A particle is moving eastwards with a velocity of $5\,\text{m/s}$. In $10\,\text{s}$ the velocity changes to $5\,\text{m/s}$ northwards. The average acceleration in this time is/are

Step 1: Identify the initial and final velocity vectors.

Let the unit vectors along the East and North directions be $\hat{i}$ and $\hat{j}$ respectively.

$$\vec{u} = 5\hat{i}\,\text{m/s}$$ $$\vec{v} = 5\hat{j}\,\text{m/s}$$ $$\Delta t = 10\,\text{s}$$

Step 2: Find the change in velocity.

The change in velocity is the difference between the final and initial velocities.

$$\Delta \vec{v} = \vec{v} - \vec{u}$$ $$\Delta \vec{v} = -5\hat{i} + 5\hat{j}\,\text{m/s}$$

Step 3: Calculate the magnitude and direction of the change in velocity.

The vector has equal magnitude components pointing West and North, indicating a North-West direction.

$$|\Delta \vec{v}| = \sqrt{(-5)^{2} + 5^{2}}$$ $$|\Delta \vec{v}| = 5\sqrt{2}\,\text{m/s}$$

Step 4: Calculate the average acceleration.

Divide the change in velocity by the time interval.

$$\vec{a}_{\text{avg}} = \frac{\Delta \vec{v}}{\Delta t}$$ $$|\vec{a}_{\text{avg}}| = \frac{5\sqrt{2}}{10}$$ $$|\vec{a}_{\text{avg}}| = \frac{1}{\sqrt{2}}\,\text{m/s}^{2}$$

The direction of the average acceleration is parallel to the change in velocity, which is North-West.

Answer: $\dfrac{1}{\sqrt{2}}\,\text{m/s}^{2}$ towards North-West.