 # A mosquito is moving with a velocity Question:

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

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

2. (2) $\tan ^{-1}\left(\frac{\sqrt{85}}{6}\right)$ from $y$-axis

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

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

Correct Option: , 2

Solution:

(2)

Given :

$\vec{v}=0.5 t^{2} \hat{i}+3 t \hat{j}+9 \hat{k}$

$\vec{v}$ at=2 $=2 \hat{i}+6 \hat{j}+9 \hat{k}$

$\therefore$ Angle made by direction of motion of mosquito will be, $\cos ^{-1} \frac{2}{11}$ ( from

$x$-axis $)=\tan ^{-1} \frac{\sqrt{117}}{2}$

$\cos ^{-1} \frac{6}{11}($ from $y$-axis $)=\tan ^{-1} \frac{\sqrt{85}}{6}$

$\cos ^{-1} \frac{9}{11}($ from z-axis $)=\tan ^{-1} \frac{\sqrt{40}}{9}$  