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Weird kinematics question

  1. Jan 12, 2008 #1
    1. The problem statement, all variables and given/known data

    (3) A motorist is approaching a green traffic light with speed [itex]v_0[/itex] when the light turns
    to amber.
    (a) If his reaction time is [itex]\tau[/itex], during which he makes his decision to stop and applies his foot to the
    brake, and if the maximum braking deceleration is a, what is the minimum distance [itex]s_{min}[/itex] from the
    intersection at the moment the light turns to amber in which he can bring his car to a stop?
    (b) If the amber light remains on for a time t before turning red, what is the maximum distance
    [itex]s_{max}[/itex] from the intersection at the moment the light turns to amber such that he can continue into the
    intersection at speed [itex]v_0[/itex] without running the red light?
    (c) Show that if his initial speed [itex]v_0[/itex] is greater than

    [itex]v_{0_{max}} = 2a(t- \tau )[/itex]

    there will be a range of distance from the intersection such that he can neither stop in time nor
    continue through the intersection without running the red light.

    2. Relevant equations

    kinematics equations

    3. The attempt at a solution

    A using [itex]vf^2=v_0^2 +2(-a)d[/itex] is [itex] s_{min}=\frac{v_0^2}{2a} +v_0\tau[/itex]

    B is simply [itex]v_0 t[/itex]

    C i'm almost clueless. i can sub the given expression into both of my derived equations for [itex]s_{min}[/itex] and [itex]s_{max}[/itex] and all i get is that they both equal [itex]2at(t-\tau)[/itex] which i can't see how to use to prove that [itex] s_{max}<s<s_{min}[/itex].

    This is actually an intermediate mechanics question but seems simple enough.
     
  2. jcsd
  3. Jan 12, 2008 #2

    mda

    User Avatar

    Substitute your values into the inequality for smax & smin.
    Solve for v.
     
  4. Jan 12, 2008 #3
    the implication needs to go the other way, if p is the statement about velocity and q is the statement about distance i need to prove p -> q, that would prove q->p

    even if i did do as you said, how would i "solve" for v?
     
    Last edited: Jan 12, 2008
  5. Jan 13, 2008 #4
    s > s_max or s < s_min
    v*t > s_max or v*t < s_min

    Since we are looking for the maximum initial speed, then
    vo_max*t = vo_max^2/(2a) + vo_max*r
    t = vo_max/(2a) + r
    t - r = vo_max/(2a)
    vo_max = 2a(t-r)
     
  6. Jan 13, 2008 #5
    i don't understand what you've done

    why did you equate the two?
     
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