Thread: Velocity story problem

A police car is traveling at a velocity of 19.0 m/s due north, when a car zooms by at a constant velocity of 42.0 m/s due north. After a reaction time of 0.800 s the policeman begins to pursue the speeder with an acceleration of 5.00 m/s2. Including the reaction time, how long does it take for the police car to catch up with the speeder?

Given: Vpolice = 19.0 m/s, Vcar = 42.0 m/s , Acc.police = 5.00m/s, t = unknown

I need help setting up this problem.

Hmm i haven't done these kinds of questions for awhile, so this may be wrong.
Vpolice * t + Apolice * t^2 = Vcar * t + Vcar * 0.8
Put it into the form of a quadratic, then solve for x, and it should be the positive number I think.

I did what u said and got:

19.00t + 5.00t^2 = 42t + 42*08

5t^2 - 23t - 33.6 = 0

X= 5.7655, X= -1.1655

so the answer is 5.77 ?

t is the time of acceleration for the police car

(19 m/s)(t + 0.8 s) + (1/2)(5 m/s^2)t^2 = (42 m/s)(t + 0.8 s)

solution to the problem is (t + 0.8 s)

got it thank you