The velocity of particle is `v=v_(0)+"gt"+ft^(2)`. If its position is `x=0` at `t=0` then its displacement after unit time `(t=1)` is

Options

A. (a)`v_0 +(g//2)+ (f//3) `
B. (b)` v_0 +g+f`
C. (c)` v_0+(g//2)+f`
D. (d)` v_0+2g +3f`

Explanation

To solve the problem, we need to find the displacement of a particle after 1 second, given its velocity function and initial conditions. The velocity of the particle is given by:

\[ v = v_0 + gt + ft^2 \]

Where:
- \( v_0 \) is the initial velocity,
- \( g \) is the acceleration due to gravity,
- \( f \) is a constant,
- \( t \) is the time.

### Step-by-step Solution:

1. **Understand the relationship between velocity and displacement**:
Velocity is the rate of change of displacement with respect to time. Therefore, we can express displacement \( x \) as the integral of velocity over time.

\[ x(t) = \int v(t) \, dt \]

2. **Substitute the velocity function into the integral**:
We substitute the given velocity function into the integral.

\[ x(t) = \int (v_0 + gt + ft^2) \, dt \]

3. **Integrate the velocity function**:
Now we perform the integration term by term.

\[ x(t) = \int v_0 \, dt + \int gt \, dt + \int ft^2 \, dt \]

- The integral of \( v_0 \) with respect to \( t \) is \( v_0 t \).
- The integral of \( gt \) with respect to \( t \) is \( \frac{g}{2} t^2 \).
- The integral of \( ft^2 \) with respect to \( t \) is \( \frac{f}{3} t^3 \).

Therefore, we have:

\[ x(t) = v_0 t + \frac{g}{2} t^2 + \frac{f}{3} t^3 + C \]

Here, \( C \) is the constant of integration.

4. **Apply the initial condition**:
We know that at \( t = 0 \), the position \( x = 0 \). Therefore, we can find \( C \):

\[ x(0) = v_0(0) + \frac{g}{2}(0)^2 + \frac{f}{3}(0)^3 + C = 0 \]

This implies \( C = 0 \).

5. **Final expression for displacement**:
Thus, the displacement function simplifies to:

\[ x(t) = v_0 t + \frac{g}{2} t^2 + \frac{f}{3} t^3 \]

6. **Calculate displacement at \( t = 1 \)**:
Now we substitute \( t = 1 \) into the displacement equation:

\[ x(1) = v_0(1) + \frac{g}{2}(1)^2 + \frac{f}{3}(1)^3 \]

Simplifying this gives:

\[ x(1) = v_0 + \frac{g}{2} + \frac{f}{3} \]

### Final Answer:
The displacement after 1 second is:

\[ x(1) = v_0 + \frac{g}{2} + \frac{f}{3} \]