Kinematics Study Materials for Competitive Exams

Kinematics

Motion in One Dimension Motion in Two Dimensions Relative Motion

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

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

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

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

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

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 projected horizontally with a speed of 10 m/s?

A. 20 m
B. 30 m
C. 40 m
D. 50 m

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 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

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 long does it take to reach the ground?

A. 3 s
B. 4 s
C. 5 s
D. 6 s

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

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

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

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

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

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

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?

A. 1 s
B. 2 s
C. 3 s
D. 4 s

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

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? (Assume g = 10 m/s²)

A. 1 s
B. 2 s
C. 3 s
D. 4 s

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

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

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

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

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

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

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

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

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

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

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

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

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

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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Kinematics

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