SOLUTION: Two cars leave a city on the same road, one driving 9 mph faster than the other. After 4 hours, the car traveling faster stops for lunch. After 4 hours and 30 minutes, the car trav
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Question 949598: Two cars leave a city on the same road, one driving 9 mph faster than the other. After 4 hours, the car traveling faster stops for lunch. After 4 hours and 30 minutes, the car traveling slower has not passed the faster's cars position, but stops for lunch too. Assuming that the person in the faster car is still eating lunch, the cars are now 18 miles apart. How fast is each car driving?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Two cars leave a city on the same road, one driving 9 mph faster than the other.
After 4 hours, the car traveling faster stops for lunch.
After 4 hours and 30 minutes, the car traveling slower has not passed the faster's cars position, but stops for lunch too.
Assuming that the person in the faster car is still eating lunch, the cars are now 18 miles apart.
How fast is each car driving?
:
let s = the speed of the slower car
then
(s+9) = speed of the faster
:
Find the distance each has traveled when they stopped for lunch
dist = time * speed
4(s+9) = the faster car distance
4.5s = the slower car distance
:
4(s+9) = 4.5s + 18
4s + 36 = 4.5s + 18
36 - 18 = 4.5s - 4s
18 = .5s
s = 36 mph is the speed of the slower car
and
36 + 9 = 45 mph is the faster car
:
:
confirm this by finding the actual distance each car drove
4 * 45 = 180 mi the fast car
4.5*36 = 162 mi the slow car
-------------------------------
differs: 18 mi they are apart
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