Question

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

• A. ${\tan }^{-1}\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)fromx-axis$
• B. ${\tan }^{-1}\left(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)fromx-axis$
• C. ${\tan }^{-1}\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)fromy-axis$
• D. ${\tan }^{-1}\left(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)fromy-axis$
• E. ${\tan }^{-1}\left(\raisebox{1ex}{$\sqrt{117}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)fromx-axis$
• F. ${\tan }^{-1}\left(\raisebox{1ex}{$\sqrt{85}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)fromy-axis$

Answer 1

Answer: (E), (F)

${\tan }^{-1}\left(\raisebox{1ex}{$\sqrt{85}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)fromy-axis$

Step 1: Given Data and assumptions:

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

Let, $\alpha$ is angle from x-axis

$\beta$ is angle fom y-axis

Step 2: Formula used:

Direction cosine from x-axis $=\frac{\overrightarrow{v}.\hat{i}}{\left|\overrightarrow{v}\right|}$

Direction cosine from y-axis $=\frac{\overrightarrow{v}.\hat{j}}{\left|\overrightarrow{v}\right|}$

Step 3: Calculating the Direction:

Velocity of a mosquito after $2s$,

$\overrightarrow{v_{t=2}}=0.5\times 2^2\hat{i}+3\times 2\hat{j}+9\hat{k}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

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

$\left|\overrightarrow{v_{t=2}}\right|=\sqrt{2^2+6^2+9^2}$

$=\sqrt{121}$

$=11m/s$

The direction of Mosquito from x-axis,

$\cos \alpha =\frac{\left(2\hat{i}+6\hat{j}+9\hat{k}\right).\left(\hat{i}\right)}{11}fromx-axis$

$=\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$11$}\right.\right)fromx-axis$

$\therefore \alpha ={\tan }^{-1}\left(\raisebox{1ex}{$\sqrt{117}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)fromx-axis$

Direction of mosquito from y-axis,

$\cos \beta =\frac{\overrightarrow{v}.\hat{j}}{\left|\stackrel{\rightharpoonup }{v}\right|}fromy-axis$

$=\frac{\left(2\hat{i}+6\hat{j}+9\hat{k}\right).\left(\hat{j}\right)}{11}$

$=\left(\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$11$}\right.\right)fromy-axis$

$\therefore \beta ={\tan }^{-1}\left(\raisebox{1ex}{$\sqrt{85}$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)fromy-axis$

Hence, option E and option F are the correct answer.