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 unifrom 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 will follow these steps:

### Step 1: Identify the given values
- Mass of the car, \( m \)
- Initial velocity, \( u = 0 \) (since the car starts from rest)
- Final velocity, \( v \) (along east)
- Time, \( t = 2 \) seconds

### Step 2: Use the equation of motion to find acceleration
We can use the first equation of motion:
\[ v = u + at \]
Substituting the known values:
\[ v = 0 + a \cdot 2 \]
This simplifies to:
\[ v = 2a \]
From this, we can solve for acceleration \( a \):
\[ a = \frac{v}{2} \]

### Step 3: Calculate the force exerted on the car
Using Newton's second law of motion, the force \( F \) exerted on the car is given by:
\[ F = ma \]
Substituting the expression for acceleration \( a \):
\[ F = m \cdot \left(\frac{v}{2}\right) \]
This simplifies to:
\[ F = \frac{mv}{2} \]

### Step 4: State the direction of the force
Since the car is moving towards the east, the force exerted on the car is also directed towards the east.

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

---