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, we need to determine how the distance moved by a body, when moved by a machine delivering constant power, is related to time \( t \). Let's go through the solution step by step.

### Step 1: Understand the relationship between power, work, and time
Power \( P \) is defined as the rate at which work is done. Mathematically, this can be expressed as:
\[ P = \frac{W}{t} \]
where \( W \) is the work done and \( t \) is the time taken. Rearranging this gives us:
\[ W = P \cdot t \]

### 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. If we assume the object starts from rest, the work done can be expressed as:
\[ W = \Delta KE = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the object and \( v \) is its final velocity.

### Step 3: Equate the two expressions for work
From Step 1, we have:
\[ W = P \cdot t \]
From Step 2, we have:
\[ W = \frac{1}{2} m v^2 \]
Setting these two expressions for work equal to each other gives:
\[ P \cdot t = \frac{1}{2} m v^2 \]

### Step 4: Solve for velocity \( v \)
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = \frac{2Pt}{m} \]
Taking the square root gives us the velocity:
\[ v = \sqrt{\frac{2Pt}{m}} \]

### Step 5: Relate velocity to displacement
Velocity \( v \) is defined as the rate of change of displacement \( s \) with respect to time \( t \):
\[ v = \frac{ds}{dt} \]
Substituting the expression for \( v \) from Step 4:
\[ \frac{ds}{dt} = \sqrt{\frac{2Pt}{m}} \]

### Step 6: Separate variables and integrate
To find displacement \( s \), we can separate the variables:
\[ ds = \sqrt{\frac{2P}{m}} \cdot \sqrt{t} \, dt \]
Now, we integrate both sides:
\[ s = \int ds = \sqrt{\frac{2P}{m}} \int \sqrt{t} \, dt \]
The integral of \( \sqrt{t} \) is:
\[ \int \sqrt{t} \, dt = \frac{2}{3} t^{3/2} + C \]
Thus, we have:
\[ s = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} t^{3/2} + C \]

### Step 7: Conclude the relationship
Ignoring the constant of integration (since we are interested in the proportionality), we find that:
\[ s \propto t^{3/2} \]
This means that the distance moved by the body is proportional to \( t^{3/2} \).

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

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