Q1. 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:

Given:
Diameter of track = 200 m
Radius = 100 m
Time for one round = 40 s
Total time = 2 min 20 s = 140 s

Solution:

Circumference of track = 2 × 3.14 × 100 = 628 m

Number of rounds = 140 ÷ 40 = 3.5 rounds

Distance covered = 3.5 × 628 = 2198 m

Displacement = Diameter of track = 200 m

Answer:
Distance covered = 2198 m
Displacement = 200 m

Q2. 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:

Given:

• Distance from A to B = 300 m
• Time taken from A to B = 2 minutes 30 seconds = 150 seconds
• Distance from B to C = 100 m (back towards A)
• Time taken from B to C = 1 minute = 60 seconds

(a) From A to B

Average speed = Total distance / Total time
= 300 m / 150 s
= 2 m/s

Average velocity = Displacement / Total time
Displacement from A to B = 300 m (straight line)
= 300 m / 150 s
= 2 m/s

(b) From A to C

Total distance = distance from A to B + distance from B to C
= 300 m + 100 m
= 400 m

Total time = 150 s + 60 s = 210 s

Displacement = distance from A to C in a straight line
Since C is 100 m back from B towards A, displacement from A to C = 300 m – 100 m = 200 m (in the direction from A to B)

Average speed = Total distance / Total time
= 400 m / 210 s
≈ 1.90 m/s

Average velocity = Displacement / Total time
= 200 m / 210 s
≈ 0.95 m/s

Final answers:

• (a) Average speed = 2 m/s, Average velocity = 2 m/s
• (b) Average speed ≈ 1.90 m/s, Average velocity ≈ 0.95 m/s

Q3. 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:

Given:

• Speed going to school = 20 km/h
• Speed returning = 30 km/h
• Distance for both trips =
  $d$

   

Formula:
Average speed for whole trip = $\frac{2 \times \text{speed}_1 \times \text{speed}_2}{\text{speed}_1 + \text{speed}_2}$

Calculation:= $\frac{2 \times 20 \times 30}{20 + 30}$

= $\frac{1200}{50}$

= 24 km/h

Final answer:
Average speed for Abdul’s whole trip = 24 km/h

Q4. 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:

Given:

• Initial velocity,
  $=\mathrm{0<span\; style="font-family:\; -apple-system,\; system-ui,\; BlinkMacSystemFont,\; \text{'}Segoe\; UI\text{'},\; Helvetica,\; Arial,\; sans-serif,\; \text{'}Apple\; Color\; Emoji\text{'},\; \text{'}Segoe\; UI\; Emoji\text{'},\; \text{'}Segoe\; UI\; Symbol\text{'};\; font-size:\; inherit;">(starting\; from\; rest)</span>}$
• Acceleration,
  $a = 3.0 \, \text{m/s}^2$
• Time,
  $t = 8.0 \, \text{s}$

Distance travelled, $s = ut + \frac{1}{2}at^2$

Substitution:  $s = 0 \times 8.0 + \frac{1}{2} \times 3.0 \times (8.0)^2$

$s = 0 + \frac{1}{2} \times 3.0 \times 64$

$s = 1.5 \times 64$

$s = 96 \, \text{m}$

Final answer:
The motorboat travels 96 meters during this time.

Q5. A driver of a car travelling at 52 km h⁻¹ applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 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?
Answer:

Given:

• Initial speed of Car 1,
  $u_1 = 52 \, \text{km/h} = \frac{52 \times 1000}{3600} = 14.44 \, \text{m/s}$

   

• Time to stop,
  $t_1 = 5 \, \text{s}$

   

• Initial speed of Car 2,
  $u_2 = 3 \, \text{km/h} = \frac{3 \times 1000}{3600} = 0.83 \, \text{m/s}$

   

• Time to stop,
  $t_2 = 10 \, \text{s}$

   

Acceleration:

$a_1 = -\frac{14.44}{5} = -2.888 \, \text{m/s}^2$

$a_2 = -\frac{0.83}{10} = -0.083 \, \text{m/s}^2$

Distance travelled:

$s_1 = 14.44 \times 5 + \frac{1}{2} \times (-2.888) \times (5)^2 = 36.1 \, \text{m}$

$s_2 = 0.83 \times 10 + \frac{1}{2} \times (-0.083) \times (10)^2 = 4.15 \, \text{m}$

Final answer:
Car 1 travelled farther after brakes were applied, covering 36.1 m compared to Car 2’s 4.15 m.

Q6. Fig 8.11 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:
(a) Which of the three is travelling the fastest?
Answer:
The steepest slope on the distance-time graph corresponds to the fastest object, which is Object A.

(b) Are all three ever at the same point on the road?
Answer:
No, the three lines do not intersect at a single point.

(c) How far has C travelled when B passes A?
Answer:
At the point where B’s graph crosses A’s, check the value on C’s graph at the same time, as indicated in the figure.

(d) How far has B travelled by the time it passes C?
Answer:
Check the value on B’s graph at the tuning where B passes C, as indicated in the figure.

Q7. 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:

Given:
Height, $h=\mathrm{20m}$

Acceleration, $a = 10$ m/s² (due to gravity)
Initial velocity, $u = 0$(since the ball is gently dropped)

Formula for final velocity after falling a distance $h$:

$v^2 = u^2 + 2 a h$

Substitution:

$v^2 = 0 + 2 \times 10 \times 20$

$v^2 = 400$

Final calculation:

$v = \sqrt{400} = 20$

m/s

Formula for time to fall distance $h$:

$v = u + a t$

Substituting: $20 = 0 + 10 \times t$

Final calculation:$t = \frac{20}{10} = 2$ seconds

Answer:

• Velocity with which the ball will strike the ground = 20 m/s
• Time taken to strike the ground = 2 seconds

Q8. The speed-time graph for a car is shown in Fig. 8.12.
(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
Answer:
Area under speed-time graph from 0 to 4 s = shape is a triangle
Area = ½ × base × height = ½ × 4 × 6 = 12 m

(b) Which part of the graph represents uniform motion of the car?
Answer:
The horizontal segment of the graph after the 4-second mark, where speed remains constant.

Q9. State which of the following situations are possible and give an example for each:
(a) an object with a constant acceleration but with zero velocity
Answer:
Possible; for example, an object thrown vertically upward momentarily has zero velocity at the highest point, but acceleration due to gravity remains constant.

(b) an object moving with an acceleration but with uniform speed
Answer:
Possible if the direction of velocity changes but speed remains constant, e.g., uniform circular motion.

(c) an object moving in a certain direction with an acceleration in the perpendicular direction.
Answer:
Possible; for example, a car turning on a circular track moves forward while the acceleration is toward the center (perpendicular to motion); also observed in circular motion.

Q10. An artificial satellite is moving in a circular orbit of radius 42,250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Answer:

Given:
Radius of orbit,

$r = 42,250$

km = 42,250 × 1000 = 4.225 × 10⁷ m
Time period,

$T = 24$

hours = 24 × 60 × 60 = 86400 seconds

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

Substitution: v = $\frac{2 \times 3.1416 \times 4.225 \times 10^7}{86400}$

Final Calculation: v =$\frac{2.654 \times 10^8}{86400} \approx 3071.6$m/s

Answer: The speed of the artificial satellite is approximately 3072 m/s.