Equations of motion solutions?

Question

I have the following equations of motion:

mx'' = 0
my'' = 0
mz'' = qE

How do I find the solutions to these for a particle which starts from the origin at t=0 with an initial velocity v=(v_x + v_y + v_z) ? (v is a vector, x,y,z components)

Answer 1

The double-prime ('') marks after the coordinate direction indicate that this is the second derivative of that coordinate value. The second time derivative of position is acceleration, so these are accelerations we're dealing with. Since accleration is change in velocity divided by change in time, an acceleration a over a time t produces a change in velocity equal to a*t.

mx'' = my'' = 0 means that the acceleration in the x and y directions is 0, so the velocities in those directions remain constant at v_x and v_y, respectively. mz'' = qE means that z'' = qE/m, the acceleration in the z direction. So the velocity in the z direction at time t would be v_z + (qE/m)*t.