The voltage generates a force over the electron, this force is the one we use to calculate the acceleration
[tex]\begin{gathered} V=\frac{Fd}{q} \\ 1485V\cdot1.602\cdot10^{-19}=Fd=2.38\cdot10^{-16} \end{gathered}[/tex][tex]\begin{gathered} Fd=m(a\cdot d) \\ m(ad)=2.38\cdot10^{-16} \\ ad=\frac{2.38\cdot10^{-16}}{9.109\cdot10^{-31}}=2.612\cdot10^{14} \end{gathered}[/tex][tex]\begin{gathered} vf^2=vi^2+2ad \\ vf=\sqrt{2\cdot2.612\cdot10^{14}} \\ vf=22.85\cdot10^6m/s \end{gathered}[/tex]First, we apply the energy equation to find the Fd result based on voltage. Then we find the acceleration times distance. Finally, we find the speed with a kinematic equation.
Now I'm going to find the percentage of the speed of light.
[tex]\frac{22.85\cdot10^6}{3\cdot10^8}=0.0762=7.62\%[/tex]