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Math Help - More Forces

  1. #1
    Junior Member cinder's Avatar
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    More Forces

    An advertisement claims that a particular automobile can "stop on a dime". What net force would actually be necessary to stop an automobile of mass 940 kg traveling initially at a speed of 38.0 km/h in a distance equal to the diameter of a dime, which is 1.8 cm?

    I'm not even sure how to begin with this. Maybe if I can find an appropriate acceleration, I can find the force needed to do this?
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  2. #2
    MHF Contributor
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    Let F = force to stop the car.

    Then, F = ma
    F = 940*a ----------------(1)

    What is "a"?
    It is deceleration, of course.

    Given:
    Initial velocity, Vo = 38km/hr = (38km/1hr)(1000m/1km)(1hr/3600sec) = 10.555.... m/sec
    Final velocity, Vt = 0
    Distance travelled, s = 1.8 cm = (1.8cm)(1m/100cm) = 0.018 m

    Distance = (average velocity) * time
    s = (1/2)(Vo +Vt)*t
    0.018 = (1/2)(10.555... +0)*t
    t = 0.018 / (10.555... /2) = 0.00341 sec.

    Velocity at t = (initial velocity) + a*t
    Vt = Vo +at
    0 = 10.555... +a(0.00341)
    a = -10.555... /0.00341 = -3095.47 m/sec/sec

    Hence,
    F = ma
    F = 940(-3095.47) = -2,909,742 newtons <---almost 3 million newtons!

    Negative, because F is against the direction of the car.
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  3. #3
    Junior Member cinder's Avatar
    Joined
    Feb 2006
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    I think I get it.

    I'll go over it again and let you know if I have any problems.

    Thanks!
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