Kinematics Study Materials for Competitive Exams

Kinematics

Motion in One Dimension Motion in Two Dimensions Relative Motion

Q. A person is walking at 2 m/s on a train moving at 10 m/s. What is the speed of the person relative to the ground?

A. 2 m/s
B. 8 m/s
C. 10 m/s
D. 12 m/s

Solution

Speed of person relative to ground = Speed of train + Speed of person = 10 m/s + 2 m/s = 12 m/s.

Correct Answer: D — 12 m/s

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Q. A person is walking at 4 km/h in a train moving at 36 km/h. What is the speed of the person relative to the ground?

A. 32 km/h
B. 40 km/h
C. 36 km/h
D. 4 km/h

Solution

Speed of person relative to ground = Speed of train + Speed of person = 36 km/h + 4 km/h = 40 km/h.

Correct Answer: B — 40 km/h

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Q. A person walks at 4 km/h and runs at 10 km/h. If he walks for 30 minutes and then runs for 30 minutes, how far does he travel?

A. 5 km
B. 7 km
C. 8 km
D. 10 km

Solution

Distance walked = 4 km/h * 0.5 h = 2 km. Distance run = 10 km/h * 0.5 h = 5 km. Total distance = 2 + 5 = 7 km.

Correct Answer: C — 8 km

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Q. A person walks at 4 km/h and runs at 10 km/h. If he walks for 30 minutes and then runs for 1 hour, how far does he travel?

A. 7 km
B. 8 km
C. 9 km
D. 10 km

Solution

Distance walked = 4 * 0.5 = 2 km; Distance run = 10 * 1 = 10 km; Total distance = 2 + 10 = 12 km.

Correct Answer: C — 9 km

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Q. A person walks at 4 km/h and runs at 10 km/h. If he walks for 30 minutes and then runs for 30 minutes, what is his average speed?

A. 6 km/h
B. 7 km/h
C. 8 km/h
D. 9 km/h

Solution

Total distance = (4 × 0.5) + (10 × 0.5) = 2 + 5 = 7 km. Total time = 1 hour. Average speed = 7 km/h.

Correct Answer: B — 7 km/h

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Q. A person walks at 4 km/h in a train moving at 36 km/h. What is the speed of the person relative to the ground?

A. 32 km/h
B. 36 km/h
C. 40 km/h
D. 4 km/h

Solution

Speed of person relative to ground = Speed of train + Speed of person = 36 km/h + 4 km/h = 40 km/h.

Correct Answer: C — 40 km/h

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Q. A person walks at 4 km/h in a train moving at 60 km/h. What is the speed of the person relative to the ground?

A. 64 km/h
B. 60 km/h
C. 4 km/h
D. 56 km/h

Solution

Speed of person relative to ground = Speed of train + Speed of person = 60 km/h + 4 km/h = 64 km/h.

Correct Answer: A — 64 km/h

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Q. A person walks at 4 km/h in still water. If the current of the river is 2 km/h, what is the speed of the person relative to the bank when walking upstream?

A. 2 km/h
B. 4 km/h
C. 6 km/h
D. 8 km/h

Solution

Speed upstream = Speed of person - Speed of current = 4 km/h - 2 km/h = 2 km/h.

Correct Answer: A — 2 km/h

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Q. A person walks at 4 km/h on a moving escalator that moves at 2 km/h. What is the speed of the person relative to a stationary observer?

A. 2 km/h
B. 4 km/h
C. 6 km/h
D. 8 km/h

Solution

Speed of person relative to observer = Speed of person + Speed of escalator = 4 km/h + 2 km/h = 6 km/h.

Correct Answer: C — 6 km/h

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Q. A plane flies 300 km north and then 400 km east. What is the angle of the resultant displacement with respect to the north?

A. 36.87 degrees
B. 45 degrees
C. 53.13 degrees
D. 60 degrees

Solution

Angle = tan^(-1)(400/300) = 53.13 degrees.

Correct Answer: C — 53.13 degrees

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Q. A plane flies at a speed of 200 m/s at an angle of 30 degrees to the horizontal. What is the vertical component of its velocity?

A. 100 m/s
B. 150 m/s
C. 200 m/s
D. 250 m/s

Solution

Vertical component (v_y) = v * sin(θ) = 200 * (√3/2) ≈ 173.2 m/s.

Correct Answer: A — 100 m/s

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Q. A plane is flying at 200 m/s in still air. If there is a headwind of 50 m/s, what is the speed of the plane relative to the ground?

A. 150 m/s
B. 200 m/s
C. 250 m/s
D. 300 m/s

Solution

Speed relative to ground = Speed of plane - Speed of wind = 200 m/s - 50 m/s = 150 m/s.

Correct Answer: A — 150 m/s

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Q. A projectile is launched at an angle of 30 degrees with an initial speed of 20 m/s. What is the maximum height reached by the projectile?

A. 5 m
B. 10 m
C. 15 m
D. 20 m

Solution

Using the formula: h = (u² * sin²θ) / (2g), where u = 20 m/s, θ = 30°, g = 9.8 m/s². h = (20² * (1/4)) / (2 * 9.8) = 5.1 m, approximately 5 m.

Correct Answer: B — 10 m

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Q. A projectile is launched at an angle of 30 degrees with an initial velocity of 20 m/s. What is the maximum height reached by the projectile?

A. 5 m
B. 10 m
C. 15 m
D. 20 m

Solution

Maximum height (H) = (u^2 * sin^2(θ)) / (2g) = (20^2 * (1/2)^2) / (2 * 9.8) = 10.2 m.

Correct Answer: B — 10 m

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Q. A projectile is launched at an angle of 30° with an initial speed of 40 m/s. What is the maximum height reached by the projectile?

A. 80 m
B. 60 m
C. 40 m
D. 20 m

Solution

Using the formula H = (u² * sin²θ) / (2g), where u = 40 m/s, θ = 30°, and g = 9.8 m/s², we find H = (40² * (1/4)) / (2*9.8) = 40.82 m, approximately 60 m.

Correct Answer: B — 60 m

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Q. A projectile is launched with a speed of 30 m/s at an angle of 60 degrees. What is the horizontal component of its velocity?

A. 15 m/s
B. 20 m/s
C. 25 m/s
D. 30 m/s

Solution

Horizontal component (v_x) = u * cos(θ) = 30 * 0.5 = 15 m/s.

Correct Answer: B — 20 m/s

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Q. A projectile is launched with a speed of 50 m/s at an angle of 30 degrees. What is the horizontal component of the velocity?

A. 25 m/s
B. 35 m/s
C. 43.3 m/s
D. 50 m/s

Solution

Horizontal component (u_x) = u * cos(θ) = 50 * (√3/2) = 43.3 m/s.

Correct Answer: C — 43.3 m/s

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Q. A projectile is launched with a speed of 50 m/s at an angle of 30 degrees. What is the time of flight?

A. 5 s
B. 10 s
C. 15 s
D. 20 s

Solution

Time of flight (T) = (2 * u * sin(θ)) / g = (2 * 50 * (√3/2)) / 9.8 ≈ 10.2 s.

Correct Answer: B — 10 s

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Q. A projectile is launched with a speed of 50 m/s at an angle of 60 degrees. What is the horizontal component of its velocity?

A. 25 m/s
B. 50 m/s
C. 43.3 m/s
D. 30 m/s

Solution

Horizontal component (v_x) = v * cos(θ) = 50 * 0.5 = 43.3 m/s.

Correct Answer: C — 43.3 m/s

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Q. A projectile is launched with an initial speed of 30 m/s at an angle of 60 degrees. What is the horizontal component of the velocity?

A. 15 m/s
B. 20 m/s
C. 25 m/s
D. 30 m/s

Solution

Horizontal component (vx) = u * cos(θ) = 30 * 0.5 = 15 m/s.

Correct Answer: C — 25 m/s

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Q. A projectile is launched with an initial speed of 30 m/s at an angle of 60 degrees. What is the horizontal component of its velocity?

A. 15 m/s
B. 20 m/s
C. 25 m/s
D. 30 m/s

Solution

Horizontal component (vx) = u * cos(θ) = 30 * 0.5 = 15 m/s.

Correct Answer: B — 20 m/s

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Q. A projectile is launched with an initial speed of 40 m/s at an angle of 30 degrees. What is the time of flight?

A. 4 s
B. 5 s
C. 6 s
D. 8 s

Solution

Time of flight (T) = (2u * sin(θ)) / g = (2 * 40 * (√3/2)) / 9.8 ≈ 6.1 s.

Correct Answer: C — 6 s

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Q. A projectile is launched with an initial speed of 40 m/s at an angle of 60 degrees. What is the horizontal component of its velocity?

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

Solution

Horizontal component (vx) = u * cos(θ) = 40 * 0.5 = 20 m/s.

Correct Answer: B — 30 m/s

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Q. A projectile is launched with an initial speed of 50 m/s at an angle of 30 degrees. What is the horizontal component of the velocity?

A. 25 m/s
B. 35 m/s
C. 43.3 m/s
D. 50 m/s

Solution

Horizontal component (vx) = u * cos(θ) = 50 * cos(30) = 50 * (√3/2) = 43.3 m/s.

Correct Answer: C — 43.3 m/s

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Q. A projectile is launched with an initial speed of 50 m/s at an angle of 60 degrees. What is the horizontal component of its velocity?

A. 25 m/s
B. 50 m/s
C. 43.3 m/s
D. 30 m/s

Solution

Horizontal component (v_x) = u * cos(θ) = 50 * 0.5 = 25 m/s.

Correct Answer: C — 43.3 m/s

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Q. A projectile is launched with an initial velocity of 30 m/s at an angle of 60 degrees. What is the horizontal component of the velocity?

A. 15 m/s
B. 20 m/s
C. 25 m/s
D. 30 m/s

Solution

Horizontal component (u_x) = u * cos(θ) = 30 * 0.5 = 15 m/s.

Correct Answer: B — 20 m/s

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Q. A projectile is launched with an initial velocity of 40 m/s at an angle of 45 degrees. What is the vertical component of the velocity?

A. 20 m/s
B. 28.28 m/s
C. 30 m/s
D. 40 m/s

Solution

Vertical component Vy = u * sin(θ) = 40 * (√2/2) = 28.28 m/s.

Correct Answer: B — 28.28 m/s

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Q. A projectile is launched with an initial velocity of 40 m/s at an angle of 45 degrees. What is the time of flight?

A. 4 s
B. 5 s
C. 6 s
D. 7 s

Solution

Time of flight (T) = (2u * sin(θ)) / g = (2*40*sin(45))/9.8 = 5.14 s.

Correct Answer: B — 5 s

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Q. A projectile is launched with an initial velocity of 40 m/s at an angle of 45 degrees. What is the vertical component of its velocity?

A. 20 m/s
B. 28.28 m/s
C. 30 m/s
D. 40 m/s

Solution

Vertical component Vy = u * sin(θ) = 40 * (√2/2) = 28.28 m/s.

Correct Answer: B — 28.28 m/s

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Q. A projectile is thrown with a speed of 20 m/s at an angle of 45 degrees. What is the vertical component of the velocity?

A. 10 m/s
B. 14.14 m/s
C. 20 m/s
D. 28.28 m/s

Solution

Vertical component (u_y) = u * sin(θ) = 20 * (√2/2) = 14.14 m/s.

Correct Answer: B — 14.14 m/s

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Kinematics

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