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 `m s^(-1)` in south-western direction
B. zero
C. `10.0 m s^(-1)` south-east direction
D. `14.14m s^(-1)` in south-western direction

Explanation

To solve the problem, we need to determine the change in velocity of the scooter after it turns right through an angle of 90 degrees while maintaining a constant speed.

### Step-by-Step Solution:

1. **Identify the Initial Velocity:**
The scooter is initially traveling due east at a speed of \(10 \, \text{ms}^{-1}\). We can represent this velocity vector as:
\[ \vec{v}_{\text{initial}} = 10 \hat{i} \, \text{ms}^{-1} \]
where \(\hat{i}\) represents the unit vector in the east (positive x) direction.

2. **Determine the Final Velocity:**
After turning right through an angle of \(90^\circ\), the scooter will be traveling due south. The final velocity vector can be represented as:
\[ \vec{v}_{\text{final}} = -10 \hat{j} \, \text{ms}^{-1} \]
where \(-\hat{j}\) represents the unit vector in the south (negative y) direction.

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

4. **Express the Change in Velocity Vector:**
The change in velocity vector can be expressed as:
\[ \Delta \vec{v} = -10 \hat{i} - 10 \hat{j} \]

5. **Calculate the Magnitude of the Change in Velocity:**
To find the magnitude of the change in velocity vector, we use the Pythagorean theorem:
\[ |\Delta \vec{v}| = \sqrt{(-10)^2 + (-10)^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2} \, \text{ms}^{-1} \]

6. **Determine the Direction of the Change in Velocity:**
The change in velocity vector \(-10 \hat{i} - 10 \hat{j}\) points in the southwest direction. The angle with respect to the negative x-axis (west) can be calculated as:
\[ \theta = \tan^{-1}\left(\frac{|\Delta v_y|}{|\Delta v_x|}\right) = \tan^{-1}\left(\frac{10}{10}\right) = 45^\circ \]
Thus, the direction of the change in velocity is \(45^\circ\) south of west, or southwest.

### Final Answer:
The change in velocity of the scooter is \(10\sqrt{2} \, \text{ms}^{-1}\) in the southwest direction, which is approximately \(14.14 \, \text{ms}^{-1}\).

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