An automobile of mass m accelerates from rest. If the engine supplies a constant power P, the velocity at time t is given by -

Options

A. `(2Pt)/(M)`
B. `sqrt((2Pt)/(m))`
C. `(Pt)/(2M)`
D. `sqrt((Pt)/(2 M))`

Explanation

To find the velocity of an automobile of mass \( m \) that accelerates from rest under constant power \( P \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between power, force, and velocity**:
The power \( P \) supplied by the engine is given by the formula:
\[ P = F \cdot v \]
where \( F \) is the force and \( v \) is the velocity.

2. **Relate force to mass and acceleration**:
According to Newton's second law, the force can also be expressed as:
\[ F = m \cdot a \]
where \( a \) is the acceleration. Since acceleration can be defined as the rate of change of velocity, we have:
\[ a = \frac{dv}{dt} \]
Therefore, we can rewrite the force as:
\[ F = m \cdot \frac{dv}{dt} \]

3. **Substitute force into the power equation**:
By substituting \( F \) in the power equation, we get:
\[ P = m \cdot \frac{dv}{dt} \cdot v \]
Rearranging this gives:
\[ P = m \cdot v \cdot \frac{dv}{dt} \]

4. **Separate variables for integration**:
We can rearrange the equation to separate the variables:
\[ v \, dv = \frac{P}{m} \, dt \]

5. **Integrate both sides**:
Now, we integrate both sides. The left side integrates with respect to \( v \) from \( 0 \) to \( v \), and the right side integrates with respect to \( t \) from \( 0 \) to \( t \):
\[ \int_0^v v \, dv = \int_0^t \frac{P}{m} \, dt \]
This gives:
\[ \frac{v^2}{2} \bigg|_0^v = \frac{P}{m} t \bigg|_0^t \]
Which simplifies to:
\[ \frac{v^2}{2} = \frac{P}{m} t \]

6. **Solve for velocity \( v \)**:
Multiplying both sides by 2 gives:
\[ v^2 = \frac{2Pt}{m} \]
Taking the square root of both sides results in:
\[ v = \sqrt{\frac{2Pt}{m}} \]

### Final Answer:
The velocity of the automobile at time \( t \) is given by:
\[ v = \sqrt{\frac{2Pt}{m}} \]