Proton accelerated thru potential difference - find final speed

Homework Statement

A proton is accelerated from rest through a potential difference of 107 V. Calculate the final speed of this proton.

Homework Equations

I'm not really sure which equations to use, but I figured I give all I tried...
V=-Ed
V=PE/q
PE=-qEd
and I think this applies...v(f)^2= v(i)^2 + 2ad

The Attempt at a Solution

I really can't figure out what equation i should use. What I did was..
1)used the equation V = Ed
2) becomes V= Kc(q/d^2)d
3) becomes V=Kc(q/d)
4)107= 8.99x10^9 (1.602x10^-19)/d
5)Solved for d and found it was 1.34598 x 10^-11
6) used the equation v(f)^2= v(i)^2 + 2ad
7) became v = sqrt (2a (1.34598x10^-11)
....but I don't know what "a" is so I can't solve. Any help would be greatly appreciated. I gave this a shot and really couldn't figure it out. Thanks, everyone.

Homework Statement

A proton is accelerated from rest through a potential difference of 107 V. Calculate the final speed of this proton.

Homework Equations

I'm not really sure which equations to use, but I figured I give all I tried...
V=-Ed
V=PE/q
PE=-qEd
and I think this applies...v(f)^2= v(i)^2 + 2ad

The Attempt at a Solution

I really can't figure out what equation i should use. What I did was..
1)used the equation V = Ed
2) becomes V= Kc(q/d^2)d
3) becomes V=Kc(q/d)
4)107= 8.99x10^9 (1.602x10^-19)/d
5)Solved for d and found it was 1.34598 x 10^-11
6) used the equation v(f)^2= v(i)^2 + 2ad
7) became v = sqrt (2a (1.34598x10^-11)
....but I don't know what "a" is so I can't solve. Any help would be greatly appreciated. I gave this a shot and really couldn't figure it out. Thanks, everyone.

Try energy method.
Find relationship between initial energy, and final energy.
What is the initial energy?, and what's the final energy?

Yes, by far the easiest method is to just use energy conservation.

Proton KE = potential difference

Don't forget that 1eV = 1.60E-19 J

Since the potential difference is given (107 V), you have to find the kinetic energy that it would take for that to happen. Set up a ratio like this: (1 eV)/(1.60E-19 J)=(107 V)/(x). Solve for x. This value, x, is the kinetic energy (in J) that this would take.

Now, use 1/2mv^2=KE. Since you have the KE to be x, solve for v.

v=sqrt(2KE/m)

Now, use 1/2mv^2=KE. Since you have the KE to be x, solve for v.

v=sqrt(2KE/m)

what is "m"?
what is "sqrt"?

what is "m"?
what is "sqrt"?

m: meters
sqrt: square root of