Proton in a uniform electric field

1. The problem statement, all variables and given/known data

A proton is projected in the positive x direction into a region of uniform electric field E = (-5.80 ✕ 10^5) N/C at t = 0. The proton travels 6.50 cm as it comes to rest.



Find the initial velocity and the time it takes for the proton to stop.



2. Relevant equations







3. The attempt at a solution



The charge of the proton multiplied by the electric field must equal the mass of the proton times the acceleration



qE=ma



hence



a= (-5.556x10^13)



Constant accelecration, integrate with respect to time to get velocity



V= Vi - 5.556x10^13 t



X= 0 + Vi(t) - 2.2778 t^2



Since initial position = 0.



Using the position function to find the time it takes to stop:



.065m = (5.556x10^13)t^2 - (2.2778x10^13)t^2



solving for t, t= 4.8x10^-8 s



Plugging this back to the velocity equation where at t= 4.8x10^-8 s the final velocity = 0 therefore:



Vi=(5.556x10^13 (m/s^2))(4.8x10^-8 s)



Clearly not right, it's faster than the SOL.

Where am I going wrong?



ps. I don't like to memorize stuff so please don't refer me to the kinematic equations. I prefer to derive them by myself.