SOLUTION: The Yankee Clipper leaves the pier at 9:00 AM at 8 knots. A half hour later, The River Rover leaves the same pier in the same direction traveling at 10 knots. At what time will The

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Question 100138This question is from textbook Glenco Algebra I
: The Yankee Clipper leaves the pier at 9:00 AM at 8 knots. A half hour later, The River Rover leaves the same pier in the same direction traveling at 10 knots. At what time will The River Rover overtake The Yankee Clipper?
I'm not sure how to solve this when both time and distance are missing. Thanks. This question is from textbook Glenco Algebra I

Found 2 solutions by checkley71, stanbon:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
8T=10(T-.5)
8T=10T-5
8T-10T=-5
-2T=-5
T=-5/-2
T=2.5 HOURS AFTER 9:00 AM THEY WILL MEET.
PROOF:
8*2.5=10(2.5-.5)
20=10*2
20=20

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The Yankee Clipper leaves the pier at 9:00 AM at 8 knots. A half hour later, The River Rover leaves the same pier in the same direction traveling at 10 knots. At what time will The River Rover overtake The Yankee Clipper?
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Comment: That is what algebra is all about; if something is missing,
call it "x" and solve for it.
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Clipper DATA:
Rate = 8 knots ; Distance = x miles ; time = d/r = x/8 hrs.
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Rover DATA:
Rate = 10 knots; Distance = x miles ; time = d/r = x/10 hrs.
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EQUATION:
Cipper time = Rover time + (1/2) hr.
x/8 = x/10 + (1/2)
Multiply thru by 40 to get:
5x = 4x + 20
x = 20 miles
Clipper time = 20/8= 5/2 hrs.
9:00 plus (5/2) hrs = 11:30 AM
Rover will over take Clipper at 11:30 AM
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Cheers,
Stan H.