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10 July, 18:18

Police officer at rest at the side of the highway

notices a speeder moving at 62 km/h along a

straight level road near an elementary school. When

the speeder passes, the officer accelerates at 3.0 m/s2

in pursuit. The speeder does not notice until the

police officer catches up.

Now assume that the police officer accelerates

until the police car is moving 10.0 km/h faster

than the speeder and then moves at a constant

velocity until the police officer catches up. How

long will it take to catch the speeder?

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Answers (1)
  1. 10 July, 20:38
    0
    Let's take the police officers starting position as our origin, that is Sp (t0) = Ss (t0) = 0, and, since we know the police officers starts from a standstill, Vp (t0) = 0 and Vs (t0) = 62km/h

    For the first part of the pursuit, during the officers acceleration:

    Sp (t) = Sp (t0) + Vp (t0) * t + (3/2) * t^2

    Sp (t) = 1.5*t^2

    Vp (t) = 3*t

    We need to know how much time transpires during this first bit, that is until the officers speed is 72 km/h or 20m/s, and we can find that out by using Vp (t):

    Vp (t1) = 20 = 3*t1

    t1 = 3.33 ... s

    We now look at how much space each of them covers during this time:

    Sp (t1) = 1.5 * (t1) ^2 = 16.66 ... m

    Ss (t1) = Vs*t1 = (62/3.6) * (3.33 ...) = 57.41 m

    For the second part of the pursuit we will take the officers position as our origin:

    Sp (t) = 20*t

    Ss (t) = (57.41 - 16.66) + (17.22 ...) * t

    Now we just need to equate Sp (t) to Ss (t):

    20*t = (40.75) + (17.22 ...) * t

    t = 40.75 / (2.77 ...) = 14.67 s

    But we need to add the time in the first part of the pursuit to the time we just found:

    14.67 + 3.33 = 18s

    The pursuit lasts 18 seconds.
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