physics

A 1000 kg car traveling is at 40.0 m/ s. To avoid hitting a second car the driver slams on his brakes.
The brakes provide a constant friction force of 8000 N. ( a) Find the minimum distance that the
brakes should be applied in order to avoid a collision with the vehicle in front? ( b) What would be
the speed of collision if the distance between the two vehicles is initially 45.0 m.

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  1. F = m a
    -8000 = 1000 a
    a = -8 m/s^2
    v = 40 - 8 t
    we want v = 0
    8 t = 40
    t = 5 seconds to stop (yikes!)

    d = 40*5 - (1/2)(8)(25)
    = 200 - 100
    = 100 meters

    for part b
    how long to for 45 m?
    45 = 40 t -4 t^2
    solve quadratic for t
    then
    v = 40 - 8 t

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  2. a. F = M*a.
    a = F/M = -8000/1000 = -8m/s^2.

    V^2 = Vo^2 + 2a*d.
    0 = 40^2 - 16d
    d = 100m. to stop.

    b. V^2 = Vo^2 + 2a*d.
    V^2 = 40^2 - 16*45 =
    V = ?.

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