Relative Motion in 2D — NEET Physics MCQs with Solutions
Free NEET Physics Relative Motion in 2D MCQs with step-by-step solutions (20 questions). Part of Motion in 2D. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Relative Motion in 2D · easy · numerical
Cyclist A rides due east at 6 m/s and cyclist B rides due north at 8 m/s. The magnitude of the velocity of A relative to B is:
A. 2 m/s
B. 7 m/s
C. 14 m/s
D. 10 m/s ✓ Correct
Solution: The velocities are perpendicular, so |v_AB| = √(6² + 8²) = 10 m/s.
Q2 — Relative Motion in 2D · easy · numerical
A train moves due east at 30 km/h while a car on a nearby road moves due north at 40 km/h. The speed of the train relative to the car is:
A. 50 km/h ✓ Correct
B. 10 km/h
C. 35 km/h
D. 70 km/h
Solution: Relative speed = √(30² + 40²) = √2500 = 50 km/h.
Q3 — Relative Motion in 2D · easy · numerical
A scooter moves due east at 12 m/s while a bird flies due north at 5 m/s. The speed of the scooter relative to the bird is:
A. 10 m/s
B. 7 m/s
C. 17 m/s
D. 13 m/s ✓ Correct
Solution: Relative speed = √(12² + 5²) = √169 = 13 m/s.
Q4 — Relative Motion in 2D · easy · numerical
Car A travels east at 20 m/s and car B travels west at 10 m/s on the same road. The speed of A relative to B is:
A. 30 m/s ✓ Correct
B. 15 m/s
C. 20 m/s
D. 10 m/s
Solution: For opposite directions the speeds add: v_AB = 20 + 10 = 30 m/s.
Q5 — Relative Motion in 2D · easy · numerical
Train A moving at 25 m/s follows train B moving at 15 m/s in the same direction on parallel tracks. The speed of A relative to B is:
A. 20 m/s
B. 25 m/s
C. 10 m/s ✓ Correct
D. 40 m/s
Solution: For the same direction the speeds subtract: v_AB = 25 − 15 = 10 m/s.
Q6 — Relative Motion in 2D · easy · numerical
Scooter A moves due north at 10 m/s and scooter B moves due east at 10 m/s. The magnitude of the velocity of A relative to B is:
A. 0 m/s
B. 10 m/s
C. 10√2 ≈ 14.1 m/s ✓ Correct
D. 20 m/s
Solution: v_AB = 10ĵ − 10î, so |v_AB| = √(10² + 10²) = 10√2 ≈ 14.1 m/s.
Q7 — Relative Motion in 2D · easy · numerical
Car A moves due north at 10√3 m/s and car B moves due east at 10 m/s. The velocity of A relative to B points in the direction:
A. 30° east of north
B. 45° west of north
C. 60° west of north
D. 30° west of north ✓ Correct
Solution: v_AB = 10√3 ĵ − 10 î; tanα = 10/(10√3) = 1/√3, so it points 30° towards west from north.
Q8 — Relative Motion in 2D · easy · numerical
Two scooters leave a crossing at the same moment, one heading east at 8 m/s and the other north at 6 m/s. The distance between them after 5 s is:
A. 70 m
B. 50 m ✓ Correct
C. 25 m
D. 10 m
Solution: They are 40 m and 30 m from the crossing, so separation = √(40² + 30²) = 50 m.
Q9 — Relative Motion in 2D · easy · numerical
Two trains leave a junction simultaneously along perpendicular tracks at 60 km/h and 80 km/h. The distance between them after 30 minutes is:
A. 50 km ✓ Correct
B. 100 km
C. 70 km
D. 25 km
Solution: In 0.5 h they cover 30 km and 40 km, so separation = √(30² + 40²) = 50 km.
Q10 — Relative Motion in 2D · easy · numerical
Two cars approach a right-angle crossing along perpendicular roads. At an instant, car A is 60 m from the crossing and car B is 80 m from it. Their separation at that instant is:
A. 70 m
B. 140 m
C. 100 m ✓ Correct
D. 20 m
Solution: Separation = √(60² + 80²) = √10000 = 100 m.
Q11 — Relative Motion in 2D · easy · numerical
A person jogs due north at 3 m/s. To the person, a cyclist appears to move due east at 4 m/s. The true speed of the cyclist is:
A. 4 m/s
B. 5 m/s ✓ Correct
C. 1 m/s
D. 7 m/s
Solution: v_cyclist = v_rel + v_person = 4î + 3ĵ, so true speed = √(4² + 3²) = 5 m/s.
Q12 — Relative Motion in 2D · easy · numerical
A car moves due east at 9 m/s and a train moves due north at 12 m/s. The speed of the car relative to the train is:
A. 15 m/s ✓ Correct
B. 3 m/s
C. 7.5 m/s
D. 21 m/s
Solution: Relative speed = √(9² + 12²) = √225 = 15 m/s.
Q13 — Relative Motion in 2D · medium · numerical
Car B is 60 m due north of car A. At the same moment A starts moving north at 8 m/s and B starts moving east at 8 m/s. The minimum separation between them is:
A. 30 m
B. 60√2 ≈ 84.9 m
C. 30√2 ≈ 42.4 m ✓ Correct
D. 60 m
Solution: Relative to B, A moves at 8√2 m/s at 45° to the line joining them, so d_min = 60 sin45° = 60/√2 = 30√2 m.
Q14 — Relative Motion in 2D · medium · numerical
Car A is 100 m from a right-angle crossing, moving towards it at 20 m/s. Car B, on the perpendicular road, is 120 m from the crossing, moving towards it at 30 m/s. Their separation after 3 s is:
A. 70 m
B. 35 m
C. 50 m ✓ Correct
D. 60 m
Solution: After 3 s, A is 100 − 60 = 40 m and B is 120 − 90 = 30 m from the crossing; separation = √(40² + 30²) = 50 m.
Q15 — Relative Motion in 2D · medium · numerical
Train P moves due east at 72 km/h and train Q moves due north at 54 km/h. The speed of P relative to Q, in m/s, is:
A. 12.5 m/s
B. 35 m/s
C. 25 m/s ✓ Correct
D. 5 m/s
Solution: 72 km/h = 20 m/s and 54 km/h = 15 m/s; relative speed = √(20² + 15²) = 25 m/s.
Q16 — Relative Motion in 2D · medium · numerical
Bird A flies due east at 4 m/s and bird B flies due north at 4√3 m/s. The velocity of A relative to B points south of east at an angle of:
A. 30°
B. 45°
C. 15°
D. 60° ✓ Correct
Solution: v_AB = 4î − 4√3ĵ; tanα = 4√3/4 = √3, so α = 60° south of east.
Q17 — Relative Motion in 2D · medium · numerical
Two cars leave a crossing at the same moment along perpendicular highways at 50 km/h and 120 km/h. The distance between them increases at the rate of:
A. 70 km/h
B. 170 km/h
C. 85 km/h
D. 130 km/h ✓ Correct
Solution: Starting from the same point on perpendicular roads, the separation grows at the relative speed √(50² + 120²) = 130 km/h.
Q18 — Relative Motion in 2D · medium · numerical
To a passenger in a train moving due north at 24 m/s, a car on a nearby road appears to move due east at 10 m/s. The true speed of the car is:
A. 24 m/s
B. 34 m/s
C. 14 m/s
D. 26 m/s ✓ Correct
Solution: v_car = v_rel + v_train = 10î + 24ĵ, so true speed = √(10² + 24²) = 26 m/s.
Q19 — Relative Motion in 2D · medium · numerical
Two birds leave the same tree at the same moment; one flies due east at 5 m/s and the other due north at 5 m/s. Their separation after 4 s is:
A. 20 m
B. 20√3 ≈ 34.6 m
C. 40 m
D. 20√2 ≈ 28.3 m ✓ Correct
Solution: Each covers 20 m along perpendicular directions, so separation = √(20² + 20²) = 20√2 ≈ 28.3 m.
Q20 — Relative Motion in 2D · medium · numerical
Two trains approach a right-angle level crossing on perpendicular tracks. Train A is 200 m away moving at 20 m/s; train B is 150 m away moving at 15 m/s. Their separation after 5 s is:
A. 175 m
B. 25 m
C. 125 m ✓ Correct
D. 100 m
Solution: After 5 s, A is 200 − 100 = 100 m and B is 150 − 75 = 75 m from the crossing; separation = √(100² + 75²) = 125 m.