Electric Potential of a Proton Question

[blablasometin]

A proton's speed as it passes point A is 53,000 m/s. It follows the trajectory shown in the figure below, in which V1 = 15 V and V3 = 5 V. What is the proton's speed at point B?

Figure: http://www.webassign.net/knight/p29-44alt.gif

Homework Statement

Mass of proton: 1.67e-27
Charge of proton: 1.6e-19
Difference in Electric Potential
velocity at point A: 53000m/s
V1 = 15V
V2 = 10V
V3 = 5V

Homework Equations

KE=1/2mv^2
V=(kq)/r
V=Ed

The Attempt at a Solution

10q=1/2mv^2
v=sqrt(20*1.6e-19/1.67e-27)

 

[Alter Baron]

[removed]

 

[blablasometin]

never mind I got the solution:

KE(B) + PE(B) = KE(A) + PE(A)

(1/2)mvB^2 + qV(B) = (1/2)mvA^2 + qV(A)\

vB^2 = vA^2 + (2q/m)[V(A) - V(B)]

Answer: 68739.85m/s

I should have known it was a simple conservation of energy problem