Question

Simple kinematics problem that I can't figure out about a police and a speeder: police car is traveling at 95km/hr when he's passed by a speeder going 145km/hr. After three seconds, the cop accelerates at a rate of 2.50m/s^2; how much time elapses until the cop overtakes the speeder?

Answer 1

have you started working on it? where are you blocked?

Answer 2

Speeder is going 40.2778m/s.

Cop's initial v = 26.3889m/s, a= 2.5m/s^2

after 3 seconds, the speeder is 41.667m away from cop.

x=1/2at^2+V0t

41.667=1/2(2.5)t^2+26.3889t
solve for t = 1.47579 seconds

Good luck

Answer 3

Cliff: the answer cannot be 1.4 seconds; that is too small of a figure. It should be something on the order of 15-20 seconds.

This is what I have done so far:

x=Vot+1/2at^2
police Vo=26.39m/s
police a=2.5m/s^2

speeder Vo=40.28m/s
speeder a=0

The speeder is ~ 41.67m in front of the police when the police starts accelerating, so I start the calculation from that point.

police=26.39(t)+1/2(2.5)*t^2
speeder=41.67+40.28*t

Set the two equations equal to each other and solve for t:

26.39t+1.25t^2=41.67+40.28t
0=-1.25t^2+13.89t+41.67
a=-1.25, b=13.89, c=41.67, use quadratic formula

(-13.89+-sqrt(13.89^2-(4*-1.25*41.67))/(2*-1.25)

(-13.89+20.03)/-2.5=-2.456, cannot be this one

(-13.89-20.03)/-2.5=13.56

Add +3 seconds to account for the initial time before the calculation began to arrive at final answer of 16.56 seconds.

Answer 4

So during the course of posting what I'd done so far here, I found that I wrote down 23.39m/s instead of 26.39m/s for the police's Vo in my original calculations, which was the cause of my problem. Can I get points for solving my own problem? :)

Answer 5

yes you can :)