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amilapsn
 one year ago
when does the minute hand and hour hand come one on one after 12 o' clock?
amilapsn
 one year ago
when does the minute hand and hour hand come one on one after 12 o' clock?

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3dw:1433320516438:dwIt's around 1:05. You can use math to get a more precise answer.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Let'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 hourhand 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\[x = \frac{60}{11} = 5.4545\cdots\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Therefore, the time is 1 hour and 5.4545... minutes. Or around 1:05:27.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2is it possible to model the situation wid vectors

dan815
 one year ago
Best ResponseYou've already chosen the best response.0we know both angular velocity so

dan815
 one year ago
Best ResponseYou've already chosen the best response.0u can write position vectors for both and equate them together

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2t is in hours : minute hand : <cos(2pi*60*t), sin(2pi*60*t)> hour hand : <cos(2pi*1*5m), sin(2pi*1*t)> how to solve

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2wolfram 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
 one year ago
Best ResponseYou've already chosen the best response.0i think its better to write it out as arc length wrt to time

dan815
 one year ago
Best ResponseYou've already chosen the best response.0and 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
 one year ago
Best ResponseYou've already chosen the best response.3\[\rm \omega_{second}= 2\pi / T = 2\pi ~rad / 60 sec = \pi/30 ~rad ~sec^{1}\]\[\rm \omega_{hour} = 2\pi / T = 2\pi ~rad / (60\times 60)sec = \pi /1800 ~ rad~sec^{1}\]We equate the positions now? (The position of second hand can be a multiple of that of the hour hand.)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3\[\rm \omega _{hour} = \frac{2\pi}{60 \times 60 \times 24} ~rad ~ sec^{1}\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i guess we can just jump straight to arc angles since arclength/r is a constant

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3I mean this isn't any different from what I did earlier. We're equating angles and using fancier terms. :P

dan815
 one year ago
Best ResponseYou've already chosen the best response.0ya its same thing lol, oky lets try to see vectors then

dan815
 one year ago
Best ResponseYou've already chosen the best response.0same simplifcation in polar form

dan815
 one year ago
Best ResponseYou've already chosen the best response.0its not really ideal to solve in cartesian form so

dan815
 one year ago
Best ResponseYou've already chosen the best response.0thats why wolfram giving up

dan815
 one year ago
Best ResponseYou've already chosen the best response.0you solving sec and min there eh
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