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Police car problem. please help i need it in an hour!

  1. Sep 15, 2008 #1
    An unmarked police car, traveling a constant 87.0km/hr, is passed by a speeder traveling 134km/hr. Precisely 1.11s after the speeder passes, the policeman steps on the accelerator. If the police car's acceleration is 1.94m/s2, how much time elapses after the police car is passed until it overtakes the speeder (assumed moving at constant speed)?

  2. jcsd
  3. Sep 15, 2008 #2

    Andrew Mason

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    Do a distance - time graph for each of the vehicles. What is the relationship between the co-ordinates (distance and time) of the two cars when the police car overtakes the speeder?

  4. Sep 15, 2008 #3
    thank you ill try that. does anyone know how to do it using the x=xo+vot+1/2at^2 equation?
  5. Sep 15, 2008 #4


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    Rather than looking for an equation, why don't you start by figuring all the things you know?

    First of all convert everything to m/sec before you get tripped up by that.
    87 km/hr = ? m/s
    134 km/hr = ? m/s

    Next how far past the police does the speeder get before the police car starts up? You need that distance

    Starting from t=0 when the police steps on the gas, what is the distance the police will go considering his initial velocity and constant acceleration 1.94? This will be an equation that describes his X-position as a function of time.

    The speeder has his own equation that describes his position as a function of time from the same t=0. What is it? (Remember he has the 1.11 sec * his speed head start)

    You should have two equations that describe each car's position as a function of time. When they are equal to each other - guess what that = caught.
  6. Sep 15, 2008 #5
    I already converted and ive set the individual problems up for each car so many times but somehow i just get lost. I dont know what to do once i get to that point.
  7. Sep 15, 2008 #6
    Is this a calculus based course or a non-calculus based course?
  8. Sep 15, 2008 #7
    its non-calculus.
  9. Sep 15, 2008 #8
    can someone just walk me through it and explain the whole thing. that would be the most helpful.
  10. Sep 15, 2008 #9


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    Write out your equation here and we can help you.
  11. Sep 15, 2008 #10
    Since this is a non-Calculus based course, you have already been given the equations you need to solve it, as you already listed: x=xo+vot+1/2at^2

    This is a standard position equation. What you need to do is take the data given to you in the problem, create a system of equations using that position equation; one for the police car, one for the speeder, then see where they intersect. Then you will have x and t (position and time). You will have to modify the equation for the policemen to adjust for the 1.11 second delay.
  12. Sep 15, 2008 #11
    if anyone has the answer please can i have it. i keep getting so lost when i try to set the equations equal to each other and it is due really soon.
  13. Sep 16, 2008 #12


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    Sorry, can't give you the answer. I would help you solve it, if you had ever posted your work. Besides, I already showed you how to go about it.

    What joy would a grade be if it wasn't yours?
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