# Answer to Question #16704 in Mechanics | Relativity for Erco

Question #16704

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?

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?

Expert's answer

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

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

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