The initial velocity given to a particle is u and accelration is given by `a=at^((3)/(2))` . What will be the velocity of particle after time t.

Options

A. `u+at^((5)/(2))`
B. `u+(2)/(5)at^((5)/(2))`
C. `u+(5)/(2)at^((5)/(2))`
D. `u+at^((3)/(2))`

Explanation

To find the velocity of a particle after time \( t \) when the initial velocity is \( u \) and the acceleration is given by \( a = a t^{\frac{3}{2}} \), we can follow these steps:

### Step 1: Understand the relationship between acceleration, velocity, and time
The acceleration \( a \) is defined as the rate of change of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Given that \( a = a t^{\frac{3}{2}} \), we can rewrite this as:
\[ \frac{dv}{dt} = a t^{\frac{3}{2}} \]

### Step 2: Rearrange the equation for integration
We can rearrange the equation to separate the variables \( v \) and \( t \):
\[ dv = a t^{\frac{3}{2}} dt \]

### Step 3: Integrate both sides
Now, we will integrate both sides. The left side integrates with respect to \( v \) and the right side integrates with respect to \( t \):
\[ \int dv = \int a t^{\frac{3}{2}} dt \]

The left side becomes:
\[ v = \int a t^{\frac{3}{2}} dt \]

For the right side, we can factor out \( a \) (a constant) and integrate:
\[ \int t^{\frac{3}{2}} dt = \frac{t^{\frac{5}{2}}}{\frac{5}{2}} = \frac{2}{5} t^{\frac{5}{2}} \]

Thus, we have:
\[ v = a \cdot \frac{2}{5} t^{\frac{5}{2}} + C \]
where \( C \) is the constant of integration.

### Step 4: Apply initial conditions
At \( t = 0 \), the initial velocity is \( u \):
\[ v(0) = u \Rightarrow C = u \]

### Step 5: Write the final expression for velocity
Substituting \( C \) back into the equation gives us:
\[ v = u + a \cdot \frac{2}{5} t^{\frac{5}{2}} \]

### Final Answer
The velocity of the particle after time \( t \) is:
\[ v = u + \frac{2}{5} a t^{\frac{5}{2}} \]

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