Question 101608
distance(d)=rate(r) times time(t) or d=rt; r=d/t and t=d/r

Let t=number of hours that it takes the faster car to be 35 mi ahead of the slower car

Distance faster car travels=(rate of faster car)*t=55t

Distance slower car travels=(rate of slower car)*t=50t

Now the difference between these distances needs to equal 35 mi, so:

55t-50t=35
5t=35  divide both sides by 5

t=7 hours-------------ans

CK

in 7 hrs faster car travels 7*55 mi=385 mi
in 7 hrs slower car travels 7*50=350 mi

385-350=35
35=35

Hope this helps--ptaylor