## satellite73 Group Title 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? 2 years ago 2 years ago

1. Zarkon Group Title

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

2. satellite73 Group Title

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

3. Zarkon Group Title

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 $d=v_1\cdot t_1+v_2\cdot t_2$

4. satellite73 Group Title

ok maybe half an hour, but still

5. Zarkon Group Title

though $v_1=v_2=h$ h=speed of the horse

6. beginnersmind Group Title

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

7. Zarkon Group Title

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

8. satellite73 Group Title

the answer is an actual number

9. beginnersmind Group Title

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

10. Zarkon Group Title

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

11. FoolAroundMath Group Title

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

12. satellite73 Group Title

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

13. Zarkon Group Title

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

14. FoolAroundMath Group Title

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

15. Zarkon Group Title

does the wagon stop after 1 mile?

16. satellite73 Group Title

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

17. satellite73 Group Title

so it cannot travel at the same rate as the train

18. Zarkon Group Title

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

19. beginnersmind Group Title

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

20. satellite73 Group Title

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

21. FoolAroundMath Group Title

|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

22. satellite73 Group Title

@beginnersmind left hand side maybe $$\frac{1}{w}$$?

23. FoolAroundMath Group Title

edit: messed up my right's from left's.

24. satellite73 Group Title

@FoolAroundMath not 2

25. beginnersmind Group Title

@satellite Yeah. Got it, I think. Couldn't have done it without setting w to 1 though.

26. FoolAroundMath Group Title

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}$$

27. satellite73 Group Title

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