The velocity of a particle is `v = v_0 + g t + ft^2`. If its position is `x = 0` at `t = 0`, then its displacement after unit time `(t = 1)` is.

Options

A. `v_0 + g//2 + f//3`
B. `v_0 + g + f`
C. `v_0 + g//2 + f`
D. `v_0 + 2g + 3f`

Explanation

To find the displacement of the particle after unit time (t = 1) given the velocity function \( v = v_0 + gt + ft^2 \), we will follow these steps:

### Step 1: Understand the relationship between velocity and displacement
The velocity \( v \) is the derivative of displacement \( x \) with respect to time \( t \). Therefore, we can express the displacement as the integral of the velocity function over time.

### Step 2: Set up the integral for displacement
Since the position \( x = 0 \) at \( t = 0 \), the displacement \( x(t) \) at any time \( t \) can be found by integrating the velocity function from 0 to \( t \):

\[ x(t) = \int_0^t v(t') \, dt' \]

### Step 3: Substitute the velocity function into the integral
Substituting the given velocity function into the integral, we have:

\[ x(t) = \int_0^t (v_0 + gt' + ft'^2) \, dt' \]

### Step 4: Evaluate the integral
Now we can evaluate the integral:

\[ x(t) = \int_0^t v_0 \, dt' + \int_0^t gt' \, dt' + \int_0^t ft'^2 \, dt' \]

Calculating each term separately:

1. **First term**:
\[ \int_0^t v_0 \, dt' = v_0 t \]

2. **Second term**:
\[ \int_0^t gt' \, dt' = g \left[\frac{t'^2}{2}\right]_0^t = g \frac{t^2}{2} \]

3. **Third term**:
\[ \int_0^t ft'^2 \, dt' = f \left[\frac{t'^3}{3}\right]_0^t = f \frac{t^3}{3} \]

### Step 5: Combine the results
Combining all the terms, we get:

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

### Step 6: Substitute \( t = 1 \)
Now, we substitute \( t = 1 \) to find the displacement after unit time:

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

This simplifies to:

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

### Final Result
Thus, the displacement after unit time \( (t = 1) \) is:

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