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Math Help - A few more physics problems I need help with...

  1. #1
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    A few more physics problems I need help with...

    1.) A car with an initial velocity of 3.5 m/s coasts with the engine off and the transmission in neutral. If the comined effect of air resistance and rolling friction causes a deceleration of 0.50 m/s, after how long a time will the car come to a complete stop?

    2.) A car accelerates from rest at a constant rate of 2.0 m/s for 5.0 s. What is the speed of the car at the end of that time? (b) How far does the car travel in this time?

    3.) An experimental rocket car starting from rest reaches a speed of 560 km/h after a straight 400-m run on a road. Assuming that the acceleration is constant, (a) what was the time of the run, and (b) what is the magnitude of the acceleration?
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  2. #2
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    Quote Originally Posted by John 5 View Post
    1.) A car with an initial velocity of 3.5 m/s coasts with the engine off and the transmission in neutral. If the comined effect of air resistance and rolling friction causes a deceleration of 0.50 m/s, after how long a time will the car come to a complete stop?
    At a constant decelleration of 0.5 m/s^2 it will take 7s for the velocity to
    change by 3.5 m/s, which is what will be needed to bring the car to a stop.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by John 5 View Post
    1.) 2.) A car accelerates from rest at a constant rate of 2.0 m/s for 5.0 s. What is the speed of the car at the end of that time? (b) How far does the car travel in this time?
    5 seconds at an acceleration of 2 m/s^2 gives a velocity of 5x2=10m/s

    Starting from rest, under constant accelleration a, the distance travelled in
    t seconds is:

    s = (a t^2)/2 = 25 m.

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by John 5 View Post
    1.) 3.) An experimental rocket car starting from rest reaches a speed of 560 km/h after a straight 400-m run on a road. Assuming that the acceleration is constant, (a) what was the time of the run, and (b) what is the magnitude of the acceleration?
    Starting from rest, under a constant accelleration a for a time t the distance
    run is:

    s = (a t^2)/2

    and a speed of:

    v = a t

    (with every thing in m, m/s, and m/s^2)

    So in this case:

    560*1000/3600 = a t,

    and:

    400 = (a t^2)/2 = (a t) t/2 = (560*1000/3600) t/2

    so:

    t = (400*2)/(5600*1000/3600) ~= 5.14 s,

    and:

    a = (560*1000/3600)/5.14 ~= 30.26 m/s^2

    RonL
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  5. #5
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    Thanks for the quick responses... you are SUCH a great asset to tMF.
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