A scooter going due east at `10 ms^(-1)` turns right through an angle of `90^(@)`. If the speed of the scooter remain unchanged in taking turn, the change is the velocity the scooter is

Options

A. `20.0 ms^(-1)` south-east direction
B. zero
C. `10.0 ms^(-1)` in south direction
D. `14.14 ms^(-1)` in south-west direction

Answer

To solve the problem of finding the change in velocity of a scooter that turns right through an angle of 90 degrees while maintaining a constant speed, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Velocity**:
The scooter is initially moving due east with a speed of \(10 \, \text{m/s}\). We can represent this velocity vector as:
\[ \vec{v_1} = 10 \hat{i} \]
where \(\hat{i}\) is the unit vector in the east (x) direction.

2. **Identify Final Velocity**:
After turning right through an angle of \(90^\circ\), the scooter will be moving due south with the same speed of \(10 \, \text{m/s}\). We can represent this final velocity vector as:
\[ \vec{v_2} = -10 \hat{j} \]
where \(-\hat{j}\) indicates the south (negative y) direction.

3. **Calculate Change in Velocity**:
The change in velocity (\(\Delta \vec{v}\)) is given by the difference between the final and initial velocity vectors:
\[ \Delta \vec{v} = \vec{v_2} - \vec{v_1} \]
Substituting the values we have:
\[ \Delta \vec{v} = (-10 \hat{j}) - (10 \hat{i}) = -10 \hat{i} - 10 \hat{j} \]

4. **Magnitude of Change in Velocity**:
To find the magnitude of the change in velocity, we can use the Pythagorean theorem:
\[ |\Delta \vec{v}| = \sqrt{(-10)^2 + (-10)^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2} \, \text{m/s} \]

5. **Direction of Change in Velocity**:
The direction of the change in velocity can be found by calculating the angle it makes with the negative x-axis (or east direction). The change in velocity vector \(-10 \hat{i} - 10 \hat{j}\) points diagonally in the third quadrant.

### Final Answer:
The change in velocity of the scooter is:
\[ \Delta \vec{v} = -10 \hat{i} - 10 \hat{j} \quad \text{with a magnitude of } 10\sqrt{2} \, \text{m/s} \]