Basics of 2D — IISER Physics MCQs with Solutions
Free IISER 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
An insect crawls on a table from the point (2 m, 3 m) to the point (11 m, 15 m). What is the magnitude of its displacement?
A. 13 m
B. 21 m
C. 15 m ✓ Correct
D. 9 m
Solution: Δr = (11 − 2) î + (15 − 3) ĵ = 9 î + 12 ĵ (m), so |Δr| = √(9² + 12²) = √225 = 15 m.
Q2 — Basics of 2D · easy · numerical
A trolley moves from the point (4 m, −3 m) to the point (19 m, 17 m) in 5 s. What is the magnitude of its average velocity?
A. 7 m/s
B. 3 m/s
C. 5 m/s ✓ Correct
D. 4 m/s
Solution: Displacement = (15, 20) m with magnitude √(15² + 20²) = 25 m; average velocity = 25 m ÷ 5 s = 5 m/s.
Q3 — Basics of 2D · easy · numerical
A drone flies with velocity components 16 m/s towards the east and 30 m/s towards the north. What is its speed?
A. 23 m/s
B. 46 m/s
C. 30 m/s
D. 34 m/s ✓ Correct
Solution: The components are perpendicular, so speed = √(16² + 30²) = √1156 = 34 m/s.
Q4 — Basics of 2D · easy · numerical
A ship sails 21 km due east and then 28 km due north. What is the magnitude of its net displacement?
A. 28 km
B. 49 km
C. 35 km ✓ Correct
D. 25 km
Solution: The two legs are perpendicular, so |Δr| = √(21² + 28²) = √1225 = 35 km.
Q5 — Basics of 2D · easy · numerical
The position of a probe is given by r(t) = 10t î + 4t² ĵ (metres, seconds). What is its speed at t = 3 s?
A. 26 m/s ✓ Correct
B. 34 m/s
C. 14 m/s
D. 22 m/s
Solution: v = dr/dt = 10 î + 8t ĵ; at t = 3 s, v = (10, 24) m/s, so speed = √(100 + 576) = 26 m/s.
Q6 — Basics of 2D · easy · numerical
The position of a sphere is given by r(t) = 4.5t² î + 6t² ĵ (metres, seconds). What is the magnitude of its acceleration?
A. 10.5 m/s²
B. 21 m/s²
C. 7.5 m/s²
D. 15 m/s² ✓ Correct
Solution: a = d²r/dt² = 9 î + 12 ĵ (m/s²), so |a| = √(81 + 144) = √225 = 15 m/s².
Q7 — Basics of 2D · easy · numerical
A ball is rolled off a ledge horizontally at 7.5 m/s and falls with a constant downward acceleration of 10 m/s². What is the magnitude of its displacement 2 s after leaving the ledge?
A. 35 m
B. 25 m ✓ Correct
C. 20 m
D. 15 m
Solution: Horizontal: x = 7.5 × 2 = 15 m; vertical: y = ½ × 10 × 2² = 20 m. Displacement = √(15² + 20²) = 25 m.
Q8 — Basics of 2D · easy · numerical
A robot starts from rest and moves with constant acceleration a = 2.1 î + 2.8 ĵ (m/s²). What is its speed after 10 s?
A. 24.5 m/s
B. 35 m/s ✓ Correct
C. 49 m/s
D. 28 m/s
Solution: v = at = (21, 28) m/s after 10 s, so speed = √(441 + 784) = √1225 = 35 m/s.
Q9 — Basics of 2D · medium · numerical
A ball moving at 20 m/s towards the north is deflected so that 4 s later it is moving at 20 m/s towards the east. What is the magnitude of its average acceleration over this interval?
A. 5 m/s²
B. 5√2 ≈ 7.1 m/s² ✓ Correct
C. 10 m/s²
D. 0 m/s²
Solution: The initial and final velocities are perpendicular, so |Δv| = √(20² + 20²) = 20√2 m/s; average acceleration = 20√2 ÷ 4 = 5√2 ≈ 7.1 m/s².
Q10 — Basics of 2D · medium · numerical
A robot moves with velocity v = 15 î + 15√3 ĵ (m/s). What angle does its velocity make with the x-axis?
A. 53°
B. 30°
C. 60° ✓ Correct
D. 45°
Solution: tan θ = (15√3)/15 = √3, so θ = 60°.
Q11 — Basics of 2D · medium · numerical
The velocity of a drone is v(t) = 20 î + 5t ĵ (m/s, seconds). At what time is its speed equal to 25 m/s?
A. 4 s
B. 3 s ✓ Correct
C. 2 s
D. 5 s
Solution: Speed² = 20² + (5t)² = 625 gives 25t² = 225, so t = 3 s (the second component is then 15 m/s).
Q12 — Basics of 2D · medium · numerical
A trolley starts at the origin and its position is r(t) = 2t² î + 6t ĵ (metres, seconds). How far from the origin is it at t = 3 s?
A. 12√2 ≈ 17.0 m
B. 18√2 ≈ 25.5 m ✓ Correct
C. 36 m
D. 18 m
Solution: At t = 3 s, x = 2 × 9 = 18 m and y = 6 × 3 = 18 m, so distance = √(18² + 18²) = 18√2 ≈ 25.5 m.
Q13 — Basics of 2D · medium · numerical
The position of a sphere is given by r(t) = 21t î + 7t² ĵ (metres, seconds). What is its speed at t = 2 s?
A. 35 m/s ✓ Correct
B. 28 m/s
C. 25 m/s
D. 49 m/s
Solution: v = 21 î + 14t ĵ; at t = 2 s, v = (21, 28) m/s, so speed = √(441 + 784) = 35 m/s.
Q14 — Basics of 2D · medium · numerical
The position of a robot is r(t) = t² î + 4t ĵ (metres, seconds). What is the magnitude of its average velocity over the first 4 s?
A. 4√2 ≈ 5.7 m/s ✓ Correct
B. 8 m/s
C. 4 m/s
D. 16√2 ≈ 22.6 m/s
Solution: Displacement from t = 0 to t = 4 s is (16, 16) m with magnitude 16√2 m; average velocity = 16√2 ÷ 4 = 4√2 ≈ 5.7 m/s.
Q15 — Basics of 2D · medium · numerical
A ship is moving at 15 m/s towards the east; 5 s later it is moving at 20 m/s towards the north. What is the magnitude of its average acceleration?
A. 1 m/s²
B. 7 m/s²
C. 4 m/s²
D. 5 m/s² ✓ Correct
Solution: Δv = (0, 20) − (15, 0) = (−15, 20) m/s with magnitude √(225 + 400) = 25 m/s; average acceleration = 25 ÷ 5 = 5 m/s².
Q16 — Basics of 2D · medium · numerical
A probe moves with velocity components vₓ = 16 m/s and v_y = 12 m/s. What angle does its velocity make with the x-axis?
A. 53°
B. 30°
C. 37° ✓ Correct
D. 45°
Solution: tan θ = 12/16 = 0.75, so θ = tan⁻¹(0.75) ≈ 37°.
Q17 — Basics of 2D · medium · numerical
A trolley moving east at a constant 16 m/s acquires a constant northward acceleration of 5 m/s². What is its speed 6 s later?
A. 23 m/s
B. 30 m/s
C. 34 m/s ✓ Correct
D. 46 m/s
Solution: The northward component grows to v_y = 5 × 6 = 30 m/s while vₓ stays 16 m/s, so speed = √(16² + 30²) = 34 m/s.
Q18 — Basics of 2D · medium · numerical
A robot has initial velocity u = 2 î + 5 ĵ (m/s) and constant acceleration a = 4 î + 2 ĵ (m/s²). What is the magnitude of its displacement after 3 s?
A. 24√2 ≈ 33.9 m ✓ Correct
B. 12√2 ≈ 17.0 m
C. 48 m
D. 24 m
Solution: s = ut + ½at²: x = 2(3) + ½(4)(9) = 24 m and y = 5(3) + ½(2)(9) = 24 m, so |s| = 24√2 ≈ 33.9 m.
Q19 — Basics of 2D · medium · numerical
A drone has a constant eastward velocity component of 12 m/s, while its northward component grows as v_y = 3t (m/s, seconds). At what time does its velocity make an angle of 45° with the east direction?
A. 2 s
B. 3 s
C. 6 s
D. 4 s ✓ Correct
Solution: At 45° the two components are equal: 3t = 12, so t = 4 s.
Q20 — Basics of 2D · medium · numerical
The position of a probe is r(t) = 5t î + (20t − 5t²) ĵ (metres, seconds). How far from the origin is it at t = 2 s?
A. 20 m
B. 30 m
C. 10√5 ≈ 22.4 m ✓ Correct
D. 10√3 ≈ 17.3 m
Solution: At t = 2 s, x = 10 m and y = 40 − 20 = 20 m, so distance = √(100 + 400) = 10√5 ≈ 22.4 m.