The velocity `v` varies as `v = alpha sqrt(x)`,where x represents the displacement. At `t = 0` it is located `x = 0` Then `x prop`

Options

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

Explanation

To solve the problem where the velocity \( v \) varies as \( v = \alpha \sqrt{x} \), we will follow these steps:

### Step 1: Relate velocity to displacement
We know that velocity \( v \) is the rate of change of displacement \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} \]
Given that \( v = \alpha \sqrt{x} \), we can equate the two expressions:
\[ \frac{dx}{dt} = \alpha \sqrt{x} \]

### Step 2: Rearrange the equation
We can rearrange the equation to separate variables:
\[ \frac{dx}{\sqrt{x}} = \alpha dt \]

### Step 3: Integrate both sides
Now we will integrate both sides. The left side requires the integral of \( x^{-1/2} \):
\[ \int \frac{dx}{\sqrt{x}} = \int \alpha dt \]
The integral of \( \frac{1}{\sqrt{x}} \) is \( 2\sqrt{x} \), and the integral of \( \alpha dt \) is \( \alpha t + C \) (where \( C \) is the constant of integration):
\[ 2\sqrt{x} = \alpha t + C \]

### Step 4: Solve for \( x \)
Now, we will solve for \( x \):
\[ \sqrt{x} = \frac{\alpha t + C}{2} \]
Squaring both sides gives:
\[ x = \left(\frac{\alpha t + C}{2}\right)^2 \]

### Step 5: Determine the constant \( C \)
At \( t = 0 \), we know that \( x = 0 \):
\[ 0 = \left(\frac{\alpha \cdot 0 + C}{2}\right)^2 \]
This implies \( C = 0 \). Therefore, we can simplify our expression for \( x \):
\[ x = \left(\frac{\alpha t}{2}\right)^2 \]

### Step 6: Final expression for \( x \)
Thus, we can express \( x \) in terms of \( t \):
\[ x = \frac{\alpha^2}{4} t^2 \]

### Conclusion
From the final expression, we see that \( x \) is proportional to \( t^2 \):
\[ x \propto t^2 \]