An army is marching in single file in a line that stretches 20km. A dispatch rider travels from the rear to the front, delivers a message, and returns to the rear. In the time the dispatch rider takes to do this, the army advances 20 km. How far did the dispatch rider travel in going from the rear to the front and back to the rear again. [Note that the rider's turnaround time at the front is instantaneous!]
2006-12-19 12:13:45 · 4 answers · asked by mascga3 1 in Science & Mathematics ➔ Mathematics
isn't this the band question worded differently?
2006-12-19 12:16:08 · answer #1 · answered by raj 7 · 0⤊ 0⤋
Let's give the army a speed. Let's say this whole thing happens in 1 hour, which is ridiculously fast, but I think the speed will end up not mattering.
At some point in time, t, the rider is at the front. At this point he has traveled 20 km plus t* 20 (because the front person has traveled t*20 from his previous position at time 0. The rider's speed is then (20+20t)/t km/hour,
During the next (1-t) time period, he travels a distance t*20 back to the back of the pack, and since this takes him 1-t in time, his speed is t*20/(1-t)
So we have one equation, 1 unknown:
(20+20t)/t = t*20/(1-t)
multply both sides by t*(1-t):
(1-t)*(20)(1+t) = t*20*t
(1-t^2)*20 = 20t^2
20 = 40t^2
t = sqrt(1/2), or about .71
Since the rider's speed is (20+20t)/t = 48, he has ridden for 1 hour at 48 km/hour, or about 48 km.
2006-12-19 20:34:13 · answer #2 · answered by firefly 6 · 0⤊ 0⤋
I believe its 60 km
He went forward 20km,
+ 20km that the army advanced,
+ 20km back to his spot.
=60 km
2006-12-19 20:19:15 · answer #3 · answered by i.heart.u 5 · 0⤊ 0⤋
i cant help you i just smoked a bunch of crack
2006-12-19 20:16:14 · answer #4 · answered by Anonymous · 0⤊ 2⤋