The position vector of a particle related to time $t$ is given by $\overrightarrow{\mathrm{r}}=\left(10 \mathrm{t} \hat{\mathrm{i}}+15 \mathrm{t}^{2} \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\right) \mathrm{m}$ The direction of net force experienced by the particle is :

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Kinematics

The position vector of a particle related to time

t

is given by

r = (10 t \hat{i} + 15 t^{2} \hat{j} + 7 \hat{k}) m

The direction of net force experienced by the particle is :

(A)Positive $y$ -axis✓ Correct
(B)Positive $x$ -axis
(C)Positive $z$ -axis
(D)In $x - y$ plane

Answer: (A)

Solution

Differentiating the position vector gives the velocity

v = \frac{d r}{d t} = 10 \hat{i} + 30 t \hat{j}

. A further derivative gives the acceleration

a = \frac{d v}{d t} = 30 \hat{j}

, which is constant and directed along

+ y

. Because

F = m a

with positive mass, the net force on the particle is along the positive

y

-axis.

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Single correct option

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