 hatever
the total distance to be traveled, the train in this
puzzle will have used up all of the time allowed for completing the
trip at 60 miles per hour during the first half at 30 miles per hour. So
the solution is -- well, there is .
-
Most people guess 90 mph, since (30 + 90) / 2 = 60.
Now, a train that does 30 mph for the first half and 90 mph for the second
half will average only 45 mph overall.
-
Here are other popular guesses: 120 (48 mph), 180 (51 mph),
240 (53 mph).
Try modifying the problem so that the train is required to
average only 59 miles per hour for the whole trip after traveling at 30
mph during the first half.
Surprise: The train would have to go 1,770 mph!
Supersonic trains are still the stuff of science
fiction.
Time
in the Numerator
The sophisticated person puts time in the numerator (see
Bucketing
Bandwidth). Examples abound. Here is a message received
in 2003 from Ketan Bhaidasna that
makes the point...
| Ben and Carl were in a 100-meter race.
When Ben crossed the finish line, Carl was only at the 90-meter mark.
Ben suggested they run another race. This time Ben would
start ten meters behind the starting line. Will Carl win or loose
-- or will the second race end in a tie?
Ketan's solution...
Carl will loose again. In the second race,
Ben started ten meters back. By the time Carl reaches the 90-meter
mark, Ben will have caught to him. Therefore, the final ten meters
will belong to the faster of the two. Since Ben is faster than Carl,
he will win the final 10 meters and the race.
|
| The analytic approach...
Let the respective velocities be VB
and VC
with the first
race ending at time T.
We have that T = 100 / VB
= 90 / VC.
Thus, VB
= 100/T and VC = 90/T.
Ben finishes the second race at 110 / VB
= 110/100 / T = 1.1T
Carl finishes the second race at 100 / VC
= 100/90/T = 1.111T,
slower by about
1%.
So then, how much of a head-start should Ben give Carl?
This question will be pertinent in the World's
Fastest Dragster. |
|
