Answer
(a) $2.21 \times 10^4 m/s^2$
(b) See the graphs.
Work Step by Step
For (a):
From Coulomb's law, we know that the force between two charges at rest is $F = \frac{1}{4 \pi \epsilon_{0}}\frac{q_1 q_2}{r^2} $.
And from Newton's second law, we know that $F = ma$.
Then clearly, $ a = F/m = \frac{1}{4 \pi \epsilon_{0}}\frac{q_1 q_2}{mr^2} = (9 \times 10^9)\frac{(1.6\times10^{-19})^2}{(1.67 \times 10^{-27}) (0.00250)^2} = 2.21 \times 10^4 m/s^2$
For (b):
We note that the protons repel each other, and the acceleration of the moving proton $a \propto \frac{1}{r^2}$. This means that as the protons get farther apart, the acceleration slowly decreases with time.
As the protons will continue to repel, the velocity will keep increasing, but at a slower rate.
