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

Options

A. 16 m/s
B. 112 m/s
C. `(112)/(5) m/s`
D. `81 m//s`

Answer

To solve the problem step by step, we will follow the given information and equations related to kinematics.

### Step 1: Understand the relationship between acceleration and velocity
We are given that the acceleration \( a \) of the particle is defined as:
\[ a = \sqrt{v} \]
where \( v \) is the velocity of the particle.

### Step 2: Relate acceleration to velocity
Acceleration can also be expressed as the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Thus, we can equate the two expressions for acceleration:
\[ \frac{dv}{dt} = \sqrt{v} \]

### Step 3: Rearrange the equation for integration
Rearranging the equation gives us:
\[ \frac{dv}{\sqrt{v}} = dt \]
This can be rewritten as:
\[ v^{-\frac{1}{2}} dv = dt \]

### Step 4: Integrate both sides
Now we will integrate both sides. The left side will be integrated with respect to \( v \), and the right side will be integrated with respect to \( t \):
\[ \int v^{-\frac{1}{2}} dv = \int dt \]

### Step 5: Perform the integration
The integral of \( v^{-\frac{1}{2}} \) is:
\[ 2v^{\frac{1}{2}} \]
And the integral of \( dt \) is:
\[ t \]
Thus, we have:
\[ 2v^{\frac{1}{2}} = t + C \]
where \( C \) is the constant of integration.

### Step 6: Determine the constant of integration
To find \( C \), we use the initial condition. At \( t = 0 \), the initial velocity \( v_0 = 4 \, \text{m/s} \):
\[ 2(4)^{\frac{1}{2}} = 0 + C \]
Calculating gives:
\[ 2(2) = C \]
So, \( C = 4 \).

### Step 7: Substitute back the constant
Now substituting \( C \) back into the equation:
\[ 2v^{\frac{1}{2}} = t + 4 \]

### Step 8: Solve for velocity \( v \)
We need to find the velocity after \( t = 4 \) seconds:
\[ 2v^{\frac{1}{2}} = 4 + 4 \]
\[ 2v^{\frac{1}{2}} = 8 \]
Dividing both sides by 2:
\[ v^{\frac{1}{2}} = 4 \]
Now squaring both sides gives:
\[ v = 16 \, \text{m/s} \]

### Final Answer
The velocity of the particle after 4 seconds is:
\[ v = 16 \, \text{m/s} \]

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