1. The velocity is given as $v(t)=v_0+gt+Ft^2.$

2. Displacement is obtained by integrating velocity with respect to time: $x(t)-x(0)=\int_0^t v(t)\,dt.$

3. Since $x(0)=0$, the displacement after time $t=1$ is $x(1)=\int_0^1 (v_0+gt+Ft^2)\,dt.$

4. Integrate term by term: $\int_0^1 v_0\,dt=v_0,$ $\int_0^1 gt\,dt=g\left[\frac{t^2}{2}\right]_0^1=\frac{g}{2},$ $\int_0^1 Ft^2\,dt=F\left[\frac{t^3}{3}\right]_0^1=\frac{F}{3}.$

5. Therefore, $x(1)=v_0+\frac{g}{2}+\frac{F}{3}.$

6. Comparing with the options:

   • A: $v_0+g+F$ ❌
   • B: $v_0+\frac{g}{2}+\frac{F}{3}$ ✅
   • C: $v_0+2g+3F$ ❌
   • D: $v_0+\frac{g}{2}+F$ ❌

Hence, the correct answer is Option B.