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Practice question

MCQ

Question

A particle’s position is given by x\=2t2+3t and y\=t2−2t (in meters and seconds). What is the magnitude of its velocity at t\=2s?

Options

Choose one · Correct answer highlighted

Explanation

Velocity: vx=dxdt=4t+3,vy=dydt=2t−2. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 10 m/s as the result, so option A is correct.

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