Suppose that a cyclist began a 476

mi ride across a state at the western edge of the state, at the same time that a car traveling toward it leaves the eastern end of the state. If the bicycle and car met after 8.5

hr and the car traveled 33.2

mph faster than the bicycle, find the average rate of each.

Question

12-04-2017
Mathematics

contestada

Suppose that a cyclist began a 476

mi ride across a state at the western edge of the state, at the same time that a car traveling toward it leaves the eastern end of the state. If the bicycle and car met after 8.5

hr and the car traveled 33.2

mph faster than the bicycle, find the average rate of each.

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jdoe0001 jdoe0001 · Answer 1 · 2017-04-12T22:41:38+02:00

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 [tex]\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}[/tex]

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

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