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What is the stopping distance?

  1. Sep 4, 2016 #1
    1. The problem statement, all variables and given/known data
    A car is stopped in a distance D by a constant friction force that is independent of the car's speed. What is the stopping distance (in terms of D) if
    (a) The car's initial speed is tripled, and
    (b) The speed is the same as it originally was but the friction force is tripled?
    (Solve using the work-kinetic energy theorem.)



    2. Relevant equations
    ##W = ΔK = KEf - KEi##
    ##W = FD##
    ##∑Fx = -f_k = ma_x##



    3. The attempt at a solution
    This one has got me really stuck. Here's what I've gotten so far.

    The work is equal to the change in kinetic energy. The final velocity is zero, so ##KEf = 0##. There is an initial velocity, so it initially has kinetic energy. ##W = ΔKE = 0 - KEi = -KEi##

    Work is also equal to force times distance, so ##-KEi = FD##

    I am unsure on how to proceed now. Any help is appreciated, thank you.
     
  2. jcsd
  3. Sep 4, 2016 #2

    haruspex

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    The question refers to the car's initial speed, so clearly that needs to feature in your equations.
     
  4. Sep 4, 2016 #3
    Okay, the equations that involve Initial speed are

    ##V_x = V_xi + a_xt##
    ##x = x_i + V_xit + .5a_xt^2##
    ##V_x^2 = Vi_x^2 + 2a_x(x - x_i)##
    ##x - x_i = ((V_xi + V_x)/(2))t##

    Okay, I assume that I need to find ##a_x## first. Going about this in the proper way seems to be difficult.
    From the first equation, the value of ##a_x## is equal to ##(-Vi_x/t)##
    From the second equation, the value of ##a_x## is equal to ##(x - x_i - Vi_xt)/(.5t^2)##
    From the third equation, the value of ##a_x## is equal to ##(V_x^2 - V^2_xi)/(2(x - x_i))##
    From Newton's second law, the value of ##a_x## is equal to ##(-f_k/m)##

    I'm still thinking about how to go about this.
     
  5. Sep 4, 2016 #4

    haruspex

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    Yes, those all involve speed, but none of them link up immediately to the relevant equations you listed originally. There is an equation that does.
     
  6. Sep 5, 2016 #5
    Would you happen to be referring to ##W = .5mV_f^2 - .5mV_i^2##?
     
  7. Sep 5, 2016 #6

    PeroK

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    Just a suggestion/observation. You seem to be letting all these equations obscure what you are trying to do. Instead of thinking in terms of equations, you could think in terms of the physics involved:

    1) The vehicle has an initial Kinetic Energy and when it has stopped it has none. All the KE has gone.

    2) Friction, acting over the distance D, must have done work equivalent to the loss in KE.

    3) If the vehicle has triple the initial velocity, what does this do to the initial KE?

    4) If the frictional force is tripled, what does this do to the work done by friction over a given distance?

    I would use these observations and questions to guide my search for the correct equations.

    The other thing that is worth pointing out is that, with a question like this, you need to be able to introduce your own variables (as they are not all explicitly given to you). What I would do is:

    Let ##v## be the initial velocity of the car, ##F## be the frictional force and ##D## the stopping distance (you are given that one, but there's no harm writing it down).

    In part a) the new initial velocity will be ##3v##, friction is still ##F## and the stopping distance we will call ##D_a##..

    In part b) the initial velocity is still ##v##, the new frictional force is ##3F## and we'll call the stopping distance ##D_b##.

    That just gets everything out in the open for starters. Now you can hunt down your equations.
     
  8. Sep 5, 2016 #7

    haruspex

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    Yes. Also, please heed PeroK's excellent advice.
     
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