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    beren lper's Avatar
    beren lper Posts: 2, Reputation: 1
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    #1

    Mar 6, 2013, 05:33 AM
    Homework help
    An unmarked police car traveling a constant 95km/h is passed by a speeder traveling 130km/h. Precisely 2.50s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.20m/s2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?
    smoothy's Avatar
    smoothy Posts: 25,490, Reputation: 2853
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    #2

    Mar 6, 2013, 05:56 AM
    Homework rules require you to actually attempt the work first...

    Tell us what you have done.. and what you have gotten first.
    beren lper's Avatar
    beren lper Posts: 2, Reputation: 1
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    #3

    Mar 6, 2013, 07:36 AM
    first I found Δx of speeder = 130km/h × (1/3.6)(t)= (36.1t)m
    then Δx of police car at first = 95km/h × (1/3.6)(2.50)= 66.0m

    Δx of police car at second = 66(t-2.50)+.5(2.20)(t-2.50)^2
    Δx of police car total = 66.0 + 66(t-2.50)+.5(2.20)(t-2.50)^2 = 36.1t

    then I found t = -4.82 or -17.4 (the problem is that t can't be negative :s) so please help

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