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

The correct option is (1/√2) m/s² toward north-west.

Let us find the average acceleration of the particle step-by-step.
Average acceleration ($a_{\text{avg}}$) is defined as the change in velocity divided by the time interval:
${\overrightarrow{a}}_{\text{avg}}=\frac{\Delta \overrightarrow{v}}{\Delta t}=\frac{{\overrightarrow{v}}_f-{\overrightarrow{v}}_i}{t}$

Let the coordinate axes be defined such that:
- The positive x-axis represents the East direction, with unit vector $\hat{i}$.
- The positive y-axis represents the North direction, with unit vector $\hat{j}$.

From the given data:
- Initial velocity, ${\overrightarrow{v}}_i=5\hat{i}\text{m/s}$ (moving eastwards)
- Final velocity, ${\overrightarrow{v}}_f=5\hat{j}\text{m/s}$ (moving northwards)
- Time interval, $t=10\text{s}$

First, let us calculate the change in velocity vector ($\Delta \overrightarrow{v}$):
$\Delta \overrightarrow{v}={\overrightarrow{v}}_f-{\overrightarrow{v}}_i=5\hat{j}-5\hat{i}=-5\hat{i}+5\hat{j}$

Next, we calculate the magnitude of the change in velocity:
$\left|\Delta \overrightarrow{v}\right|=\sqrt{{(-5)}^2+5^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}\text{m/s}$

Now, we find the magnitude of the average acceleration:
$\left|{\overrightarrow{a}}_{\text{avg}}\right|=\frac{\left|\Delta \overrightarrow{v}\right|}{t}=\frac{5\sqrt{2}}{10}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}{\text{m/s}}^2$

Finally, let us determine the direction of the acceleration.
The direction of acceleration is the same as the direction of the change in velocity vector, $\Delta \overrightarrow{v}=-5\hat{i}+5\hat{j}$.
Since the x-component is negative (West) and the y-component is positive (North) with equal magnitudes, the vector points exactly midway between West and North.
Therefore, the direction is toward the north-west.

Thus, the average acceleration is $\frac{1}{\sqrt{2}}{\text{m/s}}^2$ toward north-west.