Basics of 2D — NEET Physics MCQs with Solutions
Free NEET Physics Basics of 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 — Basics of 2D · easy · numerical
A particle moves from the point (1 m, 2 m) to the point (4 m, 6 m). The magnitude of its displacement is:
A. 3 m
B. 7 m
C. 4 m
D. 5 m ✓ Correct
Solution: Δx = 3 m and Δy = 4 m, so Δr = √(3² + 4²) = 5 m.
Q2 — Basics of 2D · easy · numerical
A bird flies from the point (5 m, -2 m) to the point (11 m, 6 m). The magnitude of its displacement is:
A. 8 m
B. 14 m
C. 6 m
D. 10 m ✓ Correct
Solution: Δx = 6 m and Δy = 8 m, so Δr = √(6² + 8²) = 10 m.
Q3 — Basics of 2D · easy · numerical
A person walks 30 m east and then 40 m north, taking a total of 10 s. The magnitude of the average velocity is:
A. 7 m/s
B. 4 m/s
C. 3 m/s
D. 5 m/s ✓ Correct
Solution: Displacement = √(30² + 40²) = 50 m; average velocity = 50/10 = 5 m/s.
Q4 — Basics of 2D · easy · numerical
At one instant a football has velocity components 9 m/s horizontally and 12 m/s vertically. Its speed at that instant is:
A. 3 m/s
B. 21 m/s
C. 12 m/s
D. 15 m/s ✓ Correct
Solution: Speed = √(9² + 12²) = √225 = 15 m/s.
Q5 — Basics of 2D · easy · numerical
The position of a particle is r = (4t î + 3t ĵ) m, where t is in seconds. The speed of the particle is:
A. 7 m/s
B. 1 m/s
C. 25 m/s
D. 5 m/s ✓ Correct
Solution: vₓ = 4 m/s and v_y = 3 m/s, so speed = √(4² + 3²) = 5 m/s.
Q6 — Basics of 2D · easy · numerical
The position of a particle is r = (5t î + 2t² ĵ) m, where t is in seconds. The magnitude of its acceleration is:
A. 9 m/s²
B. 4 m/s² ✓ Correct
C. 5 m/s²
D. 2 m/s²
Solution: aₓ = 0 and a_y = d²(2t²)/dt² = 4 m/s², so |a| = 4 m/s².
Q7 — Basics of 2D · easy · numerical
The velocity of a scooter changes from 10î m/s to 10ĵ m/s in 5 s. The magnitude of the average acceleration is:
A. 2√2 ≈ 2.8 m/s² ✓ Correct
B. 2 m/s²
C. 4 m/s²
D. 0 m/s²
Solution: Δv = 10ĵ − 10î, |Δv| = 10√2 m/s; average acceleration = 10√2/5 = 2√2 ≈ 2.8 m/s².
Q8 — Basics of 2D · easy · numerical
A particle starts from rest and moves with constant acceleration a = (3î + 4ĵ) m/s². Its speed after 2 s is:
A. 10 m/s ✓ Correct
B. 14 m/s
C. 7 m/s
D. 5 m/s
Solution: |a| = √(3² + 4²) = 5 m/s²; from rest, v = |a|t = 5 × 2 = 10 m/s.
Q9 — Basics of 2D · easy · numerical
The velocity of a particle is v = (6î + 6√3 ĵ) m/s. The angle its velocity makes with the x-axis is:
A. 60° ✓ Correct
B. 30°
C. 45°
D. 90°
Solution: tanθ = v_y/vₓ = 6√3/6 = √3, so θ = 60°.
Q10 — Basics of 2D · easy · numerical
A particle starts from the origin with x = 6t and y = 8t (both in metres, t in seconds). Its distance from the origin at t = 2 s is:
A. 14 m
B. 20 m ✓ Correct
C. 28 m
D. 10 m
Solution: At t = 2 s, x = 12 m and y = 16 m, so distance = √(12² + 16²) = 20 m.
Q11 — Basics of 2D · easy · numerical
The velocity of a cyclist makes 60° with the x-axis. If the x-component of the velocity is 10 m/s, the speed of the cyclist is:
A. 10√3 ≈ 17.3 m/s
B. 20 m/s ✓ Correct
C. 5 m/s
D. 40 m/s
Solution: v = vₓ/cos60° = 10/0.5 = 20 m/s.
Q12 — Basics of 2D · easy · numerical
A car starts at the origin and 2 s later is at a point 5 m east and 12 m north of it. The magnitude of its average velocity is:
A. 3.25 m/s
B. 13 m/s
C. 6.5 m/s ✓ Correct
D. 8.5 m/s
Solution: Displacement = √(5² + 12²) = 13 m; average velocity = 13/2 = 6.5 m/s.
Q13 — Basics of 2D · medium · numerical
A particle starts from the origin with velocity 4î m/s and constant acceleration 3ĵ m/s². The magnitude of its displacement after 2 s is:
A. 8 m
B. 6 m
C. 10 m ✓ Correct
D. 14 m
Solution: x = ut = 4 × 2 = 8 m and y = ½at² = ½ × 3 × 4 = 6 m, so s = √(8² + 6²) = 10 m.
Q14 — Basics of 2D · medium · numerical
A particle has initial velocity 5î m/s and constant acceleration 4ĵ m/s². Its speed after 3 s is:
A. 12 m/s
B. 7 m/s
C. 17 m/s
D. 13 m/s ✓ Correct
Solution: vₓ = 5 m/s and v_y = 4 × 3 = 12 m/s, so speed = √(5² + 12²) = 13 m/s.
Q15 — Basics of 2D · medium · numerical
The position of a particle is r = (3t² î + 4t² ĵ) m, where t is in seconds. The magnitude of its acceleration is:
A. 5 m/s²
B. 14 m/s²
C. 7 m/s²
D. 10 m/s² ✓ Correct
Solution: aₓ = 6 m/s² and a_y = 8 m/s² (twice each t² coefficient), so |a| = √(6² + 8²) = 10 m/s².
Q16 — Basics of 2D · medium · numerical
A car moving east at 30 m/s takes a turn and 10 s later is moving north at 40 m/s. The magnitude of its average acceleration is:
A. 7 m/s²
B. 5 m/s² ✓ Correct
C. 10 m/s²
D. 1 m/s²
Solution: Δv = 40ĵ − 30î, so |Δv| = √(30² + 40²) = 50 m/s; average acceleration = 50/10 = 5 m/s².
Q17 — Basics of 2D · medium · numerical
The velocity of a ball makes 30° with the horizontal, and its vertical component is 12 m/s. The speed of the ball is:
A. 48 m/s
B. 12√3 ≈ 20.8 m/s
C. 6 m/s
D. 24 m/s ✓ Correct
Solution: v = v_y/sin30° = 12/0.5 = 24 m/s.
Q18 — Basics of 2D · medium · numerical
A particle starts from the origin with velocity 6î m/s and constant acceleration 2ĵ m/s². At what time does its displacement make 45° with the x-axis?
A. 6 s ✓ Correct
B. 12 s
C. 3 s
D. 2 s
Solution: x = 6t and y = ½ × 2 × t² = t²; for 45°, y = x, so t² = 6t giving t = 6 s.
Q19 — Basics of 2D · medium · numerical
A particle moves with a constant x-velocity of 8 m/s while its y-coordinate is y = t² (metres, t in seconds). Its speed at t = 3 s is:
A. 14 m/s
B. 8 m/s
C. 12 m/s
D. 10 m/s ✓ Correct
Solution: v_y = dy/dt = 2t = 6 m/s at t = 3 s, so speed = √(8² + 6²) = 10 m/s.
Q20 — Basics of 2D · medium · numerical
The velocity of a football changes from (6î + 8ĵ) m/s to (6î − 8ĵ) m/s in 4 s. The magnitude of the average acceleration is:
A. 5 m/s²
B. 4 m/s² ✓ Correct
C. 0 m/s²
D. 8 m/s²
Solution: Δv = −16ĵ m/s, so |Δv| = 16 m/s; average acceleration = 16/4 = 4 m/s².