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Speeding Driver & Cop Car Velocity/Time/Acceleration Q

  1. Feb 11, 2013 #1
    So I'm new here, and I apologize if I've posted in the wrong format/place/etc., but this is a makeup exam question, and I (think) I've finished it, but I want a second opinion on the correctness if you guys wouldn't mind. :smile:

    1. The problem statement, all variables and given/known data
    A speeding driver passes a hidden police car moving at a speed of 46m/s; assume the speeder keeps going this speed. The police car immediately starts accelerating after the speeder at 7.5m/s2.
    a. How long does the police car take to catch the speeder?
    b. How far has the speeder gone when police catches him?
    c. How fast is the police car moving when it catches the speeder?

    Variables for speeder denoted with 's', for police denoted with 'p'.
    Vs = 46m/s
    as = 0m/s2
    t = ?
    Δx = ?
    ap = 7.5m/s2
    Vos = 0m/s
    Vfp = ?

    2. Relevant equations & 3. The attempt at a solution

    a. Δx = 46m/st for the speeder & Δx = 1/2(7.5m/s2)t2 for the police car.
    So, Δx = 3.75m/s2t2 - 46m/st
    Plug in to quadratic formula and come up with 12.3s.

    b. Δx = 46m/s(12.3s) = 565.8m

    c. Vfp = 7.5m/s2(12.3s) = 92.3m/s

    Thanks in advance for any comments or helpful information!
     
  2. jcsd
  3. Feb 11, 2013 #2
    I get close to your answers because one should round off after the computation is completed. Don't round off your time, then use the rounded time to compute other results.
     
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