A truck travelling due north at 20 m/s turns east and travels at the same speed. What is the change in velocity :

Options

A. `40ms^(-1)N-W`
B. `20sqrt2ms^(-1)N-W`
C. `40ms^(-1)S-W`
D. `20sqrt2 ms^(-1)S-W`

Answer

To solve the problem of finding the change in velocity of a truck that travels due north at 20 m/s and then turns east at the same speed, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial and Final Velocities:**
- The truck's initial velocity (v_initial) is 20 m/s due north.
- The truck's final velocity (v_final) is 20 m/s due east.

2. **Convert Velocities to Vector Form:**
- We can represent the north direction as the positive y-axis (j-cap) and the east direction as the positive x-axis (i-cap).
- Therefore:
- \( v_{initial} = 20 \, \text{m/s} \, \hat{j} \) (north)
- \( v_{final} = 20 \, \text{m/s} \, \hat{i} \) (east)

3. **Calculate Change in Velocity:**
- The change in velocity (Δv) is given by the formula:
\[ \Delta v = v_{final} - v_{initial} \]
- Substituting the values:
\[ \Delta v = (20 \, \hat{i}) - (20 \, \hat{j}) = 20 \, \hat{i} - 20 \, \hat{j} \]

4. **Determine the Magnitude of Change in Velocity:**
- The magnitude of the change in velocity can be calculated using the Pythagorean theorem since the two velocity vectors are perpendicular to each other:
\[ |\Delta v| = \sqrt{(20)^2 + (-20)^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \, \text{m/s} \]

5. **Determine the Direction of Change in Velocity:**
- The change in velocity vector \( \Delta v = 20 \hat{i} - 20 \hat{j} \) indicates that the change is directed towards the southeast (45 degrees from both axes).

### Final Answer:
- The change in velocity is \( 20\sqrt{2} \, \text{m/s} \) directed towards the southeast.