A body is moving with velocity $30 ext{ m/s}$ towards east. After $10 ext{ s}$ its velocity becomes $40

Question

A body is moving with velocity $30 	ext{ m/s}$ towards east. After $10 	ext{ s}$ its velocity becomes $40 	ext{ m/s}$ towards north. The average acceleration of the body is:

Accepted Answer

1. **Vector Representation:** Let East be along the positive x-axis ($\hat{i}$) and North be along the positive y-axis ($\hat{j}$).
   - Initial velocity: $\vec{v}_1 = 30\hat{i} 	ext{ m/s}$
   - Final velocity: $\vec{v}_2 = 40\hat{j} 	ext{ m/s}$
2. **Change in Velocity:** $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1 = 40\hat{j} - 30\hat{i} = -30\hat{i} + 40\hat{j} 	ext{ m/s}$.
3. **Average Acceleration Formula:** Average acceleration is defined as the change in velocity divided by the time interval .
   $$ \vec{a}_{av} = \frac{\Delta \vec{v}}{\Delta t} = \frac{-30\hat{i} + 40\hat{j}}{10} = -3\hat{i} + 4\hat{j} 	ext{ m/s}^2 $$
4. **Magnitude:** The magnitude of the average acceleration is:
   $$ |\vec{a}_{av}| = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 	ext{ m/s}^2 $$

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