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OpenStudy (mathmath333):
OpenStudy (chaise):
You can find the length of the string using:
\[a^2=b^2+c^2\]
In this case, the hypotenuse (a) is the unknown in the equation, after that, you can work out how much string she pulls in, and the hyptenuse will reduce by that amount.
After that, you can work out how far the string is away from her, by using the hypotenuse and the height of her rod.
OpenStudy (chaise):
Using the same equation.
OpenStudy (mathmath333):
\(a=3\)
OpenStudy (chaise):
Yes.
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OpenStudy (chaise):
It would be a good idea to convert all your metres to centremetres.
1m=100cm
OpenStudy (mathmath333):
\(a=300\quad cm\)
OpenStudy (chaise):
Yes, now the string reduces in length by 5cm per a second for 12 seconds, for a total of 60cm
Now a=300-60cm -> 240cm
Now recalculate the unknown length.
OpenStudy (mathmath333):
i didnt get it
OpenStudy (mathmath333):
@amistre64
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OpenStudy (mathmath333):
@phi
OpenStudy (phi):
how far did you get ?
OpenStudy (mathmath333):
i got \(a=300 ~~cm\)
OpenStudy (mathmath333):
|dw:1428931082056:dw|
OpenStudy (phi):
ok, now they want the horizontal distance after 12 seconds. She reels in 5 cm/sec * 12 sec = 60 cm
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OpenStudy (mathmath333):
ok what should i do next
OpenStudy (phi):
I guess the "new hypotenuse" is 300 -60 = 240
draw in the new hypotenuse. can you do that ?
OpenStudy (mathmath333):
|dw:1428931342653:dw|
OpenStudy (phi):
I'm thinking it looks like this:
|dw:1428931400122:dw|
OpenStudy (mathmath333):
ok, wht should i do next
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OpenStudy (phi):
one reason people are forced to study math is it makes them have to think.
So you have to try to think.
Here is the whole picture
|dw:1428931540977:dw|