A proton accelerates from rest in a uniform electric field of 610 N/C...?

Question

A proton accelerates from rest in a uniform electric field of 610 N/C. At some time later, its speed is 1.40 106 m/s. (a) What is the magnitude of the acceleration of the proton? (b) How long does it take the proton to reach this speed? (c) How far has it moved in this time interval? (d) What is its kinetic energy at the later time?

Help would be AMAZING. THANK YOU.

Answer 1

I don't remember the ratio of coulomb to Mole so I will leave that up to you to find.
1 Newton would accelerate a Kilogram 1 Meter/Second/Second
1 Newton would accelerate a gram 1000 meter/sec/sec
the mass of a coulomb of protons would be coulomb/mole grams
The speed 1.4 10^6 m/s is not even close to the speed of light so Newtonian laws of motion should give a good approximation of what happens.
A=mole/coulomb*1000

AT=V
T=V/A=1.4*10^6/A

D=1/2 ATT=.7*10^6*1.4*10^6/A=.98*10^12/A

Pretend A is about 10^8
Then D is about 10 Kilometers