A body is moved from rest along a straight line by a machine delivering constant power. The ratio of displacement and velocity `(s//v)` varies with time `t` as

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Answer

To solve the problem, we need to analyze the relationship between displacement (s), velocity (v), and time (t) when a body is moved from rest by a machine delivering constant power (P).

### Step-by-Step Solution:

1. **Understanding Power**:
Power (P) is defined as the rate of doing work. Mathematically, it can be expressed as:
\[ P = \frac{W}{t} \]
where \(W\) is the work done and \(t\) is time.

2. **Work Done**:
Since the machine delivers constant power, the work done over time \(t\) can be expressed as:
\[ W = Pt \]

3. **Kinetic Energy Relation**:
The work done on the body is converted into kinetic energy. Hence, we can write:
\[ W = \frac{1}{2} mv^2 \]
Setting the two expressions for work equal gives:
\[ Pt = \frac{1}{2} mv^2 \]

4. **Solving for Velocity**:
Rearranging the equation to solve for \(v^2\):
\[ v^2 = \frac{2Pt}{m} \]
Taking the square root gives:
\[ v = \sqrt{\frac{2P}{m}} \cdot t^{1/2} \]
Let \(k = \sqrt{\frac{2P}{m}}\), then:
\[ v = kt^{1/2} \]

5. **Finding Displacement**:
The velocity \(v\) can also be expressed as the derivative of displacement \(s\) with respect to time \(t\):
\[ v = \frac{ds}{dt} \]
Substituting our expression for \(v\):
\[ \frac{ds}{dt} = kt^{1/2} \]
Rearranging gives:
\[ ds = kt^{1/2} dt \]

6. **Integrating to Find Displacement**:
Integrating both sides to find \(s\):
\[ s = \int kt^{1/2} dt \]
The integral of \(t^{1/2}\) is:
\[ s = k \cdot \frac{2}{3} t^{3/2} + C \]
Since the body starts from rest, we can set the constant \(C = 0\):
\[ s = \frac{2k}{3} t^{3/2} \]

7. **Finding the Ratio \( \frac{s}{v} \)**:
Now, we need to find the ratio of displacement to velocity:
\[ \frac{s}{v} = \frac{\frac{2k}{3} t^{3/2}}{kt^{1/2}} \]
Simplifying gives:
\[ \frac{s}{v} = \frac{2}{3} t \]

8. **Conclusion**:
The ratio of displacement to velocity \( \frac{s}{v} \) varies linearly with time \(t\) as:
\[ \frac{s}{v} = \frac{2}{3} t \]

### Final Answer:
The ratio of displacement and velocity \( \frac{s}{v} \) varies with time \(t\) as \( \frac{s}{v} = \frac{2}{3} t \).