AIIMS2018Physics-Kinematics

AIIMS 2018 Physics Equations of Motion MCQ Question

Type: MCQ-conceptual-Medium-Class 11

A particle experiences constant acceleration for 20 sec after starting from rest. If it travels a distance s₁ in the first 10 sec and distance s₂ in the next 10 sec, then

A

s₂ = s₁

B

s₂ = 2s₁

C

s₂ = 3s₁

D

s₂ = 4s₁

Correct Answer

Option C

Detailed Explanation

In uniformly accelerated motion, the distance covered in each interval increases due to the constant acceleration. For the first 10 seconds, the distance s1s_1 can be calculated using the formula s1=12at2s_1 = \frac{1}{2} a t^2, where t=10t = 10 seconds. In the next 10 seconds, the particle travels an additional distance s2=12a(202)12a(102)=12a(400100)=12a300=3s1s_2 = \frac{1}{2} a (20^2) - \frac{1}{2} a (10^2) = \frac{1}{2} a (400 - 100) = \frac{1}{2} a \cdot 300 = 3s_1, thus s2=3s1s_2 = 3s_1. Options A and B are incorrect because they do not account for the increasing distance due to acceleration, and option D overestimates the distance by assuming a quadratic increase without proper calculation.

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