There is a man who lives on the top of a hill. He stays alone and has only an old grandfather clock in his house which tells him the time. Every day he winds the clock, else the clock will stop. One day he forgets and wakes up to see the clock has stopped at some point of time at night. Now he needs to go downhill to the village square and see the correct time in the Village Clock Tower (in this idyllic village, no one has a wristwatch, and everyone relies on the Clock tower for the time)\n\nHe has a horse which takes a fixed time to go downhill (say x) and a fixed time to go uphill (say y).  Going uphill obviously takes longer (i.e y>x). However, the man does not know the value of x and y. \n\nThe question is: How will he adjust the time on his grandfather clock to the correct time?
204168
(with the assumption that grand pa's clock is too heavy to carry to the village & no wrist watch meaning he has no tool to monitor the time goes)\nIt could not be done, no matter how many times he go forth and back and write down the time. I have tried seriously calculated every possible means.\nthe best he could do is:\n1. go to the the village and buy another well-adjusted clock to adjust the one at home. (this clock could be used as a mean to adjust the grand pa's clock in the future.)\n2. go to the village and memorize the position of the sun. the next day he could adjust the clock's hands with the sun's position (but then it won't be precise).\nor \n3. He could go to the village, \nwrite down what time it is,\ngo home to his house and go back to the village,\ncount the time lapsed (that's the x + y, or the time needed to go and come back house-village-house).\nwrite down what time it is when he left the village for the second time.\nAdjust the grand pa's clock with 'That last recorded time' + 'the time lapsed or (x+y)'.\n\nThat will make the grand pa's clock to show = 'the correct time' + x\n\n-> note that when he get home for the last time, his recorded time is already late y minutes for his trip back from the village to his house.\nbut because he added 'the lapse time' (x + y) then it is =\ncorrect-time - y + (x + y), and he get 'correct-time + x'\n\nSO, from that time on, if he has to be at the village by 5 o'clock, he should leave at least when the grand pa's clock show 5 o'clock because his clock is now x minutes early than the village's tower clock.\n\nthank you.
494380
You could solve this by figuring out the rate the hour hand turns and the minute hand turns, etc.  But it is much easier to think laterally.\n\nThe hour and minute hands of a clock line up perfectly at 12 noon, then again at regular intervals after that.  But those intervals are not an even hour.  It's a little more.\n\nBecause the hands line up *eleven* times in a 12 hour period, the interval between alignment is 12/11.\n\nIf you convert this to a time, it is exactly:\n12/11 = 1.0909... hours\n= 1 hour 5.4545... minutes\n= 1 hour 5 minutes 27.2727 seconds\n= 1 hour 5 minutes 27 3/11 seconds.\n\nSo the time when the hour and minute hands align perfectly will be at: 1:05:27 3/11\n\nFirst alignment: 1:05:27 3/11\nSecond alignment: 2:10:54 6/11\nThird alignment: 3:16:21 9/11\n...\nTenth alignment: 10:54:32 8/11\nEleventh alignment: 12:00:00\n\nThe time between one and two o'clock is:\n1 hour 5 minutes 27 3/11 seconds.
612840
First of all, lets just look at 12 hrs, as it will be the same for the next twelve:\n\nThe minute hand revolves around the clock at 1 rev per 60 minutes, the hour hand revolves around at 1 rev per 60•12 minutes.  Knowing that after 60 minutes the minute hand is back to where it started we can say that 2π≡0.  We want to rewrite the above relations to give us the positions as a value of time.  This can be done by assuming that t is time (in minutes) and using m=2π/60t (minute hand) and h=2π/(12•60)t (hour hand).  \nTherefore we want to solve m=h+2πk for some integer 0≤k≤10.  The reason that we don't have k go to 11 is because if you think about it, the hour and minute hand never cross during the 11th hour, they keep getting closer, but then at 12:00 (during the 12th hour) they meet.\n\nThus 2π/60t=2π/(12•60)t+2πk ==>> 12t=t+720k ==>>11t=720k ==>> t=720/11k.\n\nNow, this is the time after 12:00 (midnight or noon, I don't care they; are the same). \nThese times are:\n\n12:00 (k=0)\n1:60/11=1:05:300/11≈1:05:27.27 (k=1)\n2:10:600/11≈2:10:54.55\n3:16:240/11≈3:16:21.81\n4:21:540/11≈4:21:49.09\n5:27:180/11≈5:27:16.36\n6:32:480/11≈6:32:43.64\n7:38:120/11≈7:38:10.91\n8:43:420/11≈8:43:38.18\n9:49:60/11≈9:49:05.45\n10:54:360/11≈10:54:32.73\n\nI decided to give you the second values as fractions also, since they can't be given accurately by a terminating decimal.
575882
Did it not take time to go from the catcher to first base.  If so, you were not ahead of time.  You may have been going faster than the person running, but that doesn't make you ahead of time.  Learn relativity for if you were moving at a speed some what close to the speed of light, then time on earth would be moving quite faster than your reference point.
360404
Imagine the average human life of about 70 years (hypothetically).\n\nMake a time line of that persons life of 70 years.  Just for giggles we will make the time line one meter long, for every year of this person's life.  So you have a time line that is 70 meters long.  \n\nLooks good, right.  70 years is almost three quarters of a century, which in perspective to us as humans, is a relatively long time.  \n\nImagine that somone could actually live for 7000 years.  Now, make a time line that is 7000 meters long for this second person.  This is our "extended time line".  \n\nThis 7000 Meter time line or 7 Kilometers is a rather larger distance in perspective to the first 70 meter line.\n\nNow imagine laying both lines down, parallel to each other in a clear open field somewhere and imagine looking down at the two lines from the sky.  \n\nOf course the shorter line is shorter, but compared to the longer line it will appear extremely small.  The small line may even appear to have ceased to exist when compared to the longer line.  \n\nDoes this help put in into perspective?\n\nKinda like having a single specific drop of water, going to the beach and dropping it into the ocean.  Now go find that specific drop of water that you released into the ocean.  It is still there, but compared to all of the other drops of water that make up the ocean, it is all but non-existent.
420869
It depends on what watch, but some watches the little thing you pull out to set them can pull out half way and all the way, the half way will let you set the date and the all the way will let you set the time. I hope that helps :)
614149
the clock will show again same when the difference between them is 12 hrs..\nnow for every 1 hr the differece is 3 min(1 clock goes 1 min forward and other 2 min back)\nso for difference to be 12 hrs( 12*60 mins) time elapsed=\n\n 12*60/3 = 240 hrs= 10 days..\nso on 10 th of Jan 2004..they will show the same time..\nhow can they show the correct time..unless its not a day after or before the given time..i mean there will be 24 hrs diff..
561709
It's all relative. Easiest way to see the answer to this problem is to first imagine the river and log stationary. Assuming the boat keeps the same speed S, he travels an hour and then turns around and takes an hour to get back(a total of 2hrs). If the river/log has some speed v then when he travels away from the log he has speed S+v (with current) but since the log has speed v he travels at speed S away from the log (same as the stationary case) and on the way back has speed S-v (against current) and the log has speed v  and so he travels at speed v+S-v = S towards the log. So it is the same as the stationary case: one hr out and one hr back = 2hr.
210639
Zeno ignored the fact that the infinitely many smaller steps take infinitely less amounts of time to do.\n\nAs the earlier answerer pointed out, sum the power series and you get a finite value for the time taken...\n\nLike a lot of these logical tricks, there's something being forgotten or ignored to make it look impossible.
583664
1. Time is a variable and not a constant. It ran backward only in Harry Potter's Vol 3 using Time Machine\n2.The perpendicular distance is the shortest distance\n3.Why not\n4.The variable nature of time can be felf only with objects moving with a velocity that approaches that of light. The theory of Relativity cannot be proved wrong because no one has understood it completely
11717
Einstein said we are living in the past time, that means we have not "now time"!\nSo, nobody can not answer, what time is it! maybe we can ask, what time was it!!
49305
Yes and no, Time travel is almost certainly not posiible and at the same time might be. It is believed that going faster than the speed of light would allow you to go back through time, but this is supposed to be impossible, but hawking radiation does it every second as it escapes black holes so it must be possible. Although if it does go back in time wouldn't it just go back to the time when it was in the black hole and not be observable to us. So maybe.\n\nAlso their is the Granfather paradox, which says that if you go back in time and kill your grandfather, then you never could exist and go back and kill your grandfather, so you can't, but then you never killed your granfather, so you do exist and kill your grandfather, so you don't exist and can't, it goes on for as long as you want to think about it. So their is another theory to bypass that one, it says that if you leave this exact point in time, you enter another universe altogether, one that you can change without consequence, making it possible to kill your grandfather, but als killing the reality of the universe you came from. Interesting huh?
368981
The fact that he took his return trip later just complicates the problem. The question can be restated as one person leaves from the top of the mountain and one person leaves from the bottom of the mountain. They leave at the same time. Explain why, if they follow the same path, they will run into each other each other at some point. Stated this way, it is obvious, right?\n\nIf you want a more rigorous solution, here you go. This is a continuity question. Let f(t) denote the distance *from the bottom of the mountain* at time t on the trip up the mountain and let g(t) denote the distance *from the bottom of the mountain* at time t. Change your units so that your time starts at zero (so 8AM is t=0) and you make it up the mountain in time 1 (so 1PM is t=1) and your distance from the bottom to the top of the mountain is 1. With this set up, g(x) lies in the interval [0,1] for every x in the interval [0,1], and the same holds for f(x). Consider the function g(t)-f(t). The physical interpretation makes it clear that f(t) and g(t), hence g(t)-f(t), are continuous functions of t; that is, you don't just disappear at one point and reappear at another point without having travelled through every point in between. g(0)-f(0)=1-0=1>0 and g(1)-f(1)=g(1)-1<=0. The intermediate value theorem tells you that there is some time t_0 such that g(t_0)-f(t_0)=0. That is, there is some time t_0 such that g(t_0)) = f(t_0). Your deja vu moment happens at time t_0.
270648
An alternative method:\nThe times taken by A, B and C to complete one lap around the track would be 10/7, 10/8 and 10/9 mins, respectively. Now we just need the LCM of these three to find out when they will all catch up with each other. \n\nThis turns out to be 10 mins.
137562
This is where convention comes in. hehe.\n\nYou see, today "time" is defined based on a reference. We define one second as exactly 9,192,631,770 oscillations or cycles of the cesium atom's resonant frequency, replacing the old second that was defined in terms of the Earth's motions.\n\nThe "uncertainity" is due to the fact that our one second here ( where its based on the measurement of this number of oscilations )  is NOT 100% the same as compared to our measurement of one second based on the previous way.\n\nWhat was the previous way? Well, it bascially defines one second the other way around. What i mean here is that in the old days time is defined relative to the motion of the earth around the sun. Ancients defined the total time taken for the earth to finish one complete revolution around the sun as one year. Then they agreed that one year is approximately made up of 365 days. Each day is then divided into 24 hours. 1 hour is then considered as having 60 minutes where 1 minute comprises as 60 seconds.\n\nOf course to achieve the definition of time as how we have it now did not just come overnight. It took some time... hehe. I did not attempt a chronological development of the idea but just intend to shed some light on the subject. You can find further information on the link below.
415817
You can't move through something that doesn't exist! Time is not a NATURAL feature of the universe but simply a concept invented by our species to separate events into past, present and future. There is no Cosmic Master Clock that keeps track of absolute, universal time. There is no physical "past" trailing along behind the "present," nor any physical "future" waiting up ahead for us to get there.
175508
If the clock is constantly moving at constant velocity, then you're right, the people on earth will say the plane's clock is going slower, whereas the people on the plane will say the earth's clocks are slow.\n\nHowever, when the plane starts and stops in these experiments, it undergoes an acceleration, which differentiates the two frames. That's why the plane's clock is behind when it's compared to the earth clock. Basically it's because the plane changes which inertial frames it is a part of when it accelerates. Your point is similar to the twin paradox, I suggest you check out my source, that should clear things up, you sound like you have a solid enough background to understand it.