Measurement of Time and Motion
1. Calculate the speed of a car that travels 150 metres in 10 seconds. Express your answer in km/h.
Answer
Speed = Distance / Time = 150 m / 10 s = 15 m/s
To convert to km/h: 15 m/s × (3600/1000) = 15 × 3.6 = 54 km/h
54 km/h
2. A runner completes 400 metres in 50 seconds. Another runner completes the same distance in 45 seconds. Who has a greater speed and by how much?
Answer:
Speed of first runner = 400 m / 50 s = 8 m/s
Speed of second runner = 400 m / 45 s ≈ 8.89 m/s
Difference in speed = 8.89 m/s – 8 m/s = 0.89 m/s
The second runner has greater speed by 0.89 m/s.
3. A train travels at a speed of 25 m/s and covers a distance of 360 km. How much time does it take?
Answer:
Convert distance to metres: 360 km = 360 × 1000 = 360,000 m
Time = Distance / Speed = 360,000 m / 25 m/s = 14,400 s
Convert to hours: 14,400 s ÷ 3600 = 4 hours
4 hours
4. A train travels 180 km in 3 h. Find its speed in: (i) km/h (ii) m/s (iii) What distance will it travel in 4 h if it maintains the same speed throughout the journey?
Answer:
(i) Speed in km/h = Distance / Time = 180 km / 3 h = 60 km/h
(ii) Speed in m/s = 60 km/h × (1000/3600) = 60 × (5/18) ≈ 16.67 m/s
(iii) Distance in 4 h = Speed × Time = 60 km/h × 4 h = 240 km
(i) 60 km/h, (ii) 16.67 m/s, (iii) 240 km
5. The fastest galloping horse can reach the speed of approximately 18 m/s. How does this compare to the speed of a train moving at 72 km/h?
Answer
Convert train’s speed to m/s: 72 km/h × (1000/3600) = 72 × (5/18) = 20 m/s
Compare: Horse speed = 18 m/s, Train speed = 20 m/s
Train is faster by: 20 m/s – 18 m/s = 2 m/s
The train is faster than the horse by 2 m/s.
6. Distinguish between uniform and non-uniform motion using the example of a car moving on a straight highway with no traffic and a car moving in city traffic.
Answer
1. Uniform motion: A car on a straight highway with no traffic moves at a constant speed, covering equal distances in equal time intervals.
2. Non-uniform motion: A car in city traffic changes speed frequently due to stops, starts, or slowdowns, covering unequal distances in equal time intervals.
Uniform motion has constant speed (highway car), while non-uniform motion has changing speed (city traffic car).
7. Data for an object covering distances in different intervals of time are given in the following table. If the object is in uniform motion, fill in the gaps in the table.
Answer
Time (s): 0, 10, 20, 30, 50, 70
Distance (m): 0, 8, 24, 32, 40, 56
For uniform motion, speed is constant. Calculate speed: From 0 to 10 s, speed = 8 m / 10 s = 0.8 m/s
Check other intervals: 10 to 20 s = (24 – 8) / 10 = 1.6 m/s (not uniform, so table may have errors or non-uniform motion)
Assuming uniform motion, use speed = 0.8 m/s for all intervals:
- At 20 s: Distance = 0.8 × 20 = 16 m (given 24 m, incorrect)
- At 30 s: Distance = 0.8 × 30 = 24 m (given 32 m, incorrect)
Correct table assuming uniform motion (speed = 0.8 m/s):
- Time (s): 0, 10, 20, 30, 50, 70
- Distance (m): 0, 8, 16, 24, 40, 56
- The given table is not for uniform motion. For uniform motion, distances should be 0, 8, 16, 24, 40, 56 m.
8. A car covers 60 km in the first hour, 70 km in the second hour, and 50 km in the third hour. Is the motion uniform? Justify your answer. Find the average speed of the car.
Answer
Uniform motion means covering equal distances in equal time intervals.
Distances: 60 km in 1st hour, 70 km in 2nd hour, 50 km in 3rd hour.
Since 60 km, 70 km, and 50 km are unequal, the motion is not uniform.
Total distance = 60 + 70 + 50 = 180 km
Total time = 1 + 1 + 1 = 3 hours
Average speed = Total distance / Total time = 180 km / 3 h = 60 km/h
The motion is non-uniform because the car covers unequal distances in equal time intervals. The average speed is 60 km/h.
9. Which type of motion is more common in daily life-uniform or non-uniform? Provide three examples from your experience to support your answer.
Answer
Non-uniform motion is more common in daily life because objects rarely move at a constant speed for long periods.
Example 1: A car in city traffic stops and starts, changing speed frequently (non-uniform).
Example 2: A person walking to school speeds up or slows down depending on obstacles or hurry (non-uniform).
Example 3: A bicycle rider slows down at turns or speeds up on straight paths (non-uniform).
Non-uniform motion is more common. Examples: car in traffic, walking to school, and riding a bicycle.
10. Data for the motion of an object are given in the following table. State whether the speed of the object is uniform or non-uniform. Find the average speed.
Answer
Table: Time (s): 0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100; Distance (m): 0, 6, 21, 29, 35, 42, 45, 55, 60, 10, 16
Note: The distances 10 m and 16 m for 90 s and 100 s seem incorrect (decreasing distance). Assuming typo, possible correct values are 70 m and 80 m (based on pattern).
Calculate speed for each interval (assuming corrected distances 70 m at 90 s, 80 m at 100 s):
- 0-10 s: (6-0) / 10 = 0.6 m/s
- 10-20 s: (21-6) / 10 = 1.5 m/s
- 20-30 s: (29-21) / 10 = 0.8 m/s
- 30-40 s: (35-29) / 10 = 0.6 m/s
- 40-50 s: (42-35) / 10 = 0.7 m/s
- 50-60 s: (45-42) / 10 = 0.3 m/s
- 60-70 s: (55-45) / 10 = 1.0 m/s
- 70-80 s: (60-55) / 10 = 0.5 m/s
- 80-90 s: (70-60) / 10 = 1.0 m/s
- 90-100 s: (80-70) / 10 = 1.0 m/s
- Speeds vary (0.3 to 1.5 m/s), so the motion is non-uniform.
- Average speed = Total distance / Total time = 80 m / 100 s = 0.8 m/s
- The motion is non-uniform because the speed varies. The average speed is 0.8 m/s.
11. A vehicle moves along a straight line and covers a distance of 2 km. In the first 500 m, it moves with a speed of 10 m/s and in the next 500 m, it moves with a speed of 5 m/s. With what speed should it move the remaining distance so that the journey is complete in 200 s? What is the average speed of the vehicle for the entire journey?
Answer
Total distance = 2 km = 2000 m
First 500 m: Speed = 10 m/s, Time = Distance / Speed = 500 / 10 = 50 s
Next 500 m: Speed = 5 m/s, Time = 500 / 5 = 100 s
Total time for first 1000 m = 50 + 100 = 150 s
Remaining distance = 2000 – 1000 = 1000 m
Remaining time = 200 s – 150 s = 50 s
Speed for remaining 1000 m = Distance / Time = 1000 / 50 = 20 m/s
Average speed = Total distance / Total time = 2000 m / 200 s = 10 m/s
The vehicle should move at 20 m/s for the remaining distance. The average speed is 10 m/s.
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