Given velocity as a function of time

$v(t)=v_0+gt+ft^2$

We know that velocity is the time derivative of position:

$v=\frac{dx}{dt}$

Find displacement from $t=0$ to $t=1$

Since $x=0$ at $t=0$, the displacement after unit time is

$x(1)-x(0)=\int_0^1 v(t)\,dt$

Substituting $v(t)$:

$x(1)=\int_0^1 (v_0+gt+ft^2)\,dt$

Integrate term by term

$x(1)=v_0\int_0^1 dt+g\int_0^1 t\,dt+f\int_0^1 t^2\,dt$

Now,

$\int_0^1 dt=1$ $\int_0^1 t\,dt=\left[\frac{t^2}{2}\right]_0^1=\frac{1}{2}$ $\int_0^1 t^2\,dt=\left[\frac{t^3}{3}\right]_0^1=\frac{1}{3}$

Therefore,

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

Match with the options

This corresponds to:

$\boxed{\text{C: } v_0+\frac{g}{2}+\frac{f}{3}}$

Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

So, they agree.