Rational Equality Word Problem

kdaniel

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Nov 13, 2010
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Any help here would be greatly appreciated. I only need the formula from the problem, I can figure it out from there. Thanks in advance.

QUESTION:
For the annual Breakwater Sailboat Race, sailboats start singly at 1 minute intervals. Boat B is
scheduled to start immediately following Boat A. The captain of Boat B feels that no matter how
fast Boat A travels his boat can sail 1 km/h faster. What is the average speed of each boat if
Boat B overtakes Boat A at the 1st bouy, 4 km from the start?
 

lookagain

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Aug 22, 2010
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2,397
kdaniel & edit said:
I only need the formula from the problem, \(\displaystyle I \ can \ figure\ it \ out \ from \ there.\)

QUESTION:
..., sailboats start singly at 1 minute intervals. Boat B is scheduled to start immediately
following Boat A. The captain of Boat B feels ...his boat can sail 1 km/h faster. What is
the average speed of each boat if Boat B overtakes Boat A ... 4 km from the start?
\(\displaystyle Let \ t \ = \ time \ in \ minutes.\)

\(\displaystyle Let \ the \ rate \ = \ kilometers \ per\ minute.\)

\(\displaystyle Let \ the \ distance \ be \ in \ kilometers.\)



\(\displaystyle . . . . . . . . . . . . .. . . . rate \ \times \ time\ = \ distance\)
------------------------------------------------------

\(\displaystyle Boat \ A . . . \ \frac{4}{t + 1} . . . \ \ (t + 1) \ .. . . . . . . \ 4\)

\(\displaystyle Boat \ B . . . . . \ \frac{4}{t} . . . . . . . . . . \ \ t \ \ . . . . . . . . . . . . . \ 4\)


\(\displaystyle Boat \ B's\ rate \ is \ 1 \ km/hr \ faster \ than \ Boat \ B's \ rate., \ or \ \frac{1}{60} \ km/min \ \ in \ addition.\)

\(\displaystyle Boat \ A's \ rate \ + \frac{1}{60} km/min \ = \ Boat \ B's \ rate.\)


A possible equation:


\(\displaystyle \frac{4}{t + 1} + \frac{1}{60} = \frac{4}{t}\)


Now you may solve this equation . . .

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Observations/hints:

1) After you solve for t, you must plug it into which expressions to calculate the boats' rates?

2) Those rates are in km/min, so if you want rates in km/hour, you must convert those rates
to get the corresponding rates in km/hr.

3) This problem is made relatively more complicated to a solver because there is a
conversion of units needed, say, hours to minutes, or minutes to hours.


\(\displaystyle "I \ only \ need \ the \ formula \ for \ the \ problem \ ,..."\)
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,644
Another way (no need to worry about minutes/hours):
Code:
A(@ b-1 kmph).............4 km................>t hours

B(@ b   kmph).............4 km................>t - 1/60 hours
A : t = 4 / (b-1)
B : t = 4/b + 1/60
So:
4 / (b-1) = 4/b + 1/60 ; simplify:
b^2 - b - 240 = 0
(b - 16)(b + 15) = 0
b = 16 ; ok?
 

kdaniel

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Nov 13, 2010
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Thank you very much. Very helpful.
 

Denis

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Joined
Feb 17, 2004
Messages
1,644
lookagain said:
\(\displaystyle "I \ only \ need \ the \ formula \ for \ the \ problem \ ,..."\)
Thank you.
 
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