What is the speed of a proton that has been accelerated from rest through a potential difference of -1300 v ?

The lack of info given to us requires that we use an energy approach. We start with electrical energy, U = -qV, where U is potential electrical energy, q is charge (which for a proton is e, which equals 1.6*10⁻¹⁹ C) and V is our potential difference. We can then use conservation of energy, assuming all our energy is converted to motion, so kinetic energy: K = 1/2m*v², where K is kinetic energy, m is the mass of the proton (1.67*10⁻²⁷ kg) and v is the velocity of the proton. We can set these energy terms equal to each other to get -qV = 1/2m*v², then we can solve for v, v = sqrt(-2qV/m), plug in the numbers, v = sqrt(-2(1.6*10⁻¹⁹)*(-1300)/(1.67*10⁻²⁷)) = 499,100 m/s, or about 1/10,000 the speed of light, which is well under the range of worrying about special relativity, so our method was correct.