Basics of 2D — JEE Main Physics MCQs with Solutions
Free JEE Main 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
The position of a particle is given by r = (8t î + 15t ĵ) m, where t is in seconds. What is the speed of the particle?
A. 23 m/s
B. 11.5 m/s
C. 7 m/s
D. 17 m/s ✓ Correct
Solution: v = dr/dt = (8 î + 15 ĵ) m/s, so speed = √(8² + 15²) = √289 = 17 m/s.
Q2 — Basics of 2D · easy · numerical
A particle moves so that its position is r = (7t î + 12t² ĵ) m. Find its speed at t = 1 s.
A. 19 m/s
B. 31 m/s
C. 25 m/s ✓ Correct
D. 13.9 m/s
Solution: v = (7 î + 24t ĵ); at t = 1 s, v = (7, 24), so speed = √(49 + 576) = 25 m/s.
Q3 — Basics of 2D · easy · numerical
The position of a particle is r = (4t³ î + 12t² ĵ) m. What is the magnitude of its acceleration at t = 1 s?
A. 48 m/s²
B. 12√2 ≈ 17.0 m/s²
C. 24√2 ≈ 33.9 m/s² ✓ Correct
D. 24 m/s²
Solution: a = d²r/dt² = (24t î + 24 ĵ); at t = 1 s, a = (24, 24), so |a| = 24√2 ≈ 33.9 m/s².
Q4 — Basics of 2D · easy · numerical
A car moving at 15 m/s due east is later found moving at 8 m/s due north, the change taking 2 s. Find the magnitude of its average acceleration.
A. 8.5 m/s² ✓ Correct
B. 11.5 m/s²
C. 3.5 m/s²
D. 17 m/s²
Solution: Δv = (0, 8) − (15, 0), so |Δv| = √(15² + 8²) = 17 m/s; average acceleration = 17/2 = 8.5 m/s².
Q5 — Basics of 2D · easy · numerical
A particle moves with x = 12t and y = 35t² (both in metres). Find the magnitude of its average velocity between t = 0 and t = 1 s.
A. 47 m/s
B. 23.5 m/s
C. 37 m/s ✓ Correct
D. 35 m/s
Solution: Displacement in 1 s is (12, 35) m, so |Δr| = √(144 + 1225) = 37 m and average velocity = 37/1 = 37 m/s.
Q6 — Basics of 2D · easy · numerical
A particle starts with velocity u = 20 î m/s and has constant acceleration a = 7 ĵ m/s². Find its speed at t = 3 s.
A. 29 m/s ✓ Correct
B. 21 m/s
C. 41 m/s
D. 27 m/s
Solution: v = u + at = (20, 0 + 7×3) = (20, 21) m/s, so speed = √(400 + 441) = √841 = 29 m/s.
Q7 — Basics of 2D · medium · numerical
The position of a particle is r = ((t³ − 5t) î + 6t² ĵ) m. Find its speed at t = 2 s.
A. 12.5 m/s
B. 25 m/s ✓ Correct
C. 31 m/s
D. 17 m/s
Solution: v = ((3t² − 5) î + 12t ĵ); at t = 2 s, v = (12 − 5, 24) = (7, 24) m/s, so speed = √(49 + 576) = 25 m/s.
Q8 — Basics of 2D · medium · numerical
The velocity of a particle varies as v = (6t² î + 35t ĵ) m/s. Find the magnitude of its acceleration at t = 1 s.
A. 47 m/s²
B. 35 m/s²
C. 41 m/s²
D. 37 m/s² ✓ Correct
Solution: a = dv/dt = (12t î + 35 ĵ); at t = 1 s, a = (12, 35), so |a| = √(144 + 1225) = 37 m/s².
Q9 — Basics of 2D · medium · numerical
A particle moves with x = 6t and y = t² (metres, seconds). At what time does its velocity make an angle of 45° with the x-axis?
A. 1.5 s
B. 12 s
C. 6 s
D. 3 s ✓ Correct
Solution: vₓ = 6 and v_y = 2t; for 45°, v_y = vₓ, so 2t = 6, giving t = 3 s.
Q10 — Basics of 2D · medium · numerical
A particle moves with x = t² and y = 12t (metres, seconds). Find its x-coordinate at the instant its x-velocity equals its y-velocity.
A. 72 m
B. 12 m
C. 6 m
D. 36 m ✓ Correct
Solution: vₓ = 2t and v_y = 12; they are equal when 2t = 12, i.e. t = 6 s, and then x = 6² = 36 m.
Q11 — Basics of 2D · medium · numerical
A particle follows r = (20t î + (20t − 5t²) ĵ) m. Find the angle between its velocity and its acceleration at t = 2 s.
A. 60°
B. 90° ✓ Correct
C. 0°
D. 45°
Solution: At t = 2 s, v = (20, 20 − 10×2) = (20, 0) (horizontal) while a = (0, −10) (vertical), so the angle between them is 90°.
Q12 — Basics of 2D · medium · numerical
A particle starts with velocity 8 m/s along the x-axis and has a constant acceleration of 15 m/s² along the y-axis. Find the magnitude of its displacement after 2 s.
A. 17 m
B. 30 m
C. 34 m ✓ Correct
D. 46 m
Solution: s = ut + ½at² gives sₓ = 8×2 = 16 m and s_y = ½×15×4 = 30 m, so |s| = √(256 + 900) = 34 m.
Q13 — Basics of 2D · medium · numerical
A particle has initial velocity u = (21 î + 8 ĵ) m/s and constant acceleration a = 4 ĵ m/s². Find its speed at t = 3 s.
A. 29 m/s ✓ Correct
B. 33 m/s
C. 23 m/s
D. 41 m/s
Solution: v = u + at = (21, 8 + 4×3) = (21, 20) m/s, so speed = √(441 + 400) = √841 = 29 m/s.
Q14 — Basics of 2D · medium · numerical
A particle moves with x = 2t² and y = 15t (metres, seconds). Find the magnitude of its average velocity between t = 1 s and t = 3 s.
A. 15 m/s
B. 34 m/s
C. 17 m/s ✓ Correct
D. 23 m/s
Solution: Δx = 2(9 − 1) = 16 m and Δy = 15(3 − 1) = 30 m, so |Δr| = √(256 + 900) = 34 m and average velocity = 34/2 = 17 m/s.
Q15 — Basics of 2D · medium · numerical
A particle moves with x = 2t³ and y = 24t (metres, seconds). At what time does its velocity make 45° with the x-axis?
A. √2 ≈ 1.41 s
B. 4 s
C. 2 s ✓ Correct
D. 1 s
Solution: vₓ = 6t² and v_y = 24; for 45° they must be equal, so 6t² = 24 gives t = 2 s.
Q16 — Basics of 2D · medium · numerical
A particle moves with x = 10t and y = 5t² (metres, seconds). At what time does its velocity make 60° with the x-axis?
A. √3 ≈ 1.73 s ✓ Correct
B. 2√3 ≈ 3.46 s
C. 1 s
D. 3 s
Solution: vₓ = 10 and v_y = 10t; tan 60° = √3 = 10t/10, so t = √3 ≈ 1.73 s.
Q17 — Basics of 2D · medium · numerical
A particle moving at a constant speed of 20 m/s changes its direction of motion by 60° in 2 s. Find the magnitude of its average acceleration.
A. 5 m/s²
B. 10 m/s² ✓ Correct
C. 20 m/s²
D. 0 m/s²
Solution: |Δv| = 2v sin(θ/2) = 2×20×sin 30° = 20 m/s, so average acceleration = 20/2 = 10 m/s².
Q18 — Basics of 2D · medium · numerical
A particle starts from the origin with velocity 20 m/s along the x-axis and constant acceleration 10 m/s² along the y-axis. Find its distance from the origin at the instant its y-velocity equals its x-velocity.
A. 20√5 ≈ 44.7 m ✓ Correct
B. 20 m
C. 40 m
D. 60 m
Solution: v_y = 10t = 20 gives t = 2 s; then x = 20×2 = 40 m and y = ½×10×4 = 20 m, so distance = √(1600 + 400) = 20√5 ≈ 44.7 m.
Q19 — Basics of 2D · medium · numerical
The velocity of a particle is v = ((3t² − 12) î + 8t ĵ) m/s. Find its speed at the moment its velocity points along the y-axis.
A. 16 m/s ✓ Correct
B. 12 m/s
C. 20 m/s
D. 8 m/s
Solution: The velocity is along the y-axis when vₓ = 3t² − 12 = 0, i.e. t = 2 s; then v_y = 8×2 = 16 m/s, which is the speed.
Q20 — Basics of 2D · medium · numerical
A particle moves with r = ((t³ − 6t²) î + (5t² + 15t) ĵ) m. Find its speed at the instant its x-acceleration becomes zero.
A. 35 m/s
B. 23 m/s
C. 37 m/s ✓ Correct
D. 47 m/s
Solution: aₓ = 6t − 12 = 0 at t = 2 s; then vₓ = 3(4) − 12(2) = −12 m/s and v_y = 10(2) + 15 = 35 m/s, so speed = √(144 + 1225) = 37 m/s.