A particle is moving eastwards with velocity of $5m/s$. In $10sec$ the velocity changes to $5m/s$ northwards. The average acceleration in this time is

The correct option is D $\frac{1}{\sqrt{2}}m/s^2$ toward north-west
According to given question,


Velocity in vector form can be written as,
${\overrightarrow{v}}_1=5\hat{i},{\overrightarrow{v}}_2=5\hat{j}$
Therefore
${\overrightarrow{a}}_{\text{avg}}=\frac{{\overrightarrow{v}}_2-{\overrightarrow{v}}_1}{\Delta t}$

By putting the values, we get,
${\overrightarrow{a}}_{\text{avg}}=\frac{5\hat{j}-5\hat{i}}{10}=\frac{\hat{j}-\hat{i}}{2}$

Hence the magnitude is given by,
$|{\overrightarrow{a}}_{\text{avg}}|=\frac{\sqrt{1^2+(-1{)}^2}}{2}=\frac{\sqrt{2}}{2}|{\overrightarrow{a}}_{\text{avg}}|=\frac{1}{\sqrt{2}}m/s^2$ towards north-west