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Math Help - Physics 11 Questions - Halfway Solved

  1. #1
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    Physics 11 Questions - Halfway Solved

    The Latex in the Physics forums are a bit messed, so I'm crossing my fingers and hoping this works.

    In a brake test on dry asphalt, a Chevvy Camaro, travelling with an initial speed of 26.8 m/s, can stop without skidding after moving 39.3m. The mass of the Camaro, including the driver, is 1580 kg.

    a. Determine the magnitudde of the average acceleration of the car during the non-skidding braking.
    b. Calculate the magnitude of the average stopping friction force.
    c. Assume that the test is now done with skidding on dry asphalt. Determine the magnitude of the kinetic friction, the magnitude of the average acceleration, and stopping distance during the skid. Compare this situation with the non-skid test.
    d. Repeat (c) for the car skidding on ice.
    e. In the skidding tests, does the mass of the car have an effect on the average acceleration? Explain, using examples.
    a.
    {v_f}^2={v_i}^2+2a\Delta{d}
    0=(26.8m/s)^2+2(a)(39.3m)
    0=718.24+78.6a
    a=9.14m/s^2

    b.
    Using Newton's law...
    F=ma
    1580kg*9.14m/s^2=14441.2\approx1.44*10^4N

    c.
    Magnitude of kinetic friction...
    \mu{K} of rubber on dry asphalt is 1.07.
    1.07*1580kg*9.8N/kg\approx1.66*10^4N
    What is the magnitude of average acceleration? I cannot seem to derive a formula for it...and I seem to be stuck with the distance.

    d.
    Magnitude of kinetic friction...
    \mu{K} of rubber on ice is 0.005.
    0.005*1580kg*9.8N/kg\approx8.0*10^1N
    I'm having the same problem as the previous for magnitude of average acceleration and stopping distance...

    e.
    I'm pretty positive it does, but I can't be sure--I have yet to find the average acceleration.


    Any help is GREATLY appreciated! Thanks!

    -Nate
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  2. #2
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    Quote Originally Posted by nathan02079 View Post
    The Latex in the Physics forums are a bit messed, so I'm crossing my fingers and hoping this works.

    a.
    {v_f}^2={v_i}^2+2a\Delta{d}
    0=(26.8m/s)^2+2(a)(39.3m)
    0=718.24+78.6a
    a=9.14m/s^2

    ok

    b.
    Using Newton's law...
    F=ma
    1580kg*9.14m/s^2=14441.2\approx1.44*10^4N

    ok

    c.
    Magnitude of kinetic friction...
    \mu{K} of rubber on dry asphalt is 1.07.
    1.07*1580kg*9.8N/kg\approx1.66*10^4N
    What is the magnitude of average acceleration? I cannot seem to derive a formula for it...and I seem to be stuck with the distance.

    \textcolor{red}{ma = f_k}

    \textcolor{red}{a = \frac{\mu mg}{m}}

    \textcolor{red}{a = \mu \cdot g}

    d.
    Magnitude of kinetic friction...
    \mu{K} of rubber on ice is 0.005.
    0.005*1580kg*9.8N/kg\approx8.0*10^1N
    I'm having the same problem as the previous for magnitude of average acceleration and stopping distance...

    e.
    I'm pretty positive it does, but I can't be sure--I have yet to find the average acceleration.


    Any help is GREATLY appreciated! Thanks!

    -Nate
    ...
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  3. #3
    Junior Member
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    Thank you!
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