phy

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?

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asked by elin
  1. distance up the hill: 1/2 *3.0*t^2

    horizontally: distanceuphill*cosTheta
    vertically: distanceuphill*sinTheta

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    posted by bobpursley
  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...

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    posted by yara

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