Equations of Motion — Homi Bhabha Class 9 Physics MCQs with Solutions
Free Homi Bhabha Class 9 Physics Equations of Motion MCQs with step-by-step solutions (10 questions). Part of Motion & Kinematics. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Equations of Motion · medium · numerical
A body starts from rest with a uniform acceleration of 2 m/s². The distance covered in 5 s is:
A. 25 m ✓ Correct
B. 10 m
C. 50 m
D. 20 m
Solution: $s = ut + \tfrac12 a t^2 = 0 + \tfrac12 (2)(25) = 25$ m.
Q2 — Equations of Motion · hard · numerical
A body moving at 10 m/s accelerates uniformly at 2 m/s². The distance it covers in the next 3 s is:
A. 9 m
B. 30 m
C. 48 m
D. 39 m ✓ Correct
Solution: $s = ut + \tfrac12 a t^2 = (10)(3) + \tfrac12 (2)(9) = 30 + 9 = 39$ m.
Q3 — Equations of Motion · medium · numerical
A body starts from rest and accelerates at 5 m/s² over a distance of 10 m. Its final velocity is:
A. 10 m/s ✓ Correct
B. 50 m/s
C. 100 m/s
D. 5 m/s
Solution: $v^2 = u^2 + 2as = 0 + 2(5)(10) = 100 \Rightarrow v = 10$ m/s.
Q4 — Equations of Motion · medium · theory
The area enclosed under a velocity–time graph gives the:
A. force on the body
B. displacement of the body ✓ Correct
C. acceleration of the body
D. speed of the body
Solution: Area under a v–t graph equals velocity × time = displacement; the slope (not area) gives acceleration.
Q5 — Equations of Motion · hard · numerical
A car moving at 20 m/s applies brakes producing a retardation of 5 m/s². The distance it travels before stopping is:
A. 40 m ✓ Correct
B. 80 m
C. 4 m
D. 20 m
Solution: $v^2 = u^2 - 2as \Rightarrow 0 = 400 - 2(5)s \Rightarrow s = \dfrac{400}{10} = 40$ m.
Q6 — Equations of Motion · hard · numerical
The stopping distance of a vehicle (for the same braking force) is proportional to the square of its speed. If the speed is doubled, the stopping distance becomes:
A. 4 times ✓ Correct
B. unchanged
C. 8 times
D. 2 times
Solution: Since $s \propto v^2$, doubling $v$ multiplies the stopping distance by $2^2 = 4$.
Q7 — Equations of Motion · easy · theory
For a body moving with uniform acceleration, which of these is the first equation of motion?
A. s = u t + 1/2 a t squared
B. v = u + a t ✓ Correct
C. v squared = u squared + 2 a s
D. s = v t
Solution: The first equation of motion links final velocity, initial velocity, acceleration and time: v = u + a t.
Q8 — Equations of Motion · medium · numerical
A body starts from rest and moves with a uniform acceleration of 2 m/s squared. What is its velocity after 8 s?
A. 10 m/s
B. 4 m/s
C. 32 m/s
D. 16 m/s ✓ Correct
Solution: Using v = u + a t with u = 0: v = 0 + 2 x 8 = 16 m/s.
Q9 — Equations of Motion · hard · numerical
A car moving at 20 m/s is brought to rest over a distance of 50 m by uniform braking. What is the magnitude of its retardation?
A. 10 m/s squared
B. 2 m/s squared
C. 8 m/s squared
D. 4 m/s squared ✓ Correct
Solution: Using v squared = u squared + 2 a s: 0 = 20 squared + 2 x a x 50, so 0 = 400 + 100 a, giving a = minus 4 m/s squared. The magnitude of the retardation is 4 m/s squared.
Q10 — Equations of Motion · hard · numerical
A stone is dropped from rest from a height. Taking g = 10 m/s squared, how far does it fall during the first 3 seconds?
A. 45 m ✓ Correct
B. 90 m
C. 15 m
D. 30 m
Solution: Using s = u t + 1/2 g t squared with u = 0: s = 1/2 x 10 x 3 squared = 1/2 x 10 x 9 = 45 m.