A car of mass m starts from rest and acquires a velocity along east `upsilon = upsilonhati (upsilon gt 0)` in two seconds Assuming the car moves with uniform acceleration the force exerted on the car is .

Options

A. `(mv)/(2)` eastward and is exerted by the car engine
B. `(mv)/(2)` eastward and is due to the friction on the tyres exerted by the road
C. more than `(mv)/(2)` eastward exerted due to the engine and overcomes the friction of the road
D. `(mv)/(2)` exerted by the engine

Explanation

To solve the problem, we need to find the force exerted on a car of mass \( m \) that starts from rest and acquires a velocity \( \mathbf{v} = v \hat{i} \) (where \( v > 0 \)) in a time \( t = 2 \) seconds, assuming uniform acceleration.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- The car starts from rest, so the initial velocity \( u = 0 \).
- The final velocity after 2 seconds is \( v = v \hat{i} \).
- The time taken \( t = 2 \) seconds.

2. **Use the Equation of Motion:**
- We can use the first equation of motion:
\[ v = u + at \]
- Substituting the known values:
\[ v = 0 + a \cdot 2 \]
- This simplifies to:
\[ v = 2a \]

3. **Calculate Acceleration:**
- Rearranging the equation gives us:
\[ a = \frac{v}{2} \]

4. **Apply Newton's Second Law:**
- According to Newton's second law, the force \( F \) exerted on an object is given by:
\[ F = ma \]
- Substituting the expression for acceleration:
\[ F = m \cdot \frac{v}{2} \]

5. **Determine the Direction of the Force:**
- Since the car is moving eastward, the force will also be directed eastward. Thus, we can express the force as:
\[ F = \frac{mv}{2} \hat{i} \]

### Final Answer:
The force exerted on the car is:
\[ F = \frac{mv}{2} \hat{i} \quad \text{(eastward)} \]