Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.70 m/s2 , how much time passes after the police car is passed by a speeder and before the police car overtakes the speeder (assumed moving at constant speed)

t=  16.75 s

Explanation:

We will solve this exercise using the kinematic expressions

corridor that goes at constant speed, suppose that its speed is v₁ = 20 m/s, it does not appear in the statement, we start counting the time when it passes the policeman.

           x₁ = v₁ t

The policeman starts from rest, so his initial velocity is zero and he has an acceleration a = 2.70 m /s², to use the same time counter we take into account that the policeman left at = 1.00 s after passing the corridor

           x₂ = v₀ (t-t₀) + ½ a (t-t₀)²

           x₂ = ½ a (t-1)²

at the point where the two meet, the position must be the same

           x₁ = x₂

          v₁ t = ½ a (t-1)²

          (t-1)² =

           t² - 2t + 1 - \frac{2 v_1 t}{a} +1 = 0

           t² - 2(1 + ) t  +1

let's we solve the second degree equation

          t² - 2 ( 1 + ) t + 1=0

          t² - 16.81 t +1=0

          t = [ 16.81 ± ] /2

          t = [16.81 ± 16.695]/2

          t₁=  16.75 s

          t2= 0.06 s

Time t₂ is less than the reaction time of humans, so the correct answer is the first time

            t=  16.75 s

i wanna say it’s between b and d, leaning more towards b.