Physics Speed/Acceleration help

A car passes a parked cop car at 46.0 m/s. The cop car starts from rest with a uniform acceleration of 2.29 m/s2.
How much time passes before the car is overtaken by the cop car?
How far does the speeder get before being overtaken by the police car?

The displacement of the motorist is:

Δx(motorist) = v₀t + 0.5at²
= (46.0m/s)t------------------------>accel… = 0, speed is constant

Displacement for the cop is:

Δx(cop) = 0.5(2.29m/s²)t²-------------------->v₀ = 0, starts from rest

Since the displacement for both are equal, we can set these two equations equal and solve for t:

Δx(motorist) = Δx(cop)
(46.0m/s)t = 0.5(2.29m/s²)t²
t = 46.0m/s / 0.5(2.29m/s²)
= 40.2s

Then, the distance the motorist travels before being overtaken by the cop is:

Δx(motorist) = v₀t + 0.5at²
= (46.0m/s)(40.2s) + 0
= 1850m

Hope this helps.

1-You can find the time with the formula v=at so t= v/a. The cop will overtake the speeder when his speed becomes 46 m/s so ==> t=46/2.29~>20 seconds!!!

2-You can find the how far the speeder goes with this formula v=l/t so l=vt. You know the speed is 46 m/s and time 20 seconds, you find l=46 x 20 seconds->920 metres!!!