JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Kinematics

At time

t = 0

a particle starts travelling from a height

7 \overset{z}{^} cm

in a plane keeping z coordinate constant. At any instant of time it's position along the

x

and

y

directions are defined as

3 t

and

5 t^{3}

respectively. At

t = 1 s

acceleration of the particle will be

(A) $- 30 y$
(B) $30 y$ ✓ Correct
(C) $3 x + 15 y$
(D) $3 x + 15 y + 7 \overset{z}{^}$

Answer: (B)

Solution

The position vector is

r = 3 t \hat{i} + 5 t^{3} \hat{j} + 7 \hat{k}

. Taking the second derivative:

\frac{d ^{2} r}{d t ^{2}} = 0 \hat{i} + 30 t \hat{j} + 0 \hat{k}

. At

t = 1

a = 30 \hat{j}

, which is written as

30 y

in the options.

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Single correct option

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