A car of mass m has an engine which can deliver power P. The minimum time in which car can be accelerated from rest to a speed v is :-

Options

A. `(mv^(2))/(2P)`
B. `Pmv^(2)`
C. `2 Pmv^(2)`
D. `(mv^(2))/(2)P`

Answer

To solve the problem of finding the minimum time in which a car can be accelerated from rest to a speed \( v \) given that the car has a mass \( m \) and an engine that can deliver power \( P \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definition of Power**:
Power (\( P \)) is defined as the rate at which work is done or energy is transferred. Mathematically, it can be expressed as:
\[ P = \frac{W}{t} \]
where \( W \) is the work done and \( t \) is the time taken.

2. **Express Work Done in Terms of Power**:
Rearranging the formula for power gives us:
\[ W = P \cdot t \]
This equation will help us relate work done to time.

3. **Use the Work-Energy Theorem**:
According to the work-energy theorem, the work done on an object is equal to its change in kinetic energy. The kinetic energy (\( KE \)) of an object is given by:
\[ KE = \frac{1}{2} m v^2 \]
Since the car starts from rest (initial velocity \( u = 0 \)), the change in kinetic energy when it reaches speed \( v \) is:
\[ \Delta KE = KE_{final} - KE_{initial} = \frac{1}{2} m v^2 - 0 = \frac{1}{2} m v^2 \]

4. **Set Work Done Equal to Change in Kinetic Energy**:
From the work-energy theorem, we can equate the work done to the change in kinetic energy:
\[ P \cdot t = \frac{1}{2} m v^2 \]

5. **Solve for Time \( t \)**:
Rearranging the equation to solve for \( t \):
\[ t = \frac{\frac{1}{2} m v^2}{P} \]
Simplifying gives:
\[ t = \frac{m v^2}{2P} \]

### Final Answer:
Thus, the minimum time in which the car can be accelerated from rest to a speed \( v \) is:
\[ t = \frac{m v^2}{2P} \]