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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?
do you have a solution or are you actually looking for a solution?
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
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 \[d=v_1\cdot t_1+v_2\cdot t_2\]
ok maybe half an hour, but still
though \[v_1=v_2=h\] h=speed of the horse
I think I got the logic, but can't do the algebra :(
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
the answer is an actual number
I think you can get a specific answer without knowing the actual speed
that seems weird to me...unless I am not understanding the question
the answer should be 3 miles. Looking for a way to explain that other than intuition
no it is not three, although i did get three the first time
what if the horse moves at the same rate as the wagon?
oops ... its 2miles .. i'll draw a figure.. hopefully that'll help
does the wagon stop after 1 mile?
don't forget that the horse travels front to back and back to front at the same time the train goes one mile
so it cannot travel at the same rate as the train
oh ... ok...well that is different
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
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
|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
@beginnersmind left hand side maybe \(\frac{1}{w}\)?
edit: messed up my right's from left's.
@FoolAroundMath not 2
@satellite Yeah. Got it, I think. Couldn't have done it without setting w to 1 though.
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 = \(1+\sqrt{2}\)
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