Motion and kinematics MCQs
MDCAT · Physics · 6 questions
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Revision summary
PMDC MDCAT syllabus: Force and motion
Kinematics describes motion without asking what causes it. Displacement and velocity are vectors; distance and speed are scalars. On a velocity–time graph, the slope is acceleration and the area under the line is displacement. For uniform acceleration, use the three equations: v = u + at, s = ut + ½at², and v² = u² + 2as; pick the one that contains the three quantities you know and the one you need. Freely falling bodies have a = g (about 9.8 m/s², often rounded to 10 in MCQs), and air resistance is ignored unless stated. A projectile moves with constant horizontal velocity and constant downward acceleration; its range R = v² sin 2θ / g is greatest at 45°, and complementary angles (30° and 60°) give the same range. Most MDCAT numericals here take under a minute if you write u, v, a, s and t first.
Questions with answers
Written by Parcha from the syllabus. Samples until a subject reviewer checks them; report anything wrong from the practice screen.
Q1 · Parcha original · Sample
The slope of a velocity–time graph gives:
- A. Displacement
- B. Acceleration
- C. Distance
- D. Speed
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Answer: B. Acceleration
Slope = change in velocity ÷ change in time = acceleration. The area under a velocity–time graph gives displacement.
Open this question on its own pageQ2 · Parcha original · Sample
The area under a velocity–time graph represents:
- A. Acceleration
- B. Force
- C. Displacement
- D. Momentum
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Answer: C. Displacement
Velocity × time = displacement, so the area between the graph and the time axis is the displacement (area below the axis counts as negative).
Q3 · Parcha original · Sample
A car starts from rest with a uniform acceleration of 2 m/s². Its velocity after 5 s is:
- A. 10 m/s
- B. 2.5 m/s
- C. 7 m/s
- D. 25 m/s
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Answer: A. 10 m/s
v = u + at = 0 + 2 × 5 = 10 m/s.
Q4 · Parcha original · Sample
The same car (from rest, 2 m/s²) covers what distance in those 5 s?
- A. 10 m
- B. 50 m
- C. 20 m
- D. 25 m
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Answer: D. 25 m
s = ut + ½at² = 0 + ½ × 2 × 5² = 25 m. Check: average velocity (0 + 10)/2 = 5 m/s, times 5 s = 25 m.
Q5 · Parcha original · Sample
Ignoring air resistance, a projectile has its maximum horizontal range when launched at:
- A. 30°
- B. 60°
- C. 45°
- D. 90°
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Answer: C. 45°
Range R = v² sin 2θ / g, which is greatest when sin 2θ = 1, i.e. 2θ = 90°, θ = 45°. Angles of 30° and 60° give equal (smaller) ranges.
Q6 · Parcha original · Sample
A stone is dropped from rest. Taking g = 10 m/s², its speed after 2 s is:
- A. 5 m/s
- B. 20 m/s
- C. 10 m/s
- D. 40 m/s
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Answer: B. 20 m/s
v = u + gt = 0 + 10 × 2 = 20 m/s (air resistance ignored).