Question 414434
Bay City and Vernonville are 168 miles apart.
 A car leaves Bay City traveling towards Vernonville, and another car leaves
 Vernonville at the same time, traveling towards Bay City.
 The car leaving Bay City averages 10 mph more than the other, and they meet after 1 hour and 36 minutes.
 What are the average speeds of the cars?
:
Change 1 hr 36 min to 1.6 hrs, (36/60)
:
Let s = speed of the Bay City car
then
(s-10) = speed of the Vern car
:
When the two cars meet, they will have traveled a total of 168 mi
Write a distance equation; dist = time * speed
:
1.6s + 1.6(s-10) = 168
1.6s + 1.6s - 16 = 168
3.2s = 168 + 16
3.2s = 184
s = {{{184/3.2}}}
s = 57.5 mph speed of the Bay City car
then, obviously; 47.5 mph is the speed of the other car
:
:
Check this by finding the total of the distances
57.5 * 1.6 = 92 mi
47.5 * 1.6 = 76 mi
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total dist: 168 mi