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/answer is: 1/√2 m/s² toward north-west.

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

Let the positive x-axis represent the East direction and the positive y-axis represent the North direction.
Therefore:
The unit vector along East is $\hat{i}$.
The unit vector along North is $\hat{j}$.

Given:
Initial velocity, ${\overrightarrow{v}}_{\text{i}}=5\hat{i}\text{m/s}$ (eastwards)
Final velocity, ${\overrightarrow{v}}_{\text{f}}=5\hat{j}\text{m/s}$ (northwards)
Time interval, $\Delta t=10\text{s}$

Step 1: Calculate the change in velocity vector ($\Delta \overrightarrow{v}$):
$\Delta \overrightarrow{v}={\overrightarrow{v}}_{\text{f}}-{\overrightarrow{v}}_{\text{i}}=5\hat{j}-5\hat{i}=-5\hat{i}+5\hat{j}$

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

Step 3: Determine the direction of the change in velocity vector:
The vector $\Delta \overrightarrow{v}=-5\hat{i}+5\hat{j}$ has a negative x-component (Westwards) and a positive y-component (Northwards) of equal magnitudes.
Thus, the direction points exactly mid-way between North and West, which is North-West.

Step 4: Calculate the magnitude of average acceleration:
$\left|{\overrightarrow{a}}_{\text{avg}}\right|=\frac{\left|\Delta \overrightarrow{v}\right|}{\Delta t}=\frac{5\sqrt{2}}{10}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}{\text{m/s}}^2$

Since acceleration is a vector quantity in the same direction as the change in velocity, the direction of the average acceleration is also toward the North-West.
Hence, the average acceleration is $1/\sqrt{2}{\text{m/s}}^2$ toward North-West.