SOLUTION: boats are 600 miles apart. They begin traveling toward one another at the same time. Boat A is traveling at 25 MPH and boat B is traveling at 35 MPH. Use this information to anwer

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Question 565865: boats are 600 miles apart. They begin traveling toward one another at the same time. Boat A is traveling at 25 MPH and boat B is traveling at 35 MPH. Use this information to anwer the follow questions.
A. After how much time do the boats meet?
B. How far are they from where Boat A started?
C. After 12 hours how far apart are the 2 boats??

Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
boats are 600 miles apart. They begin traveling toward one another at the same time. Boat A is traveling at 25 MPH and boat B is traveling at 35 MPH. Use this information to anwer the follow questions.
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Their closing speed is 25+35 = 60 mph
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Time till they meet: 600/60 = 10 hrs.
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Boat A distance: 10*25 = 250 miles
Boat B distance: 10*35 = 350 miles
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A. After how much time do the boats meet?:; 10 hrs
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B. How far are they from where Boat A started?::250 miles
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C. After 12 hours how far apart are the 2 boats??:::
The boats have passed each other 2 hrs ago.
A boat: 25*2 = 50 miles
B boat: 35*2 = 70 miles
Total distance apart: 120 miles
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Cheers,
Stan H.
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Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


They are traveling toward each other so they are approaching each other at the sum of their speeds.

a. Distance equals rate times time, so time equals distance divided by rate. Divide the starting distance by the sum of their speeds to get the time it takes for them to meet.

b. Distance equals rate times time. Multiply the time answer to a) by the speed of boat A to get the distance traveled by A.

c. Subtract the answer to part a) from 12 to get the amount of time since the two boats passed each other. Distance equals rate times time. Multiply the sum of their speeds by the time since they passed each other.

John

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