Physics:A speeder passes a parked police car at a constant speed of 38.1 m/s

A speeder passes a parked police car at a
constant speed of 38.1 m/s. At that instant,
the police car starts from rest with a uniform
acceleration of 2.21 m/s2.

a:How much time passes before the speeder
is overtaken by the police car? (in seconds)

b:How far does the speeder travel before being
overtaken by the police car? (in meters)

When the cop catches up they will both have traveled the same distance D and taken the same time T to get that far.

For the speeder

D = 38.1 . T

For the cop

D = 1/2 . a . t^2 = 0.5 . 2.21 . T^2

38.1 T = 1.105 . T^2

T = 38.1 / 1.105 = 34.5 s

D = 34.5 . 38.1 = 1314 = 1310 m