Answer:
18.03 s
Step-by-step explanation:
We have two different types of motions, the criminal moves with uniform motion while the police do it with uniformly accelerated motion. Therefore we will use the equations of these cases. We know that by the time the police reach the criminal they will have traveled the same distance.
![x=vt\\x=x_(0)+v_(0)t+(a)/(2)t^2](https://img.qammunity.org/2020/formulas/physics/high-school/f0gxbfteowhaxtpozabbj99xfenr2dlk1l.png)
The distance between the police and the criminal when the first one starts the persecution is 0, its initial speed is also zero. So:
![x=(55m/s)t\\x=(6.1m/s^2)/(2)t^2=(3.05m/s^2)t^2](https://img.qammunity.org/2020/formulas/physics/high-school/5ul1pkrrn68jri78pmtm14j7ytzievjlol.png)
Equalizing these two equations and solving for t:
![(55m/s)t=(3.05m/s^2)t^2\\(3.05m/s^2)t^2-(55m/s)t=0\\t((3.05m/s^2)t-55m/s)=0\\t=0 \\(3.05m/s^2)t-55m/s=0\\t=(55m/s)/(3.05m/s^2)=18.03 s](https://img.qammunity.org/2020/formulas/physics/high-school/29i7h9e43gan8vl11h4sd993smm3itt6pq.png)