Motion

Given:
- Diameter of circular track = 200 m, so radius $r = 100$ m
- Time for one round = 40 s
- Total time = 2 min 20 s = 140 s

Step 1: Find the number of rounds completed.
$$\text{Number of rounds} = \frac{\text{Total time}}{\text{Time per round}} = \frac{140}{40} = 3.5 \text{ rounds}$$

Step 2: Find the distance covered.

Circumference of the track (distance per round):
$$C = \pi d = \pi \times 200 = 628 \text{ m (approx.)}$$

$$\text{Distance covered} = 3.5 \times 628 = 2200 \text{ m}$$

Step 3: Find the displacement.

After 3.5 rounds, the athlete is at the diametrically opposite end of the starting point.

$$\text{Displacement} = \text{diameter} = 200 \text{ m}$$

Answer:
- Distance covered = 2200 m
- Displacement = 200 m

Given:
- A to B: distance = 300 m, time = 2 min 30 s = 150 s
- B to C: distance = 100 m (back towards A), time = 1 min = 60 s

(a) From A to B:

$$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{300}{150} = 2 \text{ m s}^{-1}$$

Displacement from A to B = 300 m (straight road)

$$\text{Average velocity} = \frac{\text{Displacement}}{\text{Time}} = \frac{300}{150} = 2 \text{ m s}^{-1} \text{ (from A to B)}$$

(b) From A to C:

Total distance = AB + BC = 300 + 100 = 400 m

Total time = 150 + 60 = 210 s

$$\text{Average speed} = \frac{400}{210} = 1.90 \text{ m s}^{-1}$$

Displacement from A to C = AB − BC = 300 − 100 = 200 m (from A towards B)

$$\text{Average velocity} = \frac{200}{210} = 0.95 \text{ m s}^{-1} \text{ (from A towards B)}$$

Answer:
- (a) Average speed = $2 \text{ m s}^{-1}$; Average velocity = $2 \text{ m s}^{-1}$ (A to B direction)
- (b) Average speed ≈ $1.90 \text{ m s}^{-1}$; Average velocity ≈ $0.95 \text{ m s}^{-1}$ (A to B direction)

Given:
- Speed from home to school, $v_1 = 20 \text{ km h}^{-1}$
- Speed from school to home, $v_2 = 30 \text{ km h}^{-1}$
- Distance one way = $d$ (same route)

Concept: Average speed = Total distance / Total time

Step 1: Find total distance.
$$\text{Total distance} = d + d = 2d$$

Step 2: Find total time.
$$t_1 = \frac{d}{v_1} = \frac{d}{20}, \quad t_2 = \frac{d}{v_2} = \frac{d}{30}$$

$$\text{Total time} = \frac{d}{20} + \frac{d}{30} = d\left(\frac{3+2}{60}\right) = \frac{5d}{60} = \frac{d}{12}$$

Step 3: Calculate average speed.
$$\text{Average speed} = \frac{2d}{d/12} = 2d \times \frac{12}{d} = 24 \text{ km h}^{-1}$$

Answer: The average speed for Abdul's entire trip is $\boxed{24 \text{ km h}^{-1}}$.

Given:
- Initial velocity, $u = 0$ (starts from rest)
- Acceleration, $a = 3.0 \text{ m s}^{-2}$
- Time, $t = 8.0 \text{ s}$

Formula used (second equation of motion):
$$s = ut + \frac{1}{2}at^2$$

Calculation:
$$s = (0)(8.0) + \frac{1}{2} \times 3.0 \times (8.0)^2$$
$$s = 0 + \frac{1}{2} \times 3.0 \times 64$$
$$s = \frac{1}{2} \times 192 = 96 \text{ m}$$

Answer: The boat travels 96 m during this time.

Note: This question refers to a speed-time graph (Fig. 7.11 in the textbook) that is not fully visible here. The standard answers based on the described graph are given below.

(a) Distance travelled during braking:

The distance travelled by the car after brakes are applied is represented by the area under the speed-time graph between the point where brakes are applied and the point where the car comes to rest. This area is a triangle (since speed decreases uniformly from 52 km h⁻¹ to 0). This triangular area should be shaded.

(b) Uniform motion:

The part of the graph where the speed-time graph is a horizontal straight line (constant speed) represents the uniform motion of the car. This is the portion before the brakes are applied, where speed remains constant at 52 km h⁻¹.