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Kinematics Problem - Police Chase

  1. Sep 6, 2009 #1
    1. The problem statement, all variables and given/known data
    A motorist travels at a constant speed of 39.0 m/s through a school zone; exceeding the posted speed limit. A policeman, waits 7.0 s before giving chase at an acceleration of 3.5 m/s2. Find the time required to catch the car, from the instant the car passes the policeman.

    2. Relevant equations
    x-x0=v0(t)+(0.5)(a)t2



    3. The attempt at a solution
    I've actually answered this problem. All I am looking for is an explanation on how I got my answer. Let me show you my work first.

    Both the speeder and the policeman get their own equation(the one mentioned above) and since we know that when both cars meet each other, they will have traveled the same distance and the time will also be the same, therefore we can make both equations equal each other. Also since we know the speeder has 0 acceleration, we can scratch the second half of the equation off for the speeder and since the police car has an initial velocity of 0, we can scratch off the first part of the policeman's equation. We then get the following equation.

    (Speeder)-----> (v-v0)t=(0.5)(a)t2 <-----(Policeman)

    From the information we have, we can also find out how much the speeder has traveled when the policecar starts accelerating.

    39m/s(7s) = 273m

    We can then add this distance to the speeder's side of the equation.

    (v-v0)t+273m=(0.5)(a)t2

    We can now plug in all the variables we know and solve for time

    (39m/s)(t) + 273m = (0.5)(3.5m/s2)(t2)

    We can simplify to get the quadratic equation:

    1.75t2 - 39t - 273 = 0

    When we solve for t, we get the two answers:
    t = 27.88s
    t = -5.96s

    The answer obviously can't be a negative number so we know it is 27.88s. We then can add 7 seconds onto this time to account for the 7 seconds the policecar waited. We then get:

    t = 34.88s (time at which policecar catches up to the speeder)

    Now this is probably a stupid question, but what I want to know is why we have to add 7 seconds at the end when we already accounted for the time the policecar waited at the beggining by adding 273 meters to the speeder's side of the equation.
     
  2. jcsd
  3. Sep 6, 2009 #2
    You said that the speeder will travel for t+7 seconds. So that means that the t you find will be 7s less than the total time. [tex]t_{total} = t + 7[/tex]
    [tex] v_o(t+7) = \frac{1}{2}at^2[/tex]
    [tex] v_ot+273 = \frac{1}{2}at^2[/tex]
    [tex] v_ot = \frac{1}{2}at^2-273[/tex]
    So that means that in time t, the police will travel 273 meters less than what he really needs to.
    Another way to do this problem would be to use this equation
    [tex]v_ot = \frac{1}{2}a(t-7)^2[/tex]
    This t is the total time. The speeder will travel for t seconds and then the poilce will travel for 7 less seconds. This way you won't have to add anything.
     
  4. Sep 6, 2009 #3
    Oh. I get it now. That makes it a lot simpler. Thanks for the help.
     
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