Question

Super hard math Problem!

A wagon train that is one mile long advances one mile at a constant rate. During the same time period, the wagon master rides his horse at a constant rate from the front of the wagon train to the rear, and then back to the front. How far did that wagon master ride?

Answer 1

let train reached a fraction of it's journey, t, when horse arrived at train's rear (covering 1-t of the train's length).
so when train go on with the rest of it's journey, 1-t, the horse had to cover 2-t in the same time frame.

letting c be train's speed and v for horse's speed,

v / c = (1 - t) / t = (2 - t) / (1 - t)
(1 - t)^2 = t(2 - t)
t^2 - 2t + 1 = 2t - t^2
2t^2 - 4t + 1 = 0
2(t - 1)^2 = 1
t = 1 +- √0.5
since 1 - t > 0,
t = 1 - √0.5

v / c = (2 - t) / (1 - t)
= (1 + √0.5) / √0.5
= √2 + 1

distance travelled by horse (and the rider, wagon master)
= (√2 + 1) * 1
= 2.41421 miles

Answer 2

OMGGGG, you saved my life! Thank you so much for your help!!!