A car starts moving from rest on a horizontal circular road of radius of curvature R at . The speed of the car is increasing at a constant rate a. The coefficient of friction between the road and the tyres of the car is The car starts slipping at time`t = t_(0)` . Then `t_(0)` is equal to:

Options

A. `(.^(4)sqrt(mu^(2) g^(2) R^(2)))/(a)`
B. `(.^(4)sqrt(R^(2) (mu^(2) g^(2) - a^(2))))/(a)`
C. `(.^(4)sqrt(R^(2) (a^(2) - mu^(2) g^(2))))/(a)`
D. `(.^(4)sqrt(R^(2) (mu^(2) g^(2) + a^(2))))/(a)`

Explanation

To solve the problem, we need to analyze the motion of the car on the circular road and the forces acting on it. Here’s the step-by-step solution:

### Step 1: Understand the motion of the car
The car starts from rest and accelerates uniformly. The speed of the car increases at a constant rate \( a \). The car is moving in a circular path of radius \( R \).

**Hint:** Remember that the car's speed increases linearly over time due to constant acceleration.

### Step 2: Identify the forces acting on the car
When the car is moving in a circle, it experiences two types of acceleration:
- **Centripetal acceleration** (\( a_c \)) due to circular motion, given by \( \frac{v^2}{R} \).
- **Tangential acceleration** (\( a_t \)) due to the increase in speed, which is equal to \( a \).

**Hint:** The total acceleration of the car can be found using the Pythagorean theorem since the two accelerations are perpendicular.

### Step 3: Set up the equation for slipping
The car will start slipping when the frictional force is equal to the required centripetal force. The maximum frictional force is given by:
\[ F_{\text{friction}} = \mu m g \]
The required centripetal force is:
\[ F_{\text{centripetal}} = m \cdot a_c = m \cdot \frac{v^2}{R} \]

At the point of slipping:
\[ \mu m g = m \cdot \frac{v^2}{R} \]

**Hint:** You can cancel the mass \( m \) from both sides since it appears in both equations.

### Step 4: Relate speed to time
The speed \( v \) of the car at time \( t_0 \) can be expressed as:
\[ v = a t_0 \]

**Hint:** Substitute this expression for \( v \) into the equation derived from the forces.

### Step 5: Substitute \( v \) into the slipping condition
Substituting \( v = a t_0 \) into the slipping condition:
\[ \mu g = \frac{(a t_0)^2}{R} \]
This simplifies to:
\[ \mu g R = a^2 t_0^2 \]

**Hint:** Rearranging this equation will help you isolate \( t_0 \).

### Step 6: Solve for \( t_0 \)
Rearranging the equation gives:
\[ t_0^2 = \frac{\mu g R}{a^2} \]
Taking the square root of both sides:
\[ t_0 = \sqrt{\frac{\mu g R}{a^2}} = \frac{\sqrt{\mu g R}}{a} \]

**Hint:** Ensure you check the units to confirm that your final expression for \( t_0 \) is dimensionally consistent.

### Final Answer
The time \( t_0 \) at which the car starts slipping is given by:
\[ t_0 = \frac{\sqrt{\mu g R}}{a} \]