Q136 / 180 · COMEDK 2012 · Physics
A particle is moving eastward with a velocity $5 \mathrm{~ms}^{-1}$. In $10 \mathrm{~s}$ the velocity changes to $5 \mathrm{~ms}^{-1}$ northwards. The average acceleration in this time is
- Correct answerA$\frac{1}{\sqrt{2}} \mathrm{~ms}^{-2}$, towards North-West
- B$\frac{1}{2} \mathrm{~ms}^{-2}$, towards North-West
- C$\frac{1}{\sqrt{2}} \mathrm{~ms}^{-2}$, towards North-East
- D$\frac{1}{2} \mathrm{~ms}^{-2}$, towards North-East
Solution
Given, $u=5 \mathrm{~ms}^{-1}$, Eastwards
$v=5 \mathrm{~ms}^{-1}$, Northwards
and $\Delta t=10 \mathrm{~s}$
Change in velocity $|\Delta v|=\sqrt{u^{2}+v^{2}}$
$$
=\sqrt{5^{2}+5^{2}}=5 \sqrt{2} \mathrm{~ms}^{-1}
$$
Direction of $\Delta v$,
$$
\tan \theta=\frac{v}{u}=\frac{-5}{5}=-1
$$
(taking eastward direction positive)
$$
\Rightarrow \quad \theta=135^{\circ}
$$
$\therefore$ Average acceleration,
$a_{\text {avg }}=\frac{\Delta v}{\Delta t}=\frac{5 \sqrt{2}}{10}=\frac{1}{\sqrt{2}} \mathrm{~ms}^{-2}$, towards North-West
Hence, the correct answer is option: 'A'.