satellite73
  • satellite73
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?
Mathematics
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SOLVED
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katieb
  • katieb
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Zarkon
  • Zarkon
do you have a solution or are you actually looking for a solution?
anonymous
  • anonymous
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
Zarkon
  • Zarkon
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 \[d=v_1\cdot t_1+v_2\cdot t_2\]

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More answers

anonymous
  • anonymous
ok maybe half an hour, but still
Zarkon
  • Zarkon
though \[v_1=v_2=h\] h=speed of the horse
beginnersmind
  • beginnersmind
I think I got the logic, but can't do the algebra :(
Zarkon
  • Zarkon
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
anonymous
  • anonymous
the answer is an actual number
beginnersmind
  • beginnersmind
I think you can get a specific answer without knowing the actual speed
Zarkon
  • Zarkon
that seems weird to me...unless I am not understanding the question
FoolAroundMath
  • FoolAroundMath
the answer should be 3 miles. Looking for a way to explain that other than intuition
anonymous
  • anonymous
no it is not three, although i did get three the first time
Zarkon
  • Zarkon
what if the horse moves at the same rate as the wagon?
FoolAroundMath
  • FoolAroundMath
oops ... its 2miles .. i'll draw a figure.. hopefully that'll help
Zarkon
  • Zarkon
does the wagon stop after 1 mile?
anonymous
  • anonymous
don't forget that the horse travels front to back and back to front at the same time the train goes one mile
anonymous
  • anonymous
so it cannot travel at the same rate as the train
Zarkon
  • Zarkon
oh ... ok...well that is different
beginnersmind
  • beginnersmind
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
anonymous
  • anonymous
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
FoolAroundMath
  • FoolAroundMath
|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
anonymous
  • anonymous
@beginnersmind left hand side maybe \(\frac{1}{w}\)?
FoolAroundMath
  • FoolAroundMath
edit: messed up my right's from left's.
anonymous
  • anonymous
@FoolAroundMath not 2
beginnersmind
  • beginnersmind
@satellite Yeah. Got it, I think. Couldn't have done it without setting w to 1 though.
FoolAroundMath
  • FoolAroundMath
Talk about half-reading 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. \[\frac{1}{w} = \frac{1}{h+w} + \frac{1}{h-w}\]\[\frac{1}{w} = \frac{2h}{h^{2}-w^{2}}\]\[h^{2}-w^{2} = 2hw\]Dividing by \(w^{2}\)on both sides of this equation \[(\frac{h}{w})^{2} - 1 = 2\frac{h}{w}\]\[t^{2}-2t-1 = 0\]\[t = \frac{2 \pm \sqrt{4-(-4)}}{2}\]\[t = 1 \pm \sqrt{2} \Rightarrow t = 1+\sqrt{2}\]We need distance traveled by horse = h*(1/w) = h/w = \(1+\sqrt{2}\)
anonymous
  • anonymous
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 \(h-1\) and so solved \(1=(h-1)(1-\frac{1}{h+1})\) since the time is one the rate is the distance more than one way to skin a cat i guess

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