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

Options

A.
B.
C.
D.

Answer

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

### Step 1: Understand the relationship between power, work, and time
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 the time taken. Rearranging gives us:
\[ W = P \cdot t \]
This is our Equation (1).

### Step 2: Apply the work-energy theorem
According to the work-energy theorem, the work done on an object is equal to the change in its kinetic energy. Since the body starts from rest, the initial kinetic energy is zero. Thus, the work done can be expressed as:
\[ W = \Delta KE = \frac{1}{2} m v^2 - 0 = \frac{1}{2} m v^2 \]
This is our Equation (2).

### Step 3: Equate the two expressions for work
From Equations (1) and (2), we can set them equal to each other:
\[ \frac{1}{2} m v^2 = P \cdot t \]
Rearranging this gives:
\[ v^2 = \frac{2Pt}{m} \]
Taking the square root of both sides, we find:
\[ v = \sqrt{\frac{2Pt}{m}} \]
This is our Equation (A).

### Step 4: Relate displacement to time
Velocity (v) is defined as the rate of change of displacement (s) with respect to time (t):
\[ v = \frac{ds}{dt} \]
Substituting Equation (A) into this gives:
\[ \frac{ds}{dt} = \sqrt{\frac{2P}{m}} \cdot t^{1/2} \]
Rearranging gives:
\[ ds = \sqrt{\frac{2P}{m}} \cdot t^{1/2} dt \]

### Step 5: Integrate to find displacement
Integrating both sides, we have:
\[ s = \int ds = \int \sqrt{\frac{2P}{m}} \cdot t^{1/2} dt \]
The constant can be factored out:
\[ s = \sqrt{\frac{2P}{m}} \cdot \int t^{1/2} dt \]
The integral of \( t^{1/2} \) is:
\[ \int t^{1/2} dt = \frac{2}{3} t^{3/2} \]
Thus, we have:
\[ s = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} t^{3/2} \]
This is our Equation (B).

### Step 6: Find the ratio of displacement to velocity
Now we need to find the ratio \( \frac{s}{v} \):
Using Equations (B) and (A):
\[ \frac{s}{v} = \frac{\sqrt{\frac{2P}{m}} \cdot \frac{2}{3} t^{3/2}}{\sqrt{\frac{2P}{m}} \cdot t^{1/2}} \]
The \( \sqrt{\frac{2P}{m}} \) terms cancel out:
\[ \frac{s}{v} = \frac{2}{3} t^{3/2 - 1/2} = \frac{2}{3} t^{1} \]
Thus, we find:
\[ \frac{s}{v} = \frac{2}{3} t \]

### Conclusion
The ratio of displacement to velocity \( \frac{s}{v} \) varies linearly with time \( t \).