A body of mass m travels in a straight line with a velocity `v = kx^(3//2)` where k is a constant. The work done in displacing the body from `x = 0` to x is proportional to:

Options

A. `x^(1//2)`
B. `x^2`
C. `x^3`
D. `x^(5//2)`

Explanation

To solve the problem, we need to determine how the work done in displacing a body from \( x = 0 \) to \( x \) is related to \( x \). The velocity of the body is given as \( v = kx^{3/2} \).

### Step 1: Understand the relationship between work done and kinetic energy
The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. Therefore, we can write:

\[ W = \Delta KE = KE_f - KE_i \]

Where \( KE_f \) is the final kinetic energy and \( KE_i \) is the initial kinetic energy.

### Step 2: Calculate the initial and final kinetic energy
At \( x = 0 \):
- The initial velocity \( v_i = k(0)^{3/2} = 0 \)
- Thus, the initial kinetic energy \( KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} m (0)^2 = 0 \)

At a position \( x \):
- The final velocity \( v_f = kx^{3/2} \)
- Therefore, the final kinetic energy \( KE_f = \frac{1}{2} m v_f^2 = \frac{1}{2} m (kx^{3/2})^2 \)

### Step 3: Substitute the expression for final kinetic energy
Substituting \( v_f \) into the expression for \( KE_f \):

\[ KE_f = \frac{1}{2} m (kx^{3/2})^2 = \frac{1}{2} m k^2 x^3 \]

### Step 4: Calculate the work done
Now, substituting the values of \( KE_f \) and \( KE_i \) into the work-energy equation:

\[ W = KE_f - KE_i = \frac{1}{2} m k^2 x^3 - 0 = \frac{1}{2} m k^2 x^3 \]

### Step 5: Determine the proportionality
From the expression for work done \( W = \frac{1}{2} m k^2 x^3 \), we can see that the work done \( W \) is proportional to \( x^3 \).

### Conclusion
Thus, the work done in displacing the body from \( x = 0 \) to \( x \) is proportional to \( x^3 \).

\[ \text{Answer: } W \propto x^3 \]