Police car catching a speeder with constant acceleration

[bigzee20]

Homework Statement

A speeder passes a parked police car at a constant speed of 27.6 m/s. At the instant, the police car starts from rest with a uniform acceleration of 2.27 m/s^2. How much t passes before the speeder is overtaken by the police car.


I tried solving for t using t=Vf-Vi /a and i get 12.15sec, which is wrong what am i doing wrong?

 

[LowlyPion]

Homework Statement

A speeder passes a parked police car at a constant speed of 27.6 m/s. At the instant, the police car starts from rest with a uniform acceleration of 2.27 m/s^2. How much t passes before the speeder is overtaken by the police car.

I tried solving for t using t=Vf-Vi /a and i get 12.15sec, which is wrong what am i doing wrong?

Observe that to catch him they must both be at the same X at the same Time.

What is the equation for the PoPo's displacement with respect to time?
And for the reckless miscreant?

When they are equal, it's book'em Danno.

 

[bigzee20]

OMMMMG I love u I GOT IT!
x=1/2a(t)^2 + Vi(t)

Cops= 1/2(2.27)(12.15)+0(12.15) = 13.79025
Speeder= 1/2(0)12.15+27.6(12.15) = 335.34
335.34/13.79025 = 24.32s

 

[LowlyPion]

OMMMMG I love u I GOT IT!
x=1/2a(t)^2 + Vi(t)

Cops= 1/2(2.27)(12.15)+0(12.15) = 13.179025
Speeder= 1/2(0)12.15+27.6(12.15) = 335.34
335.34/13.179025 = 24.32s

??

I was thinking of something considerably more direct.

1/2*a*t² = V * t

t = 2*V/a = 2*27.6/2.27 = 24.32 s