A body is moved along a straight line by a machine delivering constant power . The distance moved by the body is time `t` is proptional to

Options

A. `t^1//2`
B. `t^3//4`
C. `t^3//2`
D. `t^2`

Answer

To solve the problem of how the distance moved by a body is related to time when a machine delivers constant power, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between power, force, and velocity**:
The power \( P \) delivered by the machine can be expressed as:
\[ P = F \cdot V \]
where \( F \) is the force applied and \( V \) is the velocity of the body.

2. **Express velocity in terms of distance and time**:
The velocity \( V \) can be defined as:
\[ V = \frac{S}{t} \]
where \( S \) is the distance traveled and \( t \) is the time taken.

3. **Express 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 body and \( a \) is the acceleration.

4. **Relate acceleration to distance and time**:
The acceleration \( a \) can be expressed as:
\[ a = \frac{V}{t} = \frac{S/t}{t} = \frac{S}{t^2} \]

5. **Substitute these expressions into the power equation**:
Substituting \( F \) and \( V \) into the power equation gives:
\[ P = (m \cdot a) \cdot V = m \cdot \left(\frac{S}{t^2}\right) \cdot \left(\frac{S}{t}\right) \]
Simplifying this, we have:
\[ P = m \cdot \frac{S^2}{t^3} \]

6. **Rearranging the equation**:
From the equation above, we can rearrange it to find \( S^2 \):
\[ S^2 = \frac{P \cdot t^3}{m} \]

7. **Taking the square root**:
Taking the square root of both sides gives:
\[ S = \sqrt{\frac{P \cdot t^3}{m}} = \sqrt{\frac{P}{m}} \cdot t^{3/2} \]

8. **Establishing the proportionality**:
Since \( P \) and \( m \) are constants, we can conclude that:
\[ S \propto t^{3/2} \]

### Final Answer:
The distance moved by the body in time \( t \) is proportional to \( t^{3/2} \).

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