Question 697502
I suspect you are taking physics in high school or college.
However, to cover any possible situation, I'll add the physics "formulas" in parenthesis,
but I will show the solution as if it was a problem for a fifth grader.
A physics instructor may expect a different way to solve the problem,
but I hope I found the simplest way,
and I believe the simplest way is the best way.
 
Some painful conversions seem needed,
but the numbers were chosen so as to make this easier.
{{{126km/1hr=126*1000m/3600s=35}}}m/s
{{{162km/1hr=162*1000m/3600s=45}}}m/s
 
Accelerating from {{{0}}} to {{{45}}}m/s at {{{5}}}m/s of speed increase per second takes
{{{45/5=9}}} seconds.
({{{v-v[0]=a*t}}} <--> {{{t=(v-v[0])/a}}} but for this case {{{v[0]=0}}})
 
While accelerating, the police car has an average speed of {{{(45+0)/2=22.5}}}m/s
(because at constant acceleration, the average velocity is the average of the initial and final velocities)
so in {{{9}}}s it covers a distance of {{{9*22.5=202.5}}}m
({{{d=v*t}}})
 
The distance ({{{d}}}, in meters) covered by the bus and also by the police car can be calculated from either point of view.
({{{d=v*t}}})
 
The bus is moving for {{{t}}} seconds at {{{35}}}m/s, covering a distance of
{{{d=35*t}}} meters.
 
During those {{{t}}} seconds, the police car is stopped for 2 seconds,
covers {{{202.5}}}m for the next {{{9}}} seconds (while accelerating),
and after those  first {{{11}}} seconds,
moves at {{{45}}}m/s for the rest of the time ({{{t-11}}} seconds).
The distance covered by the police car is
{{{202.5+45(t-11)=202.5+45t-495=45t-292.5}}} meters
 
Because the same distance {{{d}}} is covered by the bus and by the car
during the {{{t}}} seconds between the time the bus passes the patrol car and the time the patrol car overtakes the bus.
{{{45t-292.5=35t}}} --> {{{45t-35t=292.5}}} --> {{{10t=292.5}}} --> {{{t=292.5/10}}} --> {{{highlight(t=29.25)}}} seconds.
 
Substituting into {{{d=35t}}} we get
{{{d=35*29.25}}} --> {{{highlight(d=1023.75)}}} meters (about 1 km).