Question

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

Answer 1

It's around 1:05. You can use math to get a more precise answer.

Answer 2

thnk you

Answer 3

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\[x = \frac{60}{11} = 5.4545\cdots\]

Answer 4

Therefore, the time is 1 hour and 5.4545... minutes.

Or around 1:05:27.

Answer 5

is it possible to model the situation wid vectors

Answer 6

ye

Answer 7

we know both angular velocity so

Answer 8

u can write position vectors for both and equate them together

Answer 9

oh wow

Answer 10

t 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

Answer 11

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

Answer 12

i think its better to write it out as arc length wrt to time

Answer 13

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

Answer 14

\[\rm \omega_{second}= 2\pi / T = 2\pi ~rad / 60 sec = \pi/30 ~rad ~sec^{-1}\]\[\rm \omega_{hour} = 2\pi / T = 2\pi ~rad / (60\times 60)sec = \pi /1800 ~ rad~sec^{-1}\]We equate the positions now? (The position of second hand can be a multiple of that of the hour hand.)