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What physics will you use to find the speed?

  1. Nov 11, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose a car starts from rest and rolls down a hill that is 50 m high. You are going to find the speed of the car at the bottom of the hill. What physics will you use to find the speed? Now find the speed.

    2. Relevant equations
    KE= 1/2mv^2
    PE= mgh
    Speed= distance x time
    3. The attempt at a solution
    I am not sure where to start since there is so little information. 50m (height) x 9.8 m/s^2 (gravity)= 500 m^2/s^ but is this PE? KE? I don't have mass so I don't think its PE, and I'm sure it's not KE.
    What I know: its initial velocity is 0, height is 50m, since it is "falling" it is under the influence of gravity 9.8 m/s^2.
    If it is falling 50 m at 9.8 m/s that's 5.102 seconds "falling"? So the speed is 5.102s x 50m = 255.01 m/s? Is this correct? Am I along the right lines?
  2. jcsd
  3. Nov 11, 2008 #2
    This is a conservation problem. Choose the bottom of the hill as h=0. This will make solving for the unknown easy, even though you can choose h=0 at the top if you wish.

    So, initially the car is at rest, so it has zero kinetic energy. Since I took h=0 at the bottom, the car has gravitational potential energy at the top. At the bottom, the car will have some final kinetic energy and zero gravitational potential energy since h=0 there.

    So simply set up your equations and solve for v:

    K_o + U_o = K_f + U_f \hfill \\
    U_g = K_f \hfill \\
    mgh = \tfrac{1}
    {2}mv^2 \hfill \\
    gh = \tfrac{1}
    {2}v^2 \hfill \\
    v = \sqrt {2(9.8\tfrac{m}
    {{s^2 }})(50m)} \hfill \\
    v \approx 31.3\tfrac{m}
    {s} \hfill \\
  4. Nov 11, 2008 #3
    I think I am getting however, how do you take out the 1/2 from the v^2 side of the equation? I don't see where it went.
  5. Nov 11, 2008 #4
    or are both sides multiplied by 2?
  6. Nov 11, 2008 #5
    Yes. Then you are taking the square root of 2gh to get v. :smile:
  7. Nov 11, 2008 #6
    Ok, I really think I have it! When you divide by 2, it becomes 2gh=v^2. The v^2 becomes v=square root sign. Then, under the square root sign you plug in the 2 (from the previous division), gravity, and the height! This is slowly making sense.
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