Question

Suppose that a cyclist began a 476

Answer 1

Well, we know the cyclist left the western part going eastwards, at the same time the car left the eastern part going westwards

the distance between them is 476 miles, and they met 8.5hrs later

let's say after 8.5hrs, the cyclist has travelled "d" miles, whilst the car has travelled the slack, or 476-d, in the same 8.5hrs

we know the rate of the car is faster... so if the cyclist rate is say "r", then the car's rate is r+33.2

thus    $\bf \begin{array}{lccclll} &distance&rate&time\\ &-----&-----&-----\\ \textit{eastbound cyclist}&d&r&8.5\\ \textit{westbound car}&476-d&r+33.2&8.5 \end{array}\\\\ -----------------------------\\\\ \begin{cases} \boxed{d}=8.5r\\\\ 476-d=(r+33.2)8.5\\ ----------\\ 476-\boxed{8.5r}=(r+33.2)8.5 \end{cases}$

solve for "r", to see how fast the cyclist was going

what about the car? well, the car's rate is r + 33.2