Question 975083
<font face="Times New Roman" size="+2">


Let *[tex \Large r] represent the rate the child runs and Let *[tex \Large r_w] represent the speed of the walkway.  Since distance equals rate times time, rate must equal distance divided by time.  When running against the walkway, the net rate is *[tex \Large r\ -\ r_w].  When running with the walkway, the net rate is *[tex \Large r\ +\ r_w].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ -\ r_w\ =\ \frac{32\text{ft}}{8\text{sec}}].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ +\ r_w\ =\ \frac{120\text{ft}}{12\text{sec}}].


Solve the two by two linear system for *[tex \Large r] and *[tex \Large r_w].


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \