An automobile engine of mass m accelerates and a constant power P is applied by the engine. The instantaneous speed of the engine will be

Options

A. `(pt)/( 2m)`
B. `(2pt)/(m)`
C. `((pt)/(2m))^(1//2)`
D. `((2pt)/(m))^(1//2)`

Answer

To find the instantaneous speed of an automobile engine that accelerates with a constant power \( P \) applied by the engine, we can follow these steps:

### Step-by-Step Solution

1. **Understand the relationship between power, force, and speed**:
The power \( P \) provided by the engine can be expressed as:
\[ P = F \cdot v \]
where \( F \) is the force exerted by the engine and \( v \) is the instantaneous speed.

2. **Express the force in terms of mass and acceleration**:
The force \( F \) can also be expressed using Newton's second law:
\[ F = m \cdot a \]
where \( m \) is the mass of the engine and \( a \) is its acceleration.

3. **Relate acceleration to the change in velocity**:
Since acceleration \( a \) is the rate of change of velocity, we can write:
\[ a = \frac{dv}{dt} \]
Substituting this into the force equation gives:
\[ F = m \cdot \frac{dv}{dt} \]

4. **Substitute the force into the power equation**:
Now substituting \( F \) into the power equation:
\[ P = m \cdot \frac{dv}{dt} \cdot v \]

5. **Rearranging the equation**:
Rearranging gives:
\[ P = m \cdot v \cdot \frac{dv}{dt} \]

6. **Separate variables for integration**:
We can separate the variables to integrate:
\[ \frac{P}{m} = v \cdot \frac{dv}{dt} \]
This can be rewritten as:
\[ \frac{P}{m} dt = v \, dv \]

7. **Integrate both sides**:
Integrating both sides, we have:
\[ \int \frac{P}{m} dt = \int v \, dv \]
The left side integrates to \( \frac{P}{m} t \), and the right side integrates to \( \frac{v^2}{2} \):
\[ \frac{P}{m} t = \frac{v^2}{2} \]

8. **Solve for \( v \)**:
Rearranging gives:
\[ v^2 = \frac{2Pt}{m} \]
Taking the square root of both sides gives:
\[ v = \sqrt{\frac{2Pt}{m}} \]

### Final Answer
The instantaneous speed of the engine is:
\[ v = \sqrt{\frac{2Pt}{m}} \]