A machine is delivering constant power to drive a body along a straight line. What is the relation between the distance travelled by the body against time ?

Options

A. `s^(2)alpha t^(3)`
B. `s^(2)alpha t^(3)`
C. `s^(3)alpha t^(2)`
D. `s alpha t^(3)`

Answer

To find the relationship between the distance traveled by a body and time when a machine delivers constant power, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Power**:
The power \( P \) delivered by the machine is given by the formula:
\[ P = \frac{W}{t} \]
where \( W \) is the work done and \( t \) is the time.

2. **Relating Work to Force and Distance**:
Work done \( W \) can also be expressed in terms of force \( F \) and distance \( s \):
\[ W = F \cdot s \]
Thus, we can rewrite the power equation as:
\[ P = \frac{F \cdot s}{t} \]

3. **Expressing Force**:
Since the machine delivers constant power, we can express the force in terms of the mass \( m \) and acceleration \( a \):
\[ F = m \cdot a \]
The acceleration can be expressed as the second derivative of distance with respect to time:
\[ a = \frac{d^2s}{dt^2} \]

4. **Substituting for Power**:
Substituting \( F \) into the power equation gives:
\[ P = \frac{m \cdot a \cdot s}{t} \]
Since \( a = \frac{d^2s}{dt^2} \), we have:
\[ P = \frac{m \cdot \frac{d^2s}{dt^2} \cdot s}{t} \]

5. **Using Velocity**:
The velocity \( v \) is defined as:
\[ v = \frac{ds}{dt} \]
Therefore, we can express acceleration as:
\[ a = \frac{dv}{dt} = \frac{d^2s}{dt^2} \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ P = m \cdot v \cdot a \]
Since \( a = \frac{dv}{dt} \), we can write:
\[ P = m \cdot v \cdot \frac{dv}{dt} \]

7. **Integrating the Equation**:
Rearranging further, we have:
\[ P \cdot dt = m \cdot v \cdot dv \]
Integrating both sides gives:
\[ P \cdot t = \frac{m}{2} v^2 \]

8. **Finding the Relationship**:
From the above equation, we can express \( v \) in terms of \( t \):
\[ v = \sqrt{\frac{2Pt}{m}} \]
Since \( v = \frac{ds}{dt} \), we can substitute:
\[ \frac{ds}{dt} = \sqrt{\frac{2Pt}{m}} \]

9. **Integrating to Find Distance**:
To find \( s \), we integrate:
\[ ds = \sqrt{\frac{2P}{m}} \cdot t^{1/2} dt \]
Integrating both sides gives:
\[ s = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} t^{3/2} + C \]
Assuming \( s = 0 \) when \( t = 0 \), we find:
\[ s = \frac{2}{3} \sqrt{\frac{2P}{m}} t^{3/2} \]

10. **Final Relation**:
Thus, we can conclude:
\[ s \propto t^{3/2} \]