SOLUTION: A)A highway patrolman traveling at the speed limit is passed by a car going 20 mph faster than the speed limit. After one minute, the patrolman speeds up to 115 mph. How long afte
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Question 1127250: A)A highway patrolman traveling at the speed limit is passed by a car going 20 mph faster than the speed limit. After one minute, the patrolman speeds up to 115 mph. How long after speeding up until the patrolman catches up with the speeding car. The speed limit is 55 mph.
B)Same question, but this time the patrolman speeds up to a speed of v mph (v>75).
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
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
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
miles.
In the same time, the patrolman covers a distance (in miles) of
.
When the patrolman catches up with the speedster,
both will have covered the same distance since the speeding car passes the patrolman.
So, .
Multiplying both sides of the equal sign times ,
the equation "simplifies' to
,
,
,
,
.
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 minute the speedster covers
miles,
while the patrolman covers
miles.
In the next minute, the speedster covers
miles,
while the patrolman covers
miles.
Total miles covered are
for the speedster, and
for the patrolman.
ANOTHER WAY TO THE SOLUTION:
In the initial first minute the speedster gains a distance of
miles.
After that, the patrolman is mph faster than the speedster,
and decreases the distance between both cars by
miles each minute.
At that rate it takes
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
miles.
In the same time, the patrolman covers a distance (in miles) of
.
When the patrolman catches up with the speedster,
both will have covered the same distance since the speeding car passes the patrolman.
So, .
Multiplying both sides of the equal sign times ,
the equation "simplifies' to
,
,
,
,
.
So, the expected answer is that the patrolman catches up with the speedster
minutes after the speeding car passes the patrolman.
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