A particle is moving eastwards with a velocity of $5 \mathrm{~ms}^{-1}$. In 10 seconds the velocity changes…
- $\frac{1}{2} \mathrm{~ms}^{-2}$ towards north
- $\frac{1}{\sqrt{2}} \mathrm{~ms}^{-2}$ towards north - east
- $\frac{1}{\sqrt{2}} \mathrm{~ms}^{-2}$ towards north - west
- zero
Solution
\text { Average acceleration }=\frac{\text { change in velocity }}{\text { time interval }}
$
$=\frac{\Delta \overrightarrow{\mathrm{v}}}{\mathrm{t}}$
$\overrightarrow{\mathrm{v}_{1}}=5 \hat{\mathrm{i}}, \overrightarrow{\mathrm{v}_{2}}=5 \hat{\mathrm{j}}$
$\Delta \overrightarrow{\mathrm{v}}=\left(\overrightarrow{\mathrm{v}}_{2}-\overrightarrow{\mathrm{v}}_{1}\right)$
$=\sqrt{\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}+2 \mathrm{v}_{1} \mathrm{v}_{2} \cos 90}$
$=\sqrt{5^{2}+5^{2}+0}$
$\left[\mathrm{As}\left|\mathrm{v}_{1}\right|=\left|\mathrm{v}_{2}\right|=5 \mathrm{~m} / \mathrm{s}\right]=5 \sqrt{2} \mathrm{~m} / \mathrm{s}$
Avg. acc. $=\frac{\Delta \overrightarrow{\mathrm{v}}}{\mathrm{t}}=\frac{5 \sqrt{2}}{10}=\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^{2}$
$\Rightarrow \tan \theta=\frac{5}{-5}=-1$
which means $\theta$ is in the second quadrant. (towards north-west)
Asked in: JEE Mains - Motion In One Dimension - Test 4