PLEASE SHOW ALL WORK! :) A particle moving in the y-z plane with a velocity vect

Question

PLEASE SHOW ALL WORK! :)

A particle moving in the y-z plane with a velocity vector v = (15y+15z) m/s enters an electric field which gives it a constant acceleration. The acceleration vector is a = (-1 y -5 z) m/s2. Ignore the effects of gravity. When the particle returns to the same z-coordinate it had when it entered the electric field, how far will it be from the point at which it entered the electric field? The particle will be m from its point of entry into the electric field when it returns to the same z-coordinate.

Explanation / Answer

t = 2 * (v_z)/(a_z) = 2 * (15/5) = 6 s

r = y = 0.5 a_y t^2 + v_y t = 0.5 * (-1) * (6*6) + 15 * 6 = 72 m

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