Question

KHW 2: 2.5-2.8 Problem 2.38 7 ot 28> Constants Part A An unmarked police car traveling a constant 95 km/h is passed by a speeder traveling 129 km/h. Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator, if the police car's acceleration is 2.60 m/s2 how much time passes after the police car is passed by a speeder and before the police car overtakes the speeder (assumed moving at constant speed)? Express your answer with the appropriate units tValue Units Submit Provide Feedback Next>

Answer 1

Define the origin to be the location where the speeder passes the police car. Start a timer at the instant that the speeder passes the police car, and find another time that both cars have the same displacement from the origin For the speeder, traveling with a constant speed, the displacement is given by the following 1m/s Ar, (129 km/h) (t)-(3583) n 3.6 km/h) For the police car, the displacement is given by two components. The first part is the distance traveled at the initially constant speed during the 1 second of reaction time. lm/s Arpp (1.00s) (95 km/h) (1.00s) 26.39 m 3.6km/h The second part of the police car displacement is that during the accelerated motion, which lasts for (t-1.00)s. So this second part of the police car displacement, using Edq is given as follows. Ár,,-Vpl (-1.00) +ta, (r-1.00)' =し(26.39 m/s)(-1.00)+4(260 m/s*)(-1.00)2]m So the total police car displacement is Δ8,-Δι,, t ΔΧ,2-(26.394 26.39 (t-1 .00) + 1.3( t-1 .00 )- ) m Now set the two displacements equal, and solve for the time 26.39 + 26.39(-1.00) +1.3(t-1.00)2=35.83 t → 1.3 t2-13.11t + 1.00 = 0 1204t12.04) 5.2 -2 8.40 x10 9.18 S The answer that is approximately 0 s corresponds to the fact that both vehicles had the same displacement of zero when the time was 0. The reason it is not exactly zero is rounding of previous values. The answer of 9.18 s is the time for the police car to overtake the speeder.

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