An automobile of mass 'm' accelerates starting from origin and initially at rest, while the engine supplies constant power P. The position is given as a function of time by :

Options

A. `((9P)/(8m))^(1/2)t^(3/2)`
B. `((8P)/(9m))^(1/2)t^(2/3)`
C. `((9m)/(8P))^(1/2)t^(3/2)`
D. `((8P)/(9m))^(1/2)t^(3/2)`

Answer

To solve the problem of an automobile of mass 'm' accelerating from rest under constant power 'P', we can follow these steps:

### Step 1: Relate Power, Force, and Velocity
We know that power (P) is given by the formula:
\[ P = F \cdot v \]
where \(F\) is the force and \(v\) is the velocity of the automobile.

### Step 2: Express Force in Terms of Mass and Acceleration
Using Newton's second law, we can express force as:
\[ F = m \cdot a \]
where \(a\) is the acceleration of the automobile.

### Step 3: Substitute Force into the Power Equation
Substituting the expression for force into the power equation gives:
\[ P = m \cdot a \cdot v \]

### Step 4: Express Acceleration in Terms of Velocity
Acceleration can be expressed as:
\[ a = \frac{dv}{dt} \]
Substituting this into the power equation, we have:
\[ P = m \cdot \frac{dv}{dt} \cdot v \]

### Step 5: Rearranging the Equation
Rearranging the equation gives:
\[ \frac{dv}{dt} = \frac{P}{m \cdot v} \]

### Step 6: Separate Variables and Integrate
We can separate the variables:
\[ v \, dv = \frac{P}{m} \, dt \]
Now, integrate both sides. The left side integrates from 0 to \(v\) and the right side from 0 to \(t\):
\[ \int_0^v v \, dv = \int_0^t \frac{P}{m} \, dt \]
This gives:
\[ \frac{v^2}{2} = \frac{P}{m} t \]

### Step 7: Solve for Velocity
Multiplying both sides by 2 gives:
\[ v^2 = \frac{2P}{m} t \]
Taking the square root, we find:
\[ v = \sqrt{\frac{2P}{m} t} \]

### Step 8: Relate Velocity to Position
We know that velocity is also defined as:
\[ v = \frac{dx}{dt} \]
Substituting for \(v\) gives:
\[ \frac{dx}{dt} = \sqrt{\frac{2P}{m} t} \]

### Step 9: Separate Variables and Integrate Again
Now, separate variables again:
\[ dx = \sqrt{\frac{2P}{m} t} \, dt \]
Integrate both sides:
\[ \int_0^x dx = \int_0^t \sqrt{\frac{2P}{m} t} \, dt \]
The left side integrates to \(x\), and the right side can be integrated using the substitution \(u = t^{3/2}\):
\[ x = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} t^{3/2} \]

### Step 10: Final Expression for Position
Thus, we have:
\[ x = \frac{2}{3} \sqrt{\frac{2P}{m}} t^{3/2} \]

### Conclusion
The position of the automobile as a function of time is:
\[ x(t) = \frac{2}{3} \sqrt{\frac{2P}{m}} t^{3/2} \]