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An unmarked police car travelling at a constant 80.0 km/h is passed by a speeder
travelling at 100.0 km/h. Precisely 1.0 s after the speeder passes, the police steps on the accelerator. If the police car’s acceleration is 2.0 m/s^2, how much time passes before the police car overtakes the speeder (assumed moving at a constant speed)?

1 Answer
P dilip_k
Mar 27, 2017

The velocity of police car

#v_p=80km"/"h=(80xx10^3)/3600m"/"s=200/9m"/"s#

The velocity of the speeder

#v_s=100km"/"h=(100xx10^3)/3600m"/"s=250/9m"/"s#

1.0 s after the speeder passes the police car the later starts accelerating @ #2m"/"s^2#.

Within this 1.0 s the speeder goes #(250/9-200/9)m=50/9m# ahead of the police car. Let the police car reaches the speeder again after #t# sec, it starts accelerating.

The distance covered by the police car during t sec after it accelerting @#a=2m"/"s^2#

#S_p=v_pxxt+1/2at^2=200/9t+1/2*2*t^2#

#=200/9t+t^2#

The distance covered by the speeder during the same t sec will be

#S_s=250/9t#

By the condition of the problem

#S_p-S_s=50/9#

#=>t^2+200/9t-250/9t=50/9#

#=>9t^2-50t-50=0#

#=>t=(50+sqrt(50^2+4*9*50))/18s=6.42s#

So if T represets the total time passed before the police car overtakes the speeder then

#T=1sec+6.42s=7.42s#

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