A thief steals an item and escapes, running at 13.5km/h. A policeman arrives at the spot of the crime after 8 minutes and immediately starts chasing the thief. 28 minutes after the policeman started to chase the thief, there is still a gap of 540m between the two. At what distance from the spot of the crime would the policeman catch up with the thief, and what is the speed (in km/h) at which the policeman ran?

Speed of the thief, v
t
​
=13.5km/h=13.5×
18
5
​
m/s=3.75m/s.

The policeman starts 8 minutes after the thief.
Total time elapsed for the thief by the checkpoint observation point (28 minutes into the chase):

Total time for thief=8+28=36 minutes=36×60=2160 seconds
Distance covered by the thief in 36 minutes:

D
t
​
=3.75×2160=8100m
At this moment, the gap between the policeman and the thief is 540m.
This means the distance covered by the policeman in 28 minutes (28×60=1680 seconds) is:

D
p
​
=D
t
​
−540=8100−540=7560m
Speed of the policeman, v
p
​
:

v
p
​
=
1680s
7560m
​
=4.5m/s
Converting to km/h:

v
p
​
=4.5×
5
18
​
=16.2km/h
Now let's find the total time t
c
​
(in seconds from the start of the policeman's chase) it takes for the policeman to catch the thief.
Initial gap when the policeman started chasing (after 8 minutes of thief running):

Initial gap=13.5km/h×
60
8
​
h=1.8km=1800m
Relative speed = v
p
​
−v
t
​
=4.5−3.75=0.75m/s
Time to catch =
0.75
1800
​
=2400 seconds=40 minutes

Total distance from the crime spot where the policeman catches the thief:

Distance=v
p
​
×time=16.2km/h×
60
40
​
h=10.8km