Solution:

Given the velocity function:

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

We can find displacement by integrating the velocity function with respect to time. The displacement \(x(t)\) is the integral of velocity:

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

Integrating:

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

Since \(x = 0\) at \(t = 0\), we know \(C = 0\). Therefore, the position function is:

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

Now, for \(t = 1\):

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

Simplifying:

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

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

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