A proton initially moves left to right along the x-axis at a speed of 2 x 10^3 m/s. It moves into an electric field, which points in the negative x direction, and travels a distance of 0.15m before coming to rest. What acceleration magnitude does the proton experience? equations E = kq/r^2 F=ma F=qE mass of a proton = 1.67x10^-27 Kg charge of a proton = 1.602 x 10^-19 C Attempt 1 qE=ma E= (9.00x10^9 N*m^2/C^2)(1.602x10^-19 C/.15m^2 ) = 6.4x 10^-8 N/C ((1.602x10^-19 C)(6.4x10^-8 N/C))/(1.67x10^-27Kg) = a a = 6.139m/s^2 actual answer = 1.33x10^7 I realize this means that i have to do something with the initial velocity, but i can't seem to find anything that ties the velocity of a proton to its acceleration in an electric field except for maybe a= (vf-vi) / (tf-ti) but I'm not given the time. What am i missing?