JEE Main 2019 — Physics solution

JEE Main2019Physics

The position vector of a particle changes with time according to the relation $r (t) = 15 t^{2} \hat{i} + (4 - 20 t^{2}) \hat{j}$ . What is the magnitude of the acceleration at t = 1?

A
50
B
100
C
40
D
25

Solution

$r = 15 t^{2} \hat{i} + 4 \hat{j} - 20 t^{2} \hat{j} \frac{d r}{d t} = 30 t \hat{i} - 40 t \hat{j} \frac{d ^{2} r}{d t ^{2}} = 30 \hat{i} - 40 \hat{j} ∴ \frac{d ^{2} r}{d t ^{2}} = 50 m / s^{2}$

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