A particle located at `x = 0` at time `t = 0`, starts moving along with the positive ` x-direction` with a velocity 'v' that varies as ` v = a sqrt(x)`. The displacement of the particle varies with time as

Options

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

Answer

To solve the problem, we need to find the displacement of a particle whose velocity varies with position according to the equation \( v = a \sqrt{x} \). The steps to derive the displacement as a function of time are as follows:

### Step 1: Write the relationship between velocity and displacement
The velocity \( v \) is defined as the rate of change of displacement \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} \]
Given that \( v = a \sqrt{x} \), we can equate the two expressions:
\[ \frac{dx}{dt} = a \sqrt{x} \]

### Step 2: Separate variables
To solve the differential equation, we can separate the variables \( x \) and \( t \):
\[ \frac{dx}{\sqrt{x}} = a \, dt \]

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

### Step 4: Determine the constant of integration
At \( t = 0 \), the particle is at \( x = 0 \):
\[ 2\sqrt{0} = a(0) + C \implies C = 0 \]
Thus, we have:
\[ 2\sqrt{x} = at \]

### Step 5: Solve for \( x \)
Now, we can solve for \( x \):
\[ \sqrt{x} = \frac{at}{2} \implies x = \left(\frac{at}{2}\right)^2 = \frac{a^2 t^2}{4} \]

### Conclusion
The displacement of the particle as a function of time is given by:
\[ x(t) = \frac{a^2}{4} t^2 \]

This shows that the displacement varies with the square of time, which indicates that the motion is uniformly accelerated.