Magnitude of acceleration from position functions

Practice question

Question

A particle’s position is given by x\=5t−t2 and y\=2t2 (in meters and seconds). What is the magnitude of its acceleration?

Options

Choose one · Correct answer highlighted

2 m/s² Correct 1.60 m/s² 2.40 m/s² 4 m/s²

Explanation

Velocity: vx=dxdt=5−2t,vy=dydt=4t. 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 2 m/s² as the result, so option A is correct.

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