Kinematics
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
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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
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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
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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 30 minutes, what is his average speed?
A.
6 km/h
B.
7 km/h
C.
8 km/h
D.
9 km/h
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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 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
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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 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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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 time of flight?
A.
5 s
B.
10 s
C.
15 s
D.
20 s
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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 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
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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 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
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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
Show solution
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
Show solution
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
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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
Show solution
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
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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
Show solution
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
Show solution
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
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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
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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
Show solution
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
Show solution
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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