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Kinematics on an inclined plane

  1. Sep 26, 2010 #1


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    1. The problem statement, all variables and given/known data
    In anticipation of a long 7° upgrade, a bus driver accelerates at a constant rate of 3ft/s2 while still on a level section of the highway. Knowing that the speed of the bus is 60mi/h as it begins to climb the grade and that the driver does not change the setting of his throttle or shift gears, determine the distance traveled by the bus up the grade when its speed has decreased to 50mi/h.

    Answer: 0.242mi (1,278ft)

    2. Relevant equations
    g = 32.3ft/s2
    θ = 7°
    Initial acceleration = 3ft/s2
    vi = 60mi/hr = 88ft/s
    vf = 50mi/hr = 73.333ft/s
    xi = 0ft
    xf = ??? ft
    3. The attempt at a solution

    [tex]\sum F_{x}=-mgsin\theta=ma[/tex]

    [tex]\sum F_{y}=N-mgcos\theta=0[/tex]

    Using the Fx equation, acceleration on the plane = -gsin(θ). The problem doesn't state anything about friction so my assumption is that friction doesn't play into the solution.

    Acceleration starts at 3ft/s2 at xi and decreases by a constant -gsin(θ) as the bus travels up the grade.

    I have a formula for acceleration when it depends on position but it produces the wrong answer.




    Last edited: Sep 26, 2010
  2. jcsd
  3. Sep 26, 2010 #2

    Doc Al

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    Staff: Mentor

    You're off by a factor of 2 in that last term:
    V2f = V2i + 2ax
  4. Sep 26, 2010 #3


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    Yeppers... That'll do it. And now the math works out and I have the correct answer. Thanks Doc!
  5. Sep 30, 2012 #4
    i'm having trouble following the math involved. Can someone set me straight?

    Vf^2 = V_0^2 + 2aD
    73.3^2 - 80^2 = 2aD
    -1022.27 = 2(3-9.81sin(7)x_f
    -511.1355 = (3-9.81(sin7))x_f
    -511.1355 = (3-1.1955)x_f
    -511.1355 = 1.8044*x_f
    x_f = -283.26 feet.

    What am i missing?
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