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 $2 \mathrm{~s}$ ?
Correct Option: , 2
(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 $)=\tan ^{-1} \frac{\sqrt{40}}{9}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.