Question 1127250
NOTE: Math problems often do not reflect real life. The patrolman could maintain speed for 1 minute, but it would take some time to accelerate to 115 mph, which is what we are supposed to assume.
 
AS A MATH PROBLEM:
As we expect the patrolman to catch up with the speedster in just a few minutes, we would define 
{{{t}}}{{{"="}}} time, in minutes, from the time the speeding car passes the patrolman.
 
Case A:
As a function of time,
the distance covered by the speedster after that moment is
{{{(55+20)(t/60)}}} miles.
In the same time, the patrolman covers a distance (in miles) of
{{{55(1/60)+115((t-1)/60)}}} .
When the patrolman catches up with the speedster,
both will have covered the same distance since the speeding car passes the patrolman.
So, {{{(55+20)(t/60)=55(1/60)+115((t-1)/60)}}} .
Multiplying both sides of the equal sign times {{{60}}} ,
the equation "simplifies' to
{{{(55+20)t=55+115(t-1)}}} ,
{{{75t=55+115t-115}}} ,
{{{0=115t-75t-115+55}}} ,
{{{0=40t-60}}} ,
{{{60=40t}}}
{{{60/40=t}}}
{{{highlight(t=1.5)}}} .
So, the expected answer is that the patrolman catches up with the speedster
1.5 minutes after the speeding car passes the patrolman.
 
A CHECK:
In the initial first {{{1}}} minute the speedster covers
{{{75(1/60)=5/4=1.25}}} miles,
while the patrolman covers
{{{55(1/60)=11/12}}} miles.
In the next {{{0.5=1/2}}} minute, the speedster covers
{{{(1/2)75(1/60)=5/8}}} miles,
while the patrolman covers
{{{(1/2)115(1/60)=115/120=23/24}}} miles.
Total miles covered are
{{{5/4+5/8=10/8+5/8=15/8}}} for the speedster, and
{{{11/12+23/24=22/24+23/24=45/24=15/8}}} for the patrolman.
 
ANOTHER WAY TO THE SOLUTION:
In the initial first {{{1}}} minute the speedster gains a distance of
{{{(75-55)(1/60)=20/60=1/3}}} miles.
After that, the patrolman is {{{115-75=40}}} mph faster than the speedster,
and decreases the distance between both cars by
{{{40(1/60)=40/60=2/3}}} miles each minute.
At that rate it takes
{{{(1/3)/(2/3)=(1/3)(3/2)=1/2}}} minutes
for the patrolman to catch up with the speedster.
 
Case B:
As a function of time,
the distance covered by the speedster after passing the patrolman is
{{{(55+20)(t/60)}}} miles.
In the same time, the patrolman covers a distance (in miles) of
{{{55(1/60)+v((t-1)/60)}}} .
When the patrolman catches up with the speedster,
both will have covered the same distance since the speeding car passes the patrolman.
So, {{{(55+20)(t/60)=55(1/60)+v((t-1)/60)}}} .
Multiplying both sides of the equal sign times {{{60}}} ,
the equation "simplifies' to
{{{(55+20)t=55+v(t-1)}}} ,
{{{75t=55+75t-75}}} ,
{{{0=vt-75t-v+55}}} ,
{{{0=(v-75)t-v+55}}} ,
{{{v-55=(v-75)t}}}
{{{(v-55)/(v-75)=t}}}
{{{highlight(t=(v-55)/(v-75))}}} .
So, the expected answer is that the patrolman catches up with the speedster
{{{(v-55)/(v-75)}}} minutes after the speeding car passes the patrolman.