Question

A particle is moving eastward with velocity $5m{s}^{-1}$ In 10 s the velocity changes to $5m{s}^{-1}$ northwards. The average acceleration in this time is

• A. $\frac{1}{\sqrt{2}}m{s}^{-2}$ towards North West
• B. $\frac{1}{2}m{s}^{-2}$ towards North West
• C. $\frac{1}{\sqrt{2}}m{s}^{-2}$ towards North East
• D. $\frac{1}{2}m{s}^{-2}$ towards North East

Answer 1

Answer: (A)

Here, $\overrightarrow{v_1}=5\hat{i}m{s}^{-1},\overrightarrow{v_2}=5\hat{j}m{s}^{-1},t=10s$
$\vec{a}=\frac{\overrightarrow{v_2}-\overrightarrow{v_1}}{1}=\frac{5\hat{j}-5\hat{i}}{10}=\frac{1}{2}(j-i)$
$|\vec{a}|=\frac{1}{2}\sqrt{(-1{)}^2+(1{)}^2}$
$=\frac{1}{2}\sqrt{2}=\frac{1}{\sqrt{2}}m{s}^{-2}$
$\tan \theta =\frac{a_y}{a_x}=\left(\frac{\frac{1}{2}}{-\frac{1}{2}}\right)=-1$
$\theta ={\tan }^{-1}\left(-1\right)=135^{\circ }$
$\therefore \vec{a}=\frac{1}{\sqrt{2}}m{s}^{-1}$ towards North West