(This is from CTK exchange)
A column of soldiers is 25 miles long and they march 25 miles a day. One morning a messenger started at the rear of the column with a message for the guy up front. The messenger began to march and gave the message to the guy up front and then returned to his position by the end of the day. Assume that the messenger marched at the same rate of speed the whole time. How many miles did the messenger march?
(In reply to
re: Forced March by Barry Knapp)
Sorry Barry, it is you that is incorrect. TomM defines little "t" as the time until they meet, not the time for the whole day (where t=1)so you don't want to include the extra "d" in the rate. Your solution is solving the question "How far does the messenger march if at the end of the day he meets the officer". Work it out, at 50 miles per day, he'll only reach the officer at the very end of the day.
The problem is a little easier to get your head around if you think of the column as standing still, and just adjust the messenger's speed relative to the column. Call the messenger's ACTUAL speed "x". Heading to the front of the line, the messenger's RELATIVE speed compared to the column is x - 25. His relative speed heading to the back of the line is x + 25. The distance both ways is 25 miles (the column is standing still). So the time to get to the front is distance over speed, or 25/(x-25) and the time to the back is 25/(x+25). The two times added together equal 1 (the whole process takes him 1 day). So you are left with the equation:
25/(x+25) + 25/(x-25) = 1
Using algebra you get this to:
0 = x² - 50x - 625
Use the quadratic and you get:
x = (50 + √5000)/2 (the negative answer, much like TomM's negative answer of t = - 1/√2 is ignored)
Simplify and you get:
x = 25 + 25 * √2 or x = 25 * (1 + √2)
This is his ACTUAL speed per day. How far does he travel in one day? His speed! 25 * (1 + √2)
re(2): Forced March
