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Speeder and Cop Question

  1. Mar 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Problem: A police car travelling at 80km/h is passed by speeder travelling at 145km/h. 3 seconds after the speeder passes, the police car starts accelerating at 2.5m/s2. How much time passes before the police overtakes the speeder (moving at a constant speed)?

    Variables:
    initial velocity of police car (vpi) 80km/h=22.222m/s
    initial velocity of speeder (vsi)=final velocity of speeder (vsf) = 40.278m/s
    acceleration of police car (ap)=2.5m/s2

    2. Relevant equations
    x=x0+v0t+1/2*a*t2

    3. The attempt at a solution
    I thought that if the distance the police car travelled equals the distance the speeder travelled, that would mean that the police overtook the speeder. So I used the distance formula/equation like this:

    time passed since the police started accelerating that the police car and speeder meets=t
    vsi*3+vsf*t=vpi*3+vpi*t+1/2*a*t2
    40.278*3+40.278t = 22.222*3+22.222*t+1/2*2.5*t2
    When I did this equation on mathway I got the answers t=13.0062276,1.4385724.
    So if I add 3 seconds to the answers I get t=16 and t=4.
    I thought the final answer would be 16 seconds because 4 seconds is too short a time.
    But it says my answer is incorrect... Could you tell me where I'm thinking wrong? Thank you!
     
  2. jcsd
  3. Mar 29, 2015 #2
    Why would you add three seconds to the answer at the end? You already took into account the 3 seconds the policeman was not accelerating.
     
  4. Mar 29, 2015 #3
    I added 3 seconds because the variable t used in the equation was the time the police car began accelerating. Not since when the speeder passes the police car.
    And the answer wasn't 13 seconds either, so I thought it was because I forgot to add 3 seconds :(
     
  5. Mar 29, 2015 #4
    Relative to the speeder, what is the initial velocity of the police car?
     
  6. Mar 29, 2015 #5
    Wait I mean relative to the initial velocity of the police car, what is the velocity of the speeder? Ignore my last post.
     
  7. Mar 29, 2015 #6
    I don't understand what you mean...? According to the question, the initial velocity of the police car is 80km/h and the initial velocity of the speeder is 145km/h.
     
  8. Mar 29, 2015 #7
    Your solution is correct. It must be a calculation error. Try recalculating.
     
  9. Mar 29, 2015 #8
    I'm constantly getting the numbers wrong:( But thank you for telling me the equations I'm using are right! I feel much better lol
     
  10. Mar 29, 2015 #9
    When I calculated, I got like 10.2 or something for t. I don't know why our answers are different. Are you using the correct quadratic formula?
     
  11. Mar 29, 2015 #10
    40.278*3+40.278t = 22.222*3+22.222*t+1/2*2.5*t2
    when I simplified this equation I got
    1.25t2-18.056t-54.168=0
    So I put this in to the quadratic formula (-b+√b2-4ac)/2a
    and the answer was 16.99....

    I'm not really good with the calculations and stuff so I mostly rely on mathway (where they do the quadratic formulas for you)
    but the values that come out when I put 40.278 and 22.222 & 80000/3600 and 145000/3600 are quite different
    so I am getting extremely confused:(

    Thank you so much for your help though.
    I'll just try again until I figure the calculations out.
     
  12. Mar 29, 2015 #11

    SammyS

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    I get t = -2.54988 and t = 16.9947 . The negative answer doesn't make sense. Throw it out.

    From the wording of the problem, it's not clear whether or not to add the three seconds.

    They may be asking for the time from the moment that the police car begins acceleration, not from the time the car passes the police car.
     
  13. Mar 30, 2015 #12
    Yes, that's right. I was making calculation mistakes while I was solving this problem. The answer is the value + 3 seconds.
    Thank you so much for your help!
     
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