Rational Equality Word Problem

[kdaniel]

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]

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?

$\displaystyle Let \ t \ = \ time \ in \ minutes.$

$\displaystyle Let \ the \ rate \ = \ kilometers \ per\ minute.$

$\displaystyle Let \ the \ distance \ be \ in \ kilometers.$



$\displaystyle . . . . . . . . . . . . .. . . . rate \ \times \ time\ = \ distance$
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$\displaystyle Boat \ A . . . \ \frac{4}{t + 1} . . . \ \ (t + 1) \ .. . . . . . . \ 4$

$\displaystyle Boat \ B . . . . . \ \frac{4}{t} . . . . . . . . . . \ \ t \ \ . . . . . . . . . . . . . \ 4$


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

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


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]

Another way (no need to worry about minutes/hours):

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]

Thank you very much. Very helpful.

 

[Denis]

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

Thank you.