A car is moving along the circle `x^(2)+y^(2)=a^(2)` in the anti-clockwise direction with a constant speed. The x-y plane is a rough horizontal stationary surface. When the car is at the point (`a cos theta, a sin theta `), the unit vector in the direction of the friction force acting on the car is

Options

A. `cos theta hati+sin thetahati`
B. `cos theta hati+sin thetahati`
C. `-cos theta hati-sin thetahatj`
D. `-cos theta hati+sin thetahatj`

Answer

To find the unit vector in the direction of the friction force acting on the car moving along the circular path defined by the equation \(x^2 + y^2 = a^2\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Position of the Car**:
The car is at the point \((a \cos \theta, a \sin \theta)\) on the circular path.

2. **Determine the Radius Vector**:
The radius vector \( \mathbf{r} \) from the origin to the point where the car is located can be expressed as:
\[ \mathbf{r} = a \cos \theta \, \hat{i} + a \sin \theta \, \hat{j} \]

3. **Identify the Direction of the Centripetal Force**:
Since the car is moving in a circular path with constant speed, there is a centripetal force acting towards the center of the circle. This force is provided by the friction force.

4. **Direction of the Friction Force**:
The direction of the friction force is opposite to the radius vector, which means it points towards the center of the circle. Therefore, the friction force vector can be expressed as:
\[ \mathbf{F}_{\text{friction}} = -\mathbf{r} = - (a \cos \theta \, \hat{i} + a \sin \theta \, \hat{j}) = -a \cos \theta \, \hat{i} - a \sin \theta \, \hat{j} \]

5. **Calculate the Magnitude of the Friction Force**:
The magnitude of the friction force vector is:
\[ |\mathbf{F}_{\text{friction}}| = \sqrt{(-a \cos \theta)^2 + (-a \sin \theta)^2} = \sqrt{a^2 \cos^2 \theta + a^2 \sin^2 \theta} = \sqrt{a^2} = a \]

6. **Find the Unit Vector**:
The unit vector in the direction of the friction force is obtained by dividing the friction force vector by its magnitude:
\[ \mathbf{u} = \frac{\mathbf{F}_{\text{friction}}}{|\mathbf{F}_{\text{friction}}|} = \frac{-a \cos \theta \, \hat{i} - a \sin \theta \, \hat{j}}{a} = -\cos \theta \, \hat{i} - \sin \theta \, \hat{j} \]

7. **Final Result**:
The unit vector in the direction of the friction force acting on the car is:
\[ \mathbf{u} = -\cos \theta \, \hat{i} - \sin \theta \, \hat{j} \]