NEETPhysics-Kinematics

NEET Physics Variable Speed MCQ Question

Type: MCQ-numerical-Medium

A particle is moving with speed V=bxV = b\sqrt{x} along positive x-axis. Calculate the speed of the particle at time t=τt = \tau (assume that the particle is at origin at t=0t = 0).

A

b2τb^2 \tau

B

b2τ2\frac{b^2 \tau}{2}

C

b2τ2\frac{b^2 \tau}{\sqrt{2}}

D

b2τ4\frac{b^2 \tau}{4}

Correct Answer

Option A

Detailed Explanation

The speed V=bxV = b\sqrt{x} implies x=V2b2x = \frac{V^2}{b^2}. At t=τt = \tau, the speed is bV2b2=b2τb\sqrt{\frac{V^2}{b^2}} = b^2 \tau.

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