A Porsche speeds past a stationary police officer while traveling110 km/h. The o

Question

A Porsche speeds past a stationary police officer while traveling110 km/h. The officer immediately beginspursuit at a constant acceleration of 12.0km/h/s (note the mixed units), just as the speeder passes theofficer.
(a) How much time will it take for the police officer to catch upto the speeder, assuming that the Porsche maintains a constantspeed?
(b) How fast will the police officer be traveling at this time?
I cannot figure out how to set up the problem. Which equationshould i use? Do I need to convert the units?

Explanation / Answer

a) Porsche v1 = 110km/h = 110*1000m/3600s = 30.56m/s police: a = 12.0km/h/s = 12.0*1000m/3600s2 = 3.33m/s2 let the distance police catch up is s, and time t wehave: police: s = (1/2)at2 Porche: s = v1*t we have: (1/2)at2 = v1*t =>t=2*v1/a = 2*30.56m/s / 3.33 m/s2 =18.35s so t = 18.35s is the time to catch up. b) the speed of the police at the time is: v = at =3.33m/s2*18.35s = 61.12 m/s let the distance police catch up is s, and time t wehave: police: s = (1/2)at2 Porche: s = v1*t we have: (1/2)at2 = v1*t =>t=2*v1/a = 2*30.56m/s / 3.33 m/s2 =18.35s so t = 18.35s is the time to catch up. b) the speed of the police at the time is: v = at =3.33m/s2*18.35s = 61.12 m/s

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