A particle starts with initial velocity 4m/s and moves such that acceleration is given by `a = sqrt(V)`
It's velocity after a displacement of `(38)/(3)` m is

Options

A. 9 m/s
B. `6^(2//3) m//s`
C. 3 m/s
D. 6 m/s

Answer

To solve the problem step by step, we will follow the concepts of kinematics and integration.

### Step 1: Understand the relationship between acceleration, velocity, and displacement.
We know that:
- Acceleration \( a = \frac{dV}{dt} \)
- Velocity \( V = \frac{dx}{dt} \)

From these, we can derive the relationship:
\[ \frac{V}{a} = \frac{dx/dt}{dV/dt} = \frac{dx}{dV} \]
This means:
\[ a = V \frac{dV}{dx} \]

### Step 2: Substitute the given acceleration into the equation.
The problem states that the acceleration \( a \) is given by:
\[ a = \sqrt{V} \]
Substituting this into our derived equation gives:
\[ V \frac{dV}{dx} = \sqrt{V} \]

### Step 3: Rearrange the equation for integration.
We can rearrange the equation:
\[ V \frac{dV}{dx} = V^{1/2} \]
Dividing both sides by \( V^{1/2} \) (assuming \( V \neq 0 \)):
\[ V^{1/2} dV = dx \]

### Step 4: Integrate both sides.
Now we will integrate both sides. The limits for \( V \) will be from the initial velocity \( 4 \, \text{m/s} \) to \( V \) (the final velocity we want to find), and the limits for \( x \) will be from \( 0 \) to \( \frac{38}{3} \, \text{m} \):
\[ \int_{4}^{V} V^{1/2} dV = \int_{0}^{\frac{38}{3}} dx \]

### Step 5: Solve the integrals.
The left side becomes:
\[ \left[ \frac{V^{3/2}}{\frac{3}{2}} \right]_{4}^{V} = \frac{2}{3} \left( V^{3/2} - 4^{3/2} \right) \]
The right side becomes:
\[ \left[ x \right]_{0}^{\frac{38}{3}} = \frac{38}{3} \]

Setting these equal gives:
\[ \frac{2}{3} \left( V^{3/2} - 8 \right) = \frac{38}{3} \]

### Step 6: Solve for \( V^{3/2} \).
Multiply both sides by \( \frac{3}{2} \):
\[ V^{3/2} - 8 = 19 \]
Thus:
\[ V^{3/2} = 27 \]

### Step 7: Solve for \( V \).
Taking both sides to the power of \( \frac{2}{3} \):
\[ V = 27^{2/3} = 9 \, \text{m/s} \]

### Final Answer:
The velocity of the particle after a displacement of \( \frac{38}{3} \) m is \( 9 \, \text{m/s} \).