Q: A body starts from rest with a uniform acceleration of 2 m/s². The distance covered in 5 s is:

25 m. s = ut + \tfrac12 a t^2 = 0 + \tfrac12 (2)(25) = 25 m.

Q: A body moving at 10 m/s accelerates uniformly at 2 m/s². The distance it covers in the next 3 s is:

39 m. s = ut + \tfrac12 a t^2 = (10)(3) + \tfrac12 (2)(9) = 30 + 9 = 39 m.

Question 1

A body starts from rest with a uniform acceleration of 2 m/s². The distance covered in 5 s is:

Accepted Answer

25 m. s = ut + 	frac12 a t^2 = 0 + 	frac12 (2)(25) = 25 m.

Question 2

A body moving at 10 m/s accelerates uniformly at 2 m/s². The distance it covers in the next 3 s is:

Accepted Answer

39 m. s = ut + 	frac12 a t^2 = (10)(3) + 	frac12 (2)(9) = 30 + 9 = 39 m.

Question 3

A body starts from rest and accelerates at 5 m/s² over a distance of 10 m. Its final velocity is:

Accepted Answer

10 m/s. v^2 = u^2 + 2as = 0 + 2(5)(10) = 100 \Rightarrow v = 10 m/s.

Question 4

The area enclosed under a velocity–time graph gives the:

Accepted Answer

displacement of the body. Area under a v–t graph equals velocity × time = displacement; the slope (not area) gives acceleration.

Question 5

A car moving at 20 m/s applies brakes producing a retardation of 5 m/s². The distance it travels before stopping is:

Accepted Answer

40 m. v^2 = u^2 - 2as \Rightarrow 0 = 400 - 2(5)s \Rightarrow s = \dfrac{400}{10} = 40 m.

Question 6

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:

Accepted Answer

4 times. Since s \propto v^2, doubling v multiplies the stopping distance by 2^2 = 4.

Equations of Motion — Homi Bhabha Exam Physics MCQs with Solutions

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