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.

Note: This question is based on a distance-time graph (Fig. 7.10) that is not fully visible in the OCR. The standard NCERT answers are provided below based on the described graph.

(a) Which is travelling the fastest?

On a distance-time graph, the slope represents speed. The object with the steepest slope is travelling the fastest.

$$\text{Speed} = \text{slope of distance-time graph}$$

From the graph, B has the steepest slope among A, B, and C.

∴ B is travelling the fastest.

(b) Are all three ever at the same point on the road?

All three objects are at the same point on the road only if their distance-time curves intersect at a single common point. From the graph, the three lines do not all meet at one point.

∴ No, all three are never at the same point on the road.

(c) How far has C travelled when B passes A?

From the graph, when B's curve intersects A's curve (i.e., B passes A), reading the position of C on the graph at that same time gives the distance C has travelled.

Based on the standard NCERT graph: C has travelled approximately 8 km when B passes A.

(d) How far has B travelled by the time it passes C?

From the graph, when B's curve intersects C's curve (i.e., B passes C), reading the distance of B at that time:

Based on the standard NCERT graph: B has travelled approximately 4 km by the time it passes C.

Given:
- Initial velocity, $u = 0$ (gently dropped)
- Distance (height), $s = 20$ m
- Acceleration, $a = 10 \text{ m s}^{-2}$

Step 1: Find the final velocity using the third equation of motion.
$$v^2 = u^2 + 2as$$
$$v^2 = (0)^2 + 2 \times 10 \times 20$$
$$v^2 = 400$$
$$v = 20 \text{ m s}^{-1}$$

Step 2: Find the time using the first equation of motion.
$$v = u + at$$
$$20 = 0 + 10 \times t$$
$$t = \frac{20}{10} = 2 \text{ s}$$

Answer:
- The ball will strike the ground with a velocity of 20 m s⁻¹.
- It will strike the ground after 2 s.

Note: This question refers to Fig. 7.11 (speed-time graph) which is not fully visible. Based on the standard NCERT graph where the car accelerates from 0 to 6 m s⁻¹ in the first 4 seconds, then moves at uniform speed.

(a) Distance in the first 4 seconds:

The distance is given by the area under the speed-time graph for the first 4 seconds.

From the graph, the car accelerates uniformly from $u = 0$ to $v = 6 \text{ m s}^{-1}$ in $t = 4$ s. This forms a triangle on the graph.

$$s = \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height}$$
$$s = \frac{1}{2} \times 4 \times 6 = 12 \text{ m}$$

The triangular area under the graph from $t = 0$ to $t = 4$ s should be shaded.

(b) Uniform motion:

The part of the graph that is a horizontal straight line (constant speed) represents uniform motion. From the standard graph, this is the portion after $t = 4$ s where the speed remains constant at 6 m s⁻¹.

Answer:
- (a) The car travels 12 m in the first 4 seconds.
- (b) The horizontal portion of the graph (after $t = 4$ s) represents uniform motion.

(a) An object with constant acceleration but zero velocity:

Possible. ✓

When an object is thrown vertically upward, at the highest point its velocity becomes zero, but the acceleration due to gravity ($g = 10 \text{ m s}^{-2}$) acts continuously downward. Thus, at that instant, velocity = 0 but acceleration = constant ($g$).

Example: A ball thrown vertically upward at its highest point.

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(b) An object moving with acceleration but with uniform speed:

Possible. ✓

Acceleration is a vector quantity — it can change direction without changing magnitude of speed. When an object moves in a circular path at constant speed, its direction changes continuously, so it has acceleration (centripetal acceleration) even though its speed is constant.

Example: A car moving along a circular road at constant speed undergoes uniform circular motion — speed is uniform but acceleration exists due to change in direction.

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(c) An object moving in a certain direction with acceleration in the perpendicular direction:

Possible. ✓

In uniform circular motion, the velocity of the object is along the tangent to the circle, while the centripetal acceleration is directed towards the centre — which is perpendicular to the velocity (and hence to the direction of motion).

Example: A stone tied to a string and whirled in a horizontal circle — the stone moves tangentially but the centripetal acceleration acts perpendicular (towards the centre).

Given:
- Radius of circular orbit, $r = 42250 \text{ km} = 42250 \times 1000 \text{ m} = 4.225 \times 10^7 \text{ m}$
- Time period, $T = 24 \text{ h} = 24 \times 3600 \text{ s} = 86400 \text{ s}$

Formula: Speed in circular motion:
$$v = \frac{\text{Circumference}}{\text{Time period}} = \frac{2\pi r}{T}$$

Calculation:
$$v = \frac{2 \times 3.14 \times 4.225 \times 10^7}{86400}$$
$$v = \frac{2 \times 3.14 \times 42250000}{86400}$$
$$v = \frac{265330000}{86400}$$
$$v \approx 3070 \text{ m s}^{-1}$$
$$v \approx 3.07 \text{ km s}^{-1}$$

Answer: The speed of the artificial satellite is approximately $\mathbf{3.07 \text{ km s}^{-1}}$ (or $3070 \text{ m s}^{-1}$).