Answer to Question #16704 in Mechanics | Relativity for Erco

Question

At t = 0, a particle leaves the origin with a velocity of 12 m/s in the positive x
direction and moves in the xy plane with a constant acceleration of
(2.0ˆi + 4.0ˆj ) m/s2. At the instant the y coordinate of the particle is 18 m, what is
the x coordinate of the particle?

Answer 1

Find time of motion from displacement in y
sy=ay*t^2/2
sy=18 m
ay = 4 m/s^2
t = sqrt(2*sy/ay)
t = sqrt(2*18/4) = sqrt(9) = 3 s

Displacement in x
sx = vx*t+ax*t^2/2
vx=12 m/s, ax=2 m/s^2, t=3s
sx = 12*3+2*3^2/2= 45 m