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Newton's Laws: Distance of a car accelerating, then slamming on the breaks.

  1. Mar 11, 2012 #1
    A car which is originally at rest accelerates down a road with a net force of 32000 N acting on a 1980 kg car. At t = 12 seconds the driver slams on the brakes to avoid hitting a deer. Calculate the distance traveled by the car.
    Coefficient of kinetic friction for the rubber on pavement is assumed to be .8


    F⃗ net=ΣF⃗ =ma⃗ (a = F/m)
    fk=μkN

    v=v0+at
    x=x0+v0t+(1/2)at2
    v2=v20+2aΔx
    a = Δv/Δt
    blablahblah

    Anyone who can tell me how to start this problem?
    FBD shows force of friction going -X, normal force y, weight -y.

    32000 = 1980a
    a = 16.16 m/s2

    v = 0 + 16.16x12
    v = 193.92 m/s

    I obviously have to do a second xvat for t=12 seconds and onward. These net force/ newtons laws problems are killing me because they involve a lot of creativity with choosing working equations, which I don't have.

    Help please!
     
  2. jcsd
  3. Mar 11, 2012 #2
    We know the force from braking is equal to Fn*uk.

    Force braking is (1980)(9.8)*.8

    acceleration from braking is therefore 9.8*.8 in the negative x direction from F=ma

    d1+d2=total distance

    d1:

    use d1=.5at^2

    d2:

    use d2=v0t+.5at^2

    where a is negative for the d2 equation and I found it above
     
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