

Answer:

We can use the formula for the average speed for the entire trip:

$$\text{Average Speed} = \frac{2 \times \text{Speed}_1 \times \text{Speed}_2}{\text{Speed}_1 + \text{Speed}_2}$$

Substituting the values:

$$\text{Average Speed} = \frac{2 \times 20 \times 30}{20 + 30} = \frac{1200}{50} = 24 \, \text{km/h}$$

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4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s–2 for 8.0 s. How far does the boat travel during this time?

Answer:

We can use the equation of motion:

$$s = ut + \frac{1}{2} a t^2$$

Given that the initial velocity $u = 0$, acceleration $a = 3.0 \, \text{m/s}^2$, and time $t = 8.0 \, \text{s}$:

$$s = 0 + \frac{1}{2} \times 3.0 \times (8.0)^2 = \frac{1}{2} \times 3.0 \times 64 = 96 \, \text{m}$$

The boat travels 96 meters during this time.

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5. A driver of a car travelling at 52 km h–1 applies the brakes. Shade the area on the graph that represents the distance travelled by the car during the period.

(b) Which part of the graph represents uniform motion of the car?

Answer:

For this question, we assume the graph refers to a velocity-time graph. The area under the graph represents the distance travelled. In the case of uniform motion, the velocity is constant, so the graph would be a horizontal line, and the area under this line represents the distance travelled during that time.

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7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s–2, with what velocity will it strike the ground? After what time will it strike the ground?

Answer:

We can use the equation of motion:

$$v^2 = u^2 + 2as$$

Where:

• Initial velocity $u = 0$,
• Acceleration $a = 10 \, \text{m/s}^2$,
• Distance $s = 20 \, \text{m}$.

Substituting the values:

$$v^2 = 0 + 2 \times 10 \times 20 = 400$$ $$v = \sqrt{400} = 20 \, \text{m/s}$$

Now, to find the time taken, we use the equation:

$$v = u + at$$

Substituting the values:

$$20 = 0 + 10 \times t$$ $$t = \frac{20}{10} = 2 \, \text{seconds}$$

Thus, the velocity when the ball strikes the ground is 20 m/s, and the time taken is 2 seconds.

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9. State which of the following situations are possible and give an example for each of these:

(a) an object with a constant acceleration but with zero velocity
(b) an object moving with an acceleration but with uniform speed
(c) an object moving in a certain direction with an acceleration in the perpendicular direction.

Answer:

• (a) Possible. Example: An object thrown vertically upward. It momentarily has zero velocity at the highest point, but its acceleration (due to gravity) remains constant.
• (b) Possible. Example: An object moving in a circular path with uniform speed. The object has acceleration towards the center (centripetal acceleration) but its speed remains constant.
• (c) Possible. Example: An object moving in a circular path, where the velocity is tangential and the acceleration is centripetal (directed towards the center of the circle).

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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.

Answer:

We can calculate the speed using the formula for the speed of an object moving in a circular orbit:

$$v = \frac{2 \pi r}{T}$$

Where:

• $r = 42250 \, \text{km} = 4.225 \times 10^7 \, \text{m}$,
• $T = 24 \, \text{hours} = 86400 \, \text{seconds}$.

Substituting the values:

$$v = \frac{2 \times \pi \times 4.225 \times 10^7}{86400} \approx 3.06 \, \text{km/s}$$

The speed of the satellite is approximately 3.06 km/s.