Question

Starting from rest, a car accelerates at 3.0 m/s^2 up a hill that is inclined 5.9 ^\circ above the horizontal.
1-How far horizontally has the car traveled in 14 s?
2-How far vertically has the car traveled in 14 s?

Answer 1

distance up the hill: 1/2 *3.0*t^2

horizontally: distanceuphill*cosTheta
vertically: distanceuphill*sinTheta

Answer 2

Vo= 0m/s
Vox = Voy = 0m/s > Xo = Xy = 0
a= 3 m/s^2

x is adjacent so Ax = 3cos(5.9)
Ay = 3sin(5.9)

Use X=Xo+Vox+1/2Axt^2 to get X1. Remember, Xo = 0...