Answer I got was 1:05 and 5.08 seconds. I have two double-sided sheets with scribbles, cuz I hate calculators, so I warn you if I'm very off......
As for showing work, no-one would understand....
5 minutes are 2,5° for the hour handle, so 1 minute would be 0.5°, so 25s would be (1/2)*25/60=5/24°
(5/24)° would be 10*5/24s for the minute handle, that would be 25/12 = 2,083333s
So the time to the 1/100 s would be 1:05:27,08
But wait, I have the solution: they never meet, just as Zenon's the paradoxon with the turtle says...
But fortunately we have the planck length, and Heisenberg's uncertainty principle, so they will meet all the same,
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after 1:05:27,083 and before 1:05:27,084.
Right?
The seconds are correct, but you are off by about twenty hundredths of a second. And yes they do meet. But you are in the right neighborhood with that concept if you would be using calculus to solve it.
But I suppose then you would have to say "during a downpour", not"in a downpour"
You’re probably right. I also nitpick words in riddles. But
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if he just about entered the tunnel and stopped his wipers for this mile or so isn’t he still driving in a downpour?
Well now that I’m typing it, I must agree with you. He’s totally driving ‘during” a downpour.
I changed the original one so that those who didn’t chew up their pencils yet, shouldn’t do it for naught.
At 12:00 the hour and minute hand are on the same spot. Exactly when, to the hundredth of a second, will they be together again?
With some help I figured out the math sort of.
Mathematicians help out here...
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So we know the hour hand will move 1/11th between 1 and 2 for every 5 minutes the clock moves. And after 1:05 it will be meeting at the first of these 1/11. So do you divide it 1/11 by 60?
With some help I figured out the math sort of.
Mathematicians help out here...
Hidden:
So we know the hour hand will move 1/11th between 1 and 2 for every 5 minutes the clock moves. And after 1:05 it will be meeting at the first of these 1/11. So do you divide it 1/11 by 60?
With some help I figured out the math sort of.
Mathematicians help out here...
Hidden:
So we know the hour hand will move 1/11th between 1 and 2 for every 5 minutes the clock moves. And after 1:05 it will be meeting at the first of these 1/11. So do you divide it 1/11 by 60?
That won’t get you the answer exactly because you need to take into account the motion of the hours hand as well.
That won’t get you the answer exactly because you need to take into account the motion of the hours hand as well.
In my mind it makes sense lol..let me try a diff way
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..the hour hand also meets the minute hand 11 times in 12 hours.since it travels at the same speed it travels 1/11 Of the distance for each hour to meet the minute.
Say after 1 it’s plus 1/11
After 2 it’s plus 2/11.
In my mind it makes sense lol..let me try a diff way
Hidden:
..the hour hand also meets the minute hand 11 times in 12 hours.since it travels at the same speed it travels 1/11 Of the distance for each hour to meet the minute.
Say after 1 it’s plus 1/11
After 2 it’s plus 2/11.
The seconds are correct, but you are off by about twenty hundredths of a second. And yes they do meet. But you are in the right neighborhood with that concept if you would be using calculus to solve it.
Right, I forgot the regression that came afterwards...
