Solve this challenging question?

Question

Between 3:00 PM and 4:00 PM, the minute hand of a clock exactly overlaps the hour hand. Many people say it is around 3:15 PM (approximately), but what is the exact time when this happens?

Show me the proof and express the result in "hh:mm:ss" format

Answer 1

The hour hand starts at 0° (measured from the x-axis), and moves in a complete circle every 12 hours x 60 minutes per hour = 720 minutes.

h(t) = 0° - t/720(360°)

The minute hand starts at 90° and moves in a complete circle every 60 minutes.

m(t) = 90° - t/60(360°)

0° - t/720(360°) = 90° - t/60(360°)
-t/2 = 90 - 6t
-t = 180 - 12t
11t = 180
t = 16 36/99 minutes = 0 hours, 16 minutes, 21 9/11 seconds

So the hands would cross at 3:16:21.81818181... pm.

(Man, Dr. Spock is getting rather emotional for a Vulcan, don't you think?)

Answer 2

Let's take the lining up of a hand at 12 o'clock as 0 degrees, and the lining up of the hand at 3 o'clock as 90 degrees, etc. Between 3:00 and 4:00, the hour hand moves from 3 to 4, and the minute hand moves through the whole clock.

The angle of the minute hand sweeps 360 degrees within the hour, in which time 60 minutes pases. So the hand moves 360/60 = 6 degrees per minute. The hour hand only moves between two numbers every hour. So it sweeps (360/12) = 30 degrees every hour, or 30/60 = 1/2 degree every minute.

At the stroke of 3:00, the minute hand is at 0 degrees and the hour hand is at 90 degrees. So at m minutes after 3:00, the minute hand is going to be pointing 6m degrees, and at the same time, the hour hand is going to be pointing 90 + (1/2)m.

To find the value of m where both of these angle measures are the same, just set the two equal to each other and solve for m:

6m = 90 + (1/2)m
(11/2)m = 90
m = 90*2/11
m = 180/11 = (176+4)/11 = 16 + 4/11

4/11 of a minute is about 22 seconds. So the hands cross at 3:16 and about 22 seconds.

Answer 3

Minute hand position = 0 + M (0 is the "12" position), M=minutes

The hour hand goes 1/12 as fast as the minute hand, so:
Hour hand position = 15+ M/12(3 is the "15" position)

We want them equal, as in the same place:

0+M = 15+ M/12

(11/12)M = 15

M= 16.3636...

0.3636... minutes = 21.8181... seconds

So, they cross at 03: 16 : 21.8181...seconds

Answer 4

Hi. It really depends on whether the minute hand is a smooth motion or moves one increment per second. 3:16:21.81818... pm

Answer 5

The time of hour/minute overlap between 3:00 PM and 4:00 PM is at exactly:

3h:16m:21.8181... (recurring)s PM,

that is, as better expressed in fractional terms,

3h:16m:21 9/11s PM or 3h:16 4/11m PM

Why is this? It's because the hands of a clock or watch line up every 12/11 hours, or every 1hour 5 5/11 minutes. The above time is simply three such successive intervals. Every line-up recurs at equal intervals of time because the hands move uniformly. Once the first line-up has happened, you can rotate the clock or watch so that the new line-up points "up", as the "12" direction did initially. That's as though you are starting exactly the same process all over again, but with the clock face having been simply rotated by the appropriate angle between overlaps.

Although the typical American student's first thought seems to be to turn to algebra to solve this and indeed every question, it is in fact QUITE UNNECESSARY. If, instead, one thinks about other (and simpler) approaches, it opens up quite remarkable connections with topics as diverse as Xeno's paradox, Geometric Series, Synodic and Sidereal Periods of Planets and many other fascinating phenomena. So, with the hope of broadening perspective, I offer the following two approaches. They can be understood and appreciated simply by observation, thinking, and basic arithmetic.

The simplest way to show that the basic overlap unit is 12/11 hours, or 1hour 5 5/11 minutes, is a very practical one. I'll call this method (a). Alternatively, you can turn that practical method into an easy thought experiment, method (b).

First, a preliminary result common to both methods (a) and (b).
Take (or imagine) an analogue watch, line up the hands at 12:00 and then rotate them forward in time. At first sight they overlap again at 1:05. However, it's a little more than that, as you can easily see, either explicitly (a) on the watch, or (b) mentally. How and why is this?:

After "1 hour" of turning, when the minute hand gets to the "12" position, the hour hand has reached the "1" position. So the minute hand, though it has gone "all the way round the dial," is now slightly behind and has not yet caught up to the hour hand again. After a further 5 minutes, when the minute hand has reached the "1" position, the hour hand has moved on slightly. They'll clearly pass just a little while later. So we should think of the first overlap as being at 1h 5 mins and a "bit."

O.K., now let's see how you can get the full and accurate answer separately in either approach (a) or approach (b).

First, method (a): If you now twirl the hands until they line up again at 12:00, you'll explicitly count that they overlapped just 11 times during that 12 hours. So the time between each overlap is 12/11 hours, obtained by merely watching and counting.

Second, method (b): Imagine repeating the hand turning that already established a recurrence time of 1h 5 mins and a "bit." The next times will be at 2h 10m and 2 "bits," then at 3h 15m and 3 "bits" ... up to 11h 55m and 11 "bits."

But clearly, this last expression must be identical with 12:00. Therefore 11 "bits" = 5 mins, or 1 "bit" = 5/11 mins. So the recurrence time is every 1h 5 5/11 mins, which is of course 12/11 hours as found from explicitly twirling the watch hands.

The particular time you were interested in is the 3h 15m and 3 "bits" time. 3 "bits" = 15/11 mins = 1 4/11 mins, so the required time is at 3h:16 4/11m. This is the basis for the three different ways in which I wrote the overlap time at the start of this answer.

There are very many other ways of establishing this particular result for clocks, and indeed for other situations far more general where the two rotational times (in this instance 1 hour and 12 hours) are not obviously comensurate. (In fact, the general case is best expressed in terms of RELATIVE ANGULAR SPEEDS on a circular race track, and how long it then takes for the "fast" rotator to GAIN 1 LAP on the "slow" rotator.)

For more information on this, the COMPLETELY GENERAL FORMULA, its connection with Xeno's Paradox, its relevance to astronomy ("overlapping of Sun, Earth, planet directions") and the development of modern tecnological society, see my "Best Answer" to the question "Non digital, traditional watch, how many times its minutes hand comes exactly over hours hand and after how?" asked by 'Ishfaq A' 4 or 5 days ago.

Live long and prosper.

P.S. A LATER ADDITION. By the way, you may have noticed the following:

I connected the specific problem of the overlap time between 3 PM and 4 PM to the more general problem involving repeated overlaps. With the basic REPEATED UNIT of 12/11 hours or 1 hour 5 5/11 minutes between overlaps established, you are now equipped to straightforwardly list EVERY overlap time (yes, ALL 11 of them), around the clock, simply by multiplying that basic unit of time by the integers 1 through 11. (In practice that means by the 9 integers 2 through 10 since you already know what 1 and 11 give you.)

The algebraic ways being suggested, as of this later writing, by 3 out of the 5 responders immediately below me, all deal SOLELY with just ONE specific overlap occasion part way through the day (between 3 and 4 PM). Using ONLY their specially tailored methods, you would have to ESTABLISH and SOLVE no fewer than 11 DIFFERENT algebraic equations in order to find all 11 overlap occasions. (Just think about it!)

With my method, you already know the 1st and 11th overlap time, and just have to multiply the basic repeated unit by 9 integers to have the rest.

You also end up with a more unifying insight, this way.

That is why it is often worthwhile to broaden one's perspective, and see a particular, specific problem as just one example of a more general class. Also, having a variety of techniques at one's command is far more useful and informative than using just one sledgehammer algebraic technique to address all mathematical problems.

Answer 6

I duno the scientific reason 4 this but i went to look at my clock as i restarted it, the answer is about 3.16pm. (I hope I'm right).