SOLUTION: 9. At noon, Shelby leaves her house and drives 50
mph to her grandmother’s house. Her brother,
Swayze, leaves at 12:30 P.M. and drives
55 mph. What is the maximum length of ti
Algebra.Com
Question 1165512: 9. At noon, Shelby leaves her house and drives 50
mph to her grandmother’s house. Her brother,
Swayze, leaves at 12:30 P.M. and drives
55 mph. What is the maximum length of time
after Shelby leaves that Shelby is at least
15 miles ahead of Swayze?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Brother starts off 25 miles behind, which is a half hour at 50 mph.
He drives 5 mph faster, so he is closing the gap at 5 mph.
After two hours he will be 15 miles behind, since he will have gained 10 miles
The maximum length of time after Shelby leaves that she is at least 15 miles ahead ENDS at 2:30 pm.
It began after she was 15 miles away.
50 mph is 5/6 mile per minute
15/(5/6)=18 minutes to go 15 miles at that speed.
So the maximum length of time is from 12:18 pm to 2:30 pm or 2h12 m.
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