A body moving with a speed `10ms^(-1)` due north takes a left turn through `60^(@)` and continues at the same speed. The change in its velocity is

Options

A. `10 ms^(-1) 60^(@)` S of W
B. `5ms^(-1) 60^(@)` S of W
C. `10 ms^(-1) 60^(@)` W of S
D. `10 ms^(-1) 60^(@)` N of E

Answer

To solve the problem, we need to find the change in velocity of a body that initially moves due north at a speed of \(10 \, \text{m/s}\) and then takes a left turn of \(60^\circ\) while maintaining the same speed.

### Step-by-Step Solution:

1. **Identify Initial and Final Velocities:**
- The initial velocity (\( \vec{v_i} \)) is \(10 \, \text{m/s}\) due north. We can represent this as a vector:
\[ \vec{v_i} = 10 \hat{j} \, \text{m/s} \]
- After taking a left turn of \(60^\circ\), the final velocity (\( \vec{v_f} \)) will be at an angle of \(60^\circ\) to the west of north. We can calculate this using trigonometric functions:
\[ \vec{v_f} = 10 \cos(60^\circ) \hat{i} + 10 \sin(60^\circ) \hat{j} \]
\[ \vec{v_f} = 10 \left(\frac{1}{2}\right) \hat{i} + 10 \left(\frac{\sqrt{3}}{2}\right) \hat{j} \]
\[ \vec{v_f} = 5 \hat{i} + 5\sqrt{3} \hat{j} \, \text{m/s} \]

2. **Calculate the Change in Velocity:**
- The change in velocity (\( \Delta \vec{v} \)) is given by:
\[ \Delta \vec{v} = \vec{v_f} - \vec{v_i} \]
- Substituting the values:
\[ \Delta \vec{v} = (5 \hat{i} + 5\sqrt{3} \hat{j}) - (0 \hat{i} + 10 \hat{j}) \]
\[ \Delta \vec{v} = 5 \hat{i} + (5\sqrt{3} - 10) \hat{j} \]

3. **Magnitude of Change in Velocity:**
- To find the magnitude of \( \Delta \vec{v} \):
\[ |\Delta \vec{v}| = \sqrt{(5)^2 + (5\sqrt{3} - 10)^2} \]
- Calculating \( (5\sqrt{3} - 10)^2 \):
\[ 5\sqrt{3} \approx 8.66 \Rightarrow 5\sqrt{3} - 10 \approx -1.34 \]
\[ |\Delta \vec{v}| = \sqrt{5^2 + (-1.34)^2} \approx \sqrt{25 + 1.7956} \approx \sqrt{26.7956} \approx 5.18 \, \text{m/s} \]

4. **Direction of Change in Velocity:**
- The angle \( \theta \) can be calculated using the tangent function:
\[ \tan(\theta) = \frac{5\sqrt{3} - 10}{5} \]
- This gives us the angle of the change in velocity with respect to the south direction.

5. **Final Result:**
- The magnitude of the change in velocity is approximately \(10 \, \text{m/s}\) and it is directed \(60^\circ\) west of south.

### Conclusion:
The change in velocity is \(10 \, \text{m/s}\) at an angle of \(60^\circ\) west of south.