Kinematics
Q. A runner completes a 400 m lap in 50 seconds. What is the average velocity of the runner?
A.
8 m/s
B.
6 m/s
C.
4 m/s
D.
2 m/s
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Solution
Average velocity = total displacement / total time. Since the runner returns to the starting point, displacement = 0. Average velocity = 0 m / 50 s = 0 m/s.
Correct Answer: B — 6 m/s
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Q. A stone is dropped from a height of 20 m. How long does it take to reach the ground?
A.
2 s
B.
1 s
C.
3 s
D.
4 s
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Solution
Using h = (1/2)gt², where h = 20 m and g = 9.8 m/s², we have 20 = (1/2)*9.8*t². Solving gives t² = 4.08, so t ≈ 2 s.
Correct Answer: A — 2 s
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Q. A stone is dropped from a height of 20 m. How long will it take to reach the ground?
A.
2 s
B.
1 s
C.
3 s
D.
4 s
Show solution
Solution
Using h = (1/2)gt², where h = 20 m and g = 9.8 m/s², we have 20 = (1/2)*9.8*t². Solving gives t² = 4.08, so t ≈ 2 s.
Correct Answer: A — 2 s
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Q. A stone is dropped from a height of 45 m. How far will it have fallen after 2 seconds?
A.
10 m
B.
20 m
C.
30 m
D.
40 m
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Solution
Distance fallen (s) = (1/2)gt² = (1/2)(9.8)(2²) = 19.6 m.
Correct Answer: C — 30 m
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Q. A stone is dropped from a height of 45 m. How far will it have fallen after 3 seconds?
A.
10.5 m
B.
20 m
C.
30 m
D.
40.5 m
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Solution
Distance fallen (s) = (1/2)gt² = (1/2)(9.8)(3²) = 44.1 m.
Correct Answer: D — 40.5 m
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Q. A stone is dropped from a height of 45 m. How far will it travel horizontally if it is thrown horizontally with a speed of 10 m/s?
A.
20 m
B.
30 m
C.
40 m
D.
50 m
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Solution
Time to fall (t) = √(2h/g) = √(2*45/9.8) ≈ 3 s; horizontal distance = v*t = 10*3 = 30 m.
Correct Answer: B — 30 m
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Q. A stone is dropped from a height of 45 m. How far will it travel horizontally if it is projected horizontally with a speed of 10 m/s?
A.
20 m
B.
30 m
C.
40 m
D.
50 m
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Solution
Time to fall t = √(2h/g) = √(90/9.8) ≈ 4.27 s; horizontal distance = 10 * 4.27 ≈ 42.7 m.
Correct Answer: C — 40 m
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Q. A stone is dropped from a height of 45 m. How long does it take to reach the ground?
A.
3 s
B.
4 s
C.
5 s
D.
6 s
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Solution
Using h = (1/2)gt², we have 45 = (1/2)(9.8)t², solving gives t ≈ 4.3 s.
Correct Answer: B — 4 s
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Q. A stone is dropped from a height of 80 m. How long does it take to hit the ground?
A.
4 s
B.
5 s
C.
6 s
D.
8 s
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Solution
Time (t) = √(2h/g) = √(2*80/9.8) = 4.04 s.
Correct Answer: A — 4 s
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Q. A stone is dropped from a height of 80 m. How long will it take to reach the ground?
A.
2 s
B.
4 s
C.
6 s
D.
8 s
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Solution
Using h = (1/2)gt², we get t = √(2h/g) = √(2*80/9.8) = 4.04 s.
Correct Answer: B — 4 s
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Q. A stone is thrown downward with an initial velocity of 5 m/s from a height of 45 m. How long will it take to hit the ground? (Assume g = 10 m/s²)
A.
3 s
B.
4 s
C.
5 s
D.
6 s
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Solution
Using the equation of motion: h = ut + 0.5gt². 45 = 5t + 0.5 * 10 * t². Solving the quadratic gives t = 3 s.
Correct Answer: B — 4 s
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Q. A stone is thrown horizontally from the top of a cliff 80 m high. How far from the base of the cliff will the stone land if it is thrown with a speed of 10 m/s?
A.
20 m
B.
40 m
C.
60 m
D.
80 m
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Solution
Time to fall = √(2h/g) = √(2*80/9.8) ≈ 4.04 s. Horizontal distance = speed * time = 10 m/s * 4.04 s ≈ 40 m.
Correct Answer: B — 40 m
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Q. A stone is thrown horizontally from the top of a cliff 80 m high. How far from the base of the cliff will it land?
A.
20 m
B.
40 m
C.
60 m
D.
80 m
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Solution
Time to fall = √(2h/g) = √(2*80/9.8) ≈ 4.04 s. Horizontal distance = horizontal speed * time. Assuming horizontal speed = 10 m/s, distance = 10 m/s * 4.04 s = 40.4 m.
Correct Answer: B — 40 m
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Q. A stone is thrown vertically upward with a speed of 20 m/s. How high will it rise before coming to rest?
A.
10 m
B.
20 m
C.
30 m
D.
40 m
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Solution
Maximum height (H) = (u²)/(2g) = (20²)/(2*9.8) ≈ 20.4 m.
Correct Answer: C — 30 m
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Q. A stone is thrown vertically upward with a speed of 20 m/s. How long will it take to reach the maximum height? (Assume g = 10 m/s²)
A.
1 s
B.
2 s
C.
3 s
D.
4 s
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Solution
Time to reach maximum height = initial velocity / g = 20 / 10 = 2 s.
Correct Answer: B — 2 s
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Q. A stone is thrown vertically upward with a speed of 20 m/s. How long will it take to reach the maximum height? (g = 10 m/s²)
A.
1 s
B.
2 s
C.
3 s
D.
4 s
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Solution
Using the formula: v = u - gt. At maximum height, v = 0. So, 0 = 20 - 10t. Solving gives t = 2 s.
Correct Answer: B — 2 s
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Q. A stone is thrown vertically upward with a speed of 20 m/s. How long will it take to reach the maximum height?
A.
1 s
B.
2 s
C.
3 s
D.
4 s
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Solution
Time to reach maximum height = initial velocity / g = 20 / 10 = 2 s.
Correct Answer: B — 2 s
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Q. A stone is thrown vertically upwards with a speed of 20 m/s. How long will it take to reach the maximum height? (g = 10 m/s²)
A.
1 s
B.
2 s
C.
3 s
D.
4 s
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Solution
Using the formula: v = u - gt. At maximum height, v = 0. So, 0 = 20 - 10t. Solving gives t = 2 s.
Correct Answer: B — 2 s
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Q. A stone is thrown vertically upwards with a speed of 20 m/s. How long will it take to reach the maximum height? (Assume g = 10 m/s²)
A.
1 s
B.
2 s
C.
3 s
D.
4 s
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Solution
Using the formula: time = (final velocity - initial velocity) / g. At maximum height, final velocity = 0. Time = (0 - 20) / -10 = 2 s.
Correct Answer: B — 2 s
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Q. A swimmer can swim at 3 km/h in still water. If he swims across a river flowing at 2 km/h, what is his resultant speed?
A.
3 km/h
B.
4 km/h
C.
5 km/h
D.
6 km/h
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Solution
Resultant speed = √(3² + 2²) = √(9 + 4) = √13 ≈ 3.6 km/h.
Correct Answer: C — 5 km/h
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Q. A swimmer can swim at 3 m/s in still water. If the river flows at 1 m/s, what is the swimmer's speed when swimming across the river?
A.
2 m/s
B.
3 m/s
C.
4 m/s
D.
5 m/s
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Solution
Speed across the river = √(3^2 - 1^2) = √(9 - 1) = √8 = 2.83 m/s (approximately 2 m/s).
Correct Answer: A — 2 m/s
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Q. A swimmer can swim at 3 m/s in still water. If the river flows at 2 m/s, what is the swimmer's speed relative to the bank when swimming upstream?
A.
1 m/s
B.
2 m/s
C.
3 m/s
D.
5 m/s
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Solution
Speed relative to bank = Speed of swimmer - Speed of river = 3 m/s - 2 m/s = 1 m/s.
Correct Answer: A — 1 m/s
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Q. A swimmer can swim at 4 km/h in still water. If he swims across a river that is 1 km wide and the current is 2 km/h, how long will it take him to reach the opposite bank?
A.
15 minutes
B.
30 minutes
C.
45 minutes
D.
1 hour
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Solution
Time = distance/speed = 1 km / 4 km/h = 0.25 hours = 15 minutes.
Correct Answer: B — 30 minutes
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Q. A train is moving at 72 km/h and a bird flies in the opposite direction at 18 km/h. What is the speed of the bird relative to the train?
A.
54 km/h
B.
72 km/h
C.
90 km/h
D.
18 km/h
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Solution
Relative speed = Speed of train + Speed of bird = 72 km/h + 18 km/h = 90 km/h.
Correct Answer: A — 54 km/h
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Q. A train is moving at 72 km/h and passes a platform in 20 seconds. What is the length of the platform if the train is 180 meters long?
A.
200 m
B.
300 m
C.
400 m
D.
500 m
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Solution
Speed of train = 72 km/h = 20 m/s. Time = 20 s. Distance = Speed × Time = 20 m/s × 20 s = 400 m. Length of platform = 400 m - 180 m = 220 m.
Correct Answer: B — 300 m
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Q. A train is moving at 72 km/h and passes a platform in 20 seconds. What is the length of the train?
A.
100 m
B.
200 m
C.
300 m
D.
400 m
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Solution
Length of train = Speed × Time = (72 km/h × (1000 m/1 km) / (3600 s/1 h)) × 20 s = 400 m.
Correct Answer: B — 200 m
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Q. A train is moving at 72 km/h and passes a platform in 30 seconds. What is the length of the platform if the train is 200 meters long?
A.
200 m
B.
300 m
C.
400 m
D.
500 m
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Solution
Speed in m/s = 72 * (1000/3600) = 20 m/s. Distance = Speed × Time = 20 m/s × 30 s = 600 m. Length of platform = 600 m - 200 m = 400 m.
Correct Answer: B — 300 m
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Q. A train is moving at 90 km/h and a bird flies at 45 km/h in the opposite direction. What is the speed of the bird relative to the train?
A.
135 km/h
B.
45 km/h
C.
90 km/h
D.
135 km/h
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Solution
Relative speed = Speed of train + Speed of bird = 90 km/h + 45 km/h = 135 km/h.
Correct Answer: A — 135 km/h
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Q. A train leaves a station and travels at 90 km/h. Another train leaves the same station 30 minutes later and travels at 120 km/h. How far from the station will they meet?
A.
90 km
B.
120 km
C.
150 km
D.
180 km
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Solution
Let the distance be d. Time taken by first train = d/90. Time taken by second train = d/120. Setting up the equation gives d = 150 km.
Correct Answer: C — 150 km
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Q. A train leaves a station at 70 km/h. Another train leaves the same station 30 minutes later at 90 km/h. How far from the station will they meet?
A.
70 km
B.
90 km
C.
100 km
D.
110 km
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Solution
Let the distance be d. Time taken by first train = d/70, second train = d/90. They meet when d/70 = d/90 + 0.5. Solving gives d = 100 km.
Correct Answer: C — 100 km
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