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JEE PHYSICS MCQ: MOTION IN A STRAIGHT LINE

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The position vector of particle changes with time according to the relation r(t)=15t2i^+(4−20t2)j^r(t) = 15{t^2}\hat i + (4 - 20{t^2})\hat j. What is the magnitude of the acceleration (in ms−2m{s^{ - \mathrm{2}}}) at t =1 ?
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Correct option: A
50
Option A (50) is correct because the magnitude of acceleration is calculated by finding the second derivative of the position vector and then taking the magnitude at t=1.

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