amilapsn
  • amilapsn
when does the minute hand and hour hand come one on one after 12 o' clock?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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ParthKohli
  • ParthKohli
|dw:1433320516438:dw|It's around 1:05. You can use math to get a more precise answer.
amilapsn
  • amilapsn
thnk you
ParthKohli
  • ParthKohli
Let's say it's 1 hr and \(x\) min. What does it mean when two hands coincide? How do we put it mathematically? It means that we can make an equation. What do we equate? The angles. We equate the angles. At 1 hour and \(x\) minutes, the angle created by the hour-hand wrt 12 is:\[30^{\circ}\left( 1+\dfrac{x}{60}\right)\]and that created by the minute hand wrt 12 is:\[360^{\circ }\cdot\dfrac{x}{60}\]Equate those two and you get\[x = \frac{60}{11} = 5.4545\cdots\]

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ParthKohli
  • ParthKohli
Therefore, the time is 1 hour and 5.4545... minutes. Or around 1:05:27.
ganeshie8
  • ganeshie8
is it possible to model the situation wid vectors
dan815
  • dan815
ye
dan815
  • dan815
we know both angular velocity so
dan815
  • dan815
u can write position vectors for both and equate them together
ParthKohli
  • ParthKohli
oh wow
ganeshie8
  • ganeshie8
t is in hours : minute hand : hour hand : how to solve
ganeshie8
  • ganeshie8
wolfram is giving up http://www.wolframalpha.com/input/?i=solve+cos%282pi*60*t%29%3Dcos%282pi*1*t%29%2C+sin%282pi*60*t%29%3Dsin%282pi*1*t%29%2C+0%3Ct%3C2
dan815
  • dan815
i think its better to write it out as arc length wrt to time
dan815
  • dan815
and equate the 2 sinusoids and u shud get 12 solutions which is easy to see from the clocks and the periods that are forming
ParthKohli
  • ParthKohli
\[\rm \omega_{second}= 2\pi / T = 2\pi ~rad / 60 sec = \pi/30 ~rad ~sec^{-1}\]\[\rm \omega_{hour} = 2\pi / T = 2\pi ~rad / (60\times 60)sec = \pi /1800 ~ rad~sec^{-1}\]We equate the positions now? (The position of second hand can be a multiple of that of the hour hand.)
dan815
  • dan815
|dw:1433321503977:dw|
ParthKohli
  • ParthKohli
Oh, whoops...
ParthKohli
  • ParthKohli
\[\rm \omega _{hour} = \frac{2\pi}{60 \times 60 \times 24} ~rad ~ sec^{-1}\]
dan815
  • dan815
i guess we can just jump straight to arc angles since arclength/r is a constant
ParthKohli
  • ParthKohli
I mean this isn't any different from what I did earlier. We're equating angles and using fancier terms. :P
dan815
  • dan815
ya its same thing lol, oky lets try to see vectors then
dan815
  • dan815
|dw:1433321910738:dw|
dan815
  • dan815
same simplifcation in polar form
dan815
  • dan815
its not really ideal to solve in cartesian form so
dan815
  • dan815
thats why wolfram giving up
ganeshie8
  • ganeshie8
https://www.desmos.com/calculator/k3awg1pnpm
dan815
  • dan815
you solving sec and min there eh
ganeshie8
  • ganeshie8
Oh right :/

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