A mosquito is moving with a velocity $\vec{v}=0.5 t^{2} \hat{i}+3 t \hat{j}+9 \hat{k} \mathrm{~m} / \mathrm{s}$ and accelerating in uniform conditions. What will be the direction of mosquito after 2 s ?

(D) $\tan ^{-1}\left(\frac{2}{3}\right)$ from y -axis

(A) $\tan ^{-1}\left(\frac{2}{3}\right)$ from $x$-axis

(B) $\tan ^{-1}\left(\frac{5}{2}\right)$ from $y$-axis

(C) $\tan ^{-1}\left(\frac{5}{2}\right)$ from $x$-axis

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Kinematics

A mosquito is moving with a velocity

v = 0.5 t^{2} \hat{i} + 3 t \hat{j} + 9 \hat{k} m / s

and accelerating in uniform conditions. What will be the direction of mosquito after 2 s ?

Answer: (D)

The conditions are uniform, so the direction asked for is that of the acceleration, which follows from differentiating the velocity:

a = \frac{d v}{d t} = t \hat{i} + 3 \hat{j}

t = 2 s

this is

a = 2 \hat{i} + 3 \hat{j} m s^{- 2}

, a vector lying in the

x y

-plane with component

2

along

x

and component

3

along

y

. Its inclination

α

to the

y

-axis satisfies

tan α = \frac{a _{x}}{a _{y}} = \frac{2}{3}

so after

2 s

the mosquito is directed at

tan^{- 1} (\frac{2}{3})

from the

y

-axis.

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option