A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity that varies as `v=psqrtx`. The displacement of the particle varies with time as (where, p is constant)

Options

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

Answer

To solve the problem step by step, we need to find the displacement \( x \) as a function of time \( t \) given that the velocity \( v \) of the particle varies with displacement \( x \) as \( v = p \sqrt{x} \).

### Step 1: Write the relationship between velocity, displacement, and time
We know that velocity \( v \) can be expressed as the derivative of displacement with respect to time:
\[ v = \frac{dx}{dt} \]

### Step 2: Substitute the expression for velocity
Given that \( v = p \sqrt{x} \), we can substitute this into the equation:
\[ \frac{dx}{dt} = p \sqrt{x} \]

### Step 3: Rearrange the equation for separation of variables
We can rearrange this equation to separate the variables \( x \) and \( t \):
\[ \frac{dx}{\sqrt{x}} = p \, dt \]

### Step 4: Integrate both sides
Now, we will integrate both sides. The left side requires the integral of \( \frac{1}{\sqrt{x}} \):
\[ \int \frac{dx}{\sqrt{x}} = \int p \, dt \]
The integral of \( \frac{1}{\sqrt{x}} \) is \( 2\sqrt{x} \), and the integral of \( p \) with respect to \( t \) is \( pt + C \) (where \( C \) is the constant of integration):
\[ 2\sqrt{x} = pt + C \]

### Step 5: Solve for \( x \)
Now, we will solve for \( x \):
\[ \sqrt{x} = \frac{pt + C}{2} \]
Squaring both sides gives:
\[ x = \left(\frac{pt + C}{2}\right)^2 \]
\[ x = \frac{(pt + C)^2}{4} \]

### Step 6: Determine the constant of integration \( C \)
We know that at \( t = 0 \), \( x = 0 \):
\[ 0 = \frac{(p \cdot 0 + C)^2}{4} \]
This implies \( C = 0 \). Therefore, we can simplify our equation for \( x \):
\[ x = \frac{(pt)^2}{4} \]
\[ x = \frac{p^2 t^2}{4} \]

### Final Result
Thus, the displacement \( x \) as a function of time \( t \) is:
\[ x = \frac{p^2}{4} t^2 \]