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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?
 one year ago
 one year ago
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?
 one year ago
 one year ago

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ZarkonBest ResponseYou've already chosen the best response.0
do you have a solution or are you actually looking for a solution?
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
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
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
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 \[d=v_1\cdot t_1+v_2\cdot t_2\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
ok maybe half an hour, but still
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
though \[v_1=v_2=h\] h=speed of the horse
 one year ago

beginnersmindBest ResponseYou've already chosen the best response.0
I think I got the logic, but can't do the algebra :(
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
well if you have an answer...tell me if I got it right... \[\frac{2h^2}{(w+h)(hw)}\] where h = speed of horse and w= speed of wagon
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
the answer is an actual number
 one year ago

beginnersmindBest ResponseYou've already chosen the best response.0
I think you can get a specific answer without knowing the actual speed
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
that seems weird to me...unless I am not understanding the question
 one year ago

FoolAroundMathBest ResponseYou've already chosen the best response.1
the answer should be 3 miles. Looking for a way to explain that other than intuition
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
no it is not three, although i did get three the first time
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
what if the horse moves at the same rate as the wagon?
 one year ago

FoolAroundMathBest ResponseYou've already chosen the best response.1
oops ... its 2miles .. i'll draw a figure.. hopefully that'll help
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
does the wagon stop after 1 mile?
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
don't forget that the horse travels front to back and back to front at the same time the train goes one mile
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
so it cannot travel at the same rate as the train
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
oh ... ok...well that is different
 one year ago

beginnersmindBest ResponseYou've already chosen the best response.0
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(hw) 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
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
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
 one year ago

FoolAroundMathBest ResponseYou've already chosen the best response.1
dw:1341539647363:dw 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
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
@beginnersmind left hand side maybe \(\frac{1}{w}\)?
 one year ago

FoolAroundMathBest ResponseYou've already chosen the best response.1
edit: messed up my right's from left's.
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
@FoolAroundMath not 2
 one year ago

beginnersmindBest ResponseYou've already chosen the best response.0
@satellite Yeah. Got it, I think. Couldn't have done it without setting w to 1 though.
 one year ago

FoolAroundMathBest ResponseYou've already chosen the best response.1
Talk about halfreading questions. I completely missed that "then back to the front" part. and it can be done without taking the speed of the wagon to be 1. \[\frac{1}{w} = \frac{1}{h+w} + \frac{1}{hw}\]\[\frac{1}{w} = \frac{2h}{h^{2}w^{2}}\]\[h^{2}w^{2} = 2hw\]Dividing by \(w^{2}\)on both sides of this equation \[(\frac{h}{w})^{2}  1 = 2\frac{h}{w}\]\[t^{2}2t1 = 0\]\[t = \frac{2 \pm \sqrt{4(4)}}{2}\]\[t = 1 \pm \sqrt{2} \Rightarrow t = 1+\sqrt{2}\]We need distance traveled by horse = h*(1/w) = h/w = \(1+\sqrt{2}\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
yeah that is what i got but slightly different method i put \(w=1\) speed of the horse as \(h\) and solved for \(h\) reasoning as follows: speed going back is \(1+h\) time is \(\frac{1}{1+h}\) leaving \(1\frac{1}{1+h}\) time for the horse to make it one mile to the front at a rate of \(h1\) and so solved \(1=(h1)(1\frac{1}{h+1})\) since the time is one the rate is the distance more than one way to skin a cat i guess
 one year ago
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