Brakes are applied to car moving with disengaged engine, bringing it to a halt after `2s` Its velocity at the momnet when the breaks are applied if the coefficient of friction between the road and the tyres is `0.4` is .

Options

A. `3.92 ms^(-1)`
B. `7.84 ms^(-1)`
C. `11.2 ms^(-1)`
D. `19.6 ms^(-1)`

Explanation

To solve the problem, we need to determine the initial velocity of the car at the moment the brakes are applied, given that it comes to a stop in 2 seconds due to the frictional force.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Car**:
When the brakes are applied, the only horizontal force acting on the car is the frictional force between the tires and the road.

2. **Calculate the Frictional Force**:
The frictional force \( F_f \) can be calculated using the formula:
\[ F_f = \mu \cdot N \]
where \( \mu \) is the coefficient of friction (0.4) and \( N \) is the normal force. For a car on a flat surface, the normal force \( N \) is equal to the weight of the car, which is \( m \cdot g \) (where \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \)).
Thus,
\[ F_f = \mu \cdot m \cdot g = 0.4 \cdot m \cdot 9.81 \]

3. **Determine the Deceleration**:
The frictional force causes a deceleration \( a \) of the car. According to Newton's second law, \( F = m \cdot a \):
\[ F_f = m \cdot a \implies 0.4 \cdot m \cdot 9.81 = m \cdot a \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ a = 0.4 \cdot 9.81 = 3.924 \, \text{m/s}^2 \]

4. **Use the Kinematic Equation**:
We know the car comes to a stop (final velocity \( v = 0 \)) in \( t = 2 \, \text{s} \). We can use the kinematic equation:
\[ v = u + at \]
where \( u \) is the initial velocity. Rearranging gives:
\[ u = v - at \]
Substituting the known values:
\[ u = 0 - (-3.924 \cdot 2) = 3.924 \cdot 2 = 7.848 \, \text{m/s} \]

5. **Final Result**:
Rounding to two decimal places, the initial velocity of the car when the brakes were applied is approximately:
\[ u \approx 7.84 \, \text{m/s} \]

### Final Answer:
The velocity at the moment when the brakes are applied is approximately **7.84 m/s**.