need help with this problem (physicS)

Question

A speeder passes a parked police car at a constant speed of 38 m/s. At that instant, the police car starts from rest with a uniform acceleration of 2.83 m/s 2 . How much time passes before the speeder is overtaken by the police car? Answer in units of s.

Answer 1

In these sorts of problems with two "actors", the core of the problem solving revolves around identifying what is common between the two actors, what variable has the same value for both actors. In this situation, it is time and distance.

For the speeder, the equation that relates his constant velocity, time, and distance is:

d=v*t

For the police car, the equation that relates his constant acceleration, time, and distance is:

d=(1/2)at²

d and t are identical in the two equations: when the police car reaches the speeder, they have travelled the same distance in the same amount of time. We now have two equations and two unknowns (d and t, because v and a are known), so we can solve:

d=38t

d=(1/2)(2.83)t²=38t

1.415t=38

t=26.855 seconds