Question 1043954
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Let *[tex \Large w] represent the speed with which Winnie walked and let *[tex \Large p] represent the speed with which Piglet walked.  The entire distance from Winnie's house to Piglet's house is *[tex \Large d] and we can divide that distance into two pieces, *[tex \Large d_1], the distance from Winnie's house to the point where they met, and *[tex \Large d_2], the distance from Piglet's house to the same point.  And we know that *[tex \Large d_1\ +\ d_2\ =\ d]. Since we know they left at the same time, the time it took each of them to get to the meeting point is the same amount of time and we will call that *[tex \Large t].


Since we know that distance equals rate times time, we can now establish some relationships between the above variables.


The distance from Piglet's house to the meeting point can be expressed in two different ways:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d_2\ =\ pt]


because *[tex \Large p] is Piglet's rate and *[tex \Large t] is the time it took Piglet to get to the meeting point, and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d_2\ =\ 8w]


because *[tex \Large w] is Winnie's rate and it took Winnie 8 minutes to get to Piglet's house after the meeting point.  But then we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{8w}{p}]


Similarly, considering the distance from Winnie's house to the meeting, we can come up with:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{18p}{w}]


And now we have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{8w}{p}\ =\ \frac{18p}{w}]


A little algebra gets us to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ =\ \frac{9}{4}p^2]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w\ =\ \frac{3}{2}p]


And now we know that Winnie walked one and one-half times as fast as Piglet.


Going back to an earlier relationship we had established:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d_2\ =\ 8w]


Now substitute *[tex \Large \frac{3}{2}p] for *[tex \Large w]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d_2\ =\ 8\left(\frac{3}{2}p\right)\ =\ 12p]


Now we know it took Piglet 12 minutes to get to the meeting place and 18 minutes to get from the meeting place to Winnie's house for a total of 30 minutes.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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