Question

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

Answer 1

do you have a solution or are you actually looking for a solution?

Answer 2

i spent an hour on it this morning, and yes i have a solution
you can google it and get one, but honestly i did not understand the yahoo answer, so i redid it differently
thought it was kind of cute because it made me think

Answer 3

I just found the time it takes to get to the back and then the time it takes to get back to the front.

then \[d=v_1\cdot t_1+v_2\cdot t_2\]

Answer 4

ok maybe half an hour, but still

Answer 5

though \[v_1=v_2=h\]
h=speed of the horse

Answer 6

I think I got the logic, but can't do the algebra :(

Answer 7

well if you have an answer...tell me if I got it right...
\[\frac{2h^2}{(w+h)(h-w)}\]

where h = speed of horse and w= speed of wagon

Answer 8

the answer is an actual number

Answer 9

I think you can get a specific answer without knowing the actual speed

Answer 10

that seems weird to me...unless I am not understanding the question

Answer 11

the answer should be 3 miles. Looking for a way to explain that other than intuition

Answer 12

no it is not three, although i did get three the first time

Answer 13

what if the horse moves at the same rate as the wagon?

Answer 14

oops ... its 2miles .. i'll draw a figure.. hopefully that'll help

Answer 15

does the wagon stop after 1 mile?

Answer 16

don't forget that the horse travels front to back and back to front at the same time the train goes one mile

Answer 17

so it cannot travel at the same rate as the train

Answer 18

oh ... ok...well that is different

Answer 19

If w is the speed of the wagon and h is the speed of the horse it takes

t=1/w time for the wagon to travel one mile.

The horse takes 1/(h+w) time to get from the start to the end and 1(h-w) to get from the end to the front. So we get

1/h = 1/(h+w)+1/(h/w). This should be solved for h/w

Answer 20

in fact that was the key to the solution, i picked 1 for the speed of the train, then solved for the speed of te horse

Answer 21

The train moves 1 mile. The initial position of the train (left rectangle) and initial position of the horse (rear of the train = extreme left) are shown.
After the train has moved 1 mile, the train's new position is drawn (The right rectangle). and the final position of the horse is shown (front of the train = extreme right).
Now its easy to see that the horse traveled 1+1 = 2miles