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If the second hand of a clock goes backwards, starting from 12:00, how long does it take until it meets with the minute hand (which is going forwards)?

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    $\begingroup$ As I wrote question is about seconds hand and minute hand. Seconds hand go backwards. And anwer should be with miliseconds. $\endgroup$ – user64076 Nov 27 '19 at 11:49
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    $\begingroup$ Does the minute hand also go backwards? $\endgroup$ – Ver Nick says Reinstate Monica Nov 27 '19 at 11:51
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    $\begingroup$ No, only seconds. $\endgroup$ – user64076 Nov 27 '19 at 11:51
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    $\begingroup$ "after what time will it meet" At what time in real life, or at what time on the clock face? $\endgroup$ – user45266 Nov 27 '19 at 22:02
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    $\begingroup$ Are you sure you mean the second hand is going backwards while the minute hand is going forwards? That sounds quite an... unnatural question? :P When I first look at the question, I think it sounds like you are rewinding the clock's second hand, and that way the minute hand will go backwards with it too, just like when you adjust time on a clock with minute hand and hour hand, when you turn back the minute hand, the hour hand goes back with it too. Just a thought :P $\endgroup$ – hellopeach Nov 28 '19 at 12:55
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The clockface is divided into $60$ segments.

The seconds hand travels at $1$ segment per second. The minutes hand at $\frac{1}{60}$ segment per second. Therefore they close in on each other at a speed of $\frac{61}{60}$ segments per second. Together they have to travel $60$ segments, which takes $60*\frac{60}{61} = 59+\frac{1}{61} = 59.016393$ seconds.

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  • $\begingroup$ Nice job! I might have done the maths in rpms instead; they are the exact same unit as "segments per second", but since the target is a full revolution, it gets rid of one factor of 60. (Because the answer is in minutes then.) $\endgroup$ – Bass Nov 27 '19 at 12:56
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Assuming it's a regular retail clock (except for the fact that the second hand runs backwards),

it's not possible to determine this to millisecond precision given the current information. All we can say that it happens somewhere between 59 and 60 seconds later. A second hand does not move continuously; it 'hops' from one second/minute mark to another. The speed of this hop determines when they meet exactly.

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    $\begingroup$ This depends entirely on the clock design. Many clocks have hands that sweep continuously without skipping between ticks. $\endgroup$ – Darrel Hoffman Nov 27 '19 at 21:50
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    $\begingroup$ Virtual-but-analog clocks (like my Apple Watch) maybe. I can assure you that the vast majority of physical clocks does not. I've never seen a continuous constant speed clock. The clocks used at train stations come close, but 1) they're not uniform 2) they go around in ~58 seconds and wait for a central signal to advance to the next minute. $\endgroup$ – Glorfindel Nov 27 '19 at 22:00
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    $\begingroup$ @Glorfindel FWIW, I've definitely seen some continuous second hands. I don't have any proof, but I'm pretty sure that research would reveal many clocks exist in both styles. $\endgroup$ – user45266 Nov 27 '19 at 22:04
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    $\begingroup$ Well I think that changed as most clocks now use a quartz drive which leads to distinct second ticks ... While older clocks that used a mechanical spring would lead to almost continuous motion of the hands $\endgroup$ – eagle275 Nov 29 '19 at 7:50
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    $\begingroup$ @Glorfindel I was looking for this answer more than the accepted one $\endgroup$ – Deunis Nov 29 '19 at 16:39
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A slightly different interpretation - using discrete positions of the hands, where a hand can only be in one position at any one time, and hands change positions "instantly" and simultaneously, I.e. a digital clock.

At 59 seconds, the second hand is one "tick" clockwise from the minute hand. At 60 seconds, the minute hand moves to the first tick, however the second hand moves from the first tick to the original location. They manage to pass by each other.

On the second rotation of the second hand though, it will enter the first tick position on the clock face, which unlike the previous rotation will not trigger the movement of the minute hand. (that is to happen on the next movement, when it reaches the top of the clock again)

It therefore takes the full 60 seconds for the first sweep, and 59 seconds of the second complete sweep before it ends up in the same location as the minute hand.

So 1m:59s, or 119s.

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    $\begingroup$ I think more people should look at this, it's a very valid consideration...and stresses once again how imprecise is human language for, well, things. $\endgroup$ – Stefano Nov 28 '19 at 18:29
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Nitpicky way of looking at it: the first time they meet will be after

exactly 0 seconds and 0 milliseconds.

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  • $\begingroup$ Omg that was so simple XD $\endgroup$ – Ver Nick says Reinstate Monica Nov 27 '19 at 21:31
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    $\begingroup$ trivial solution without value - as we "knew" already that they met exactly at 12:00 $\endgroup$ – eagle275 Nov 28 '19 at 9:20
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    $\begingroup$ One does not simply....make a question in puzzling without covering loopholes. $\endgroup$ – George Menoutis Nov 28 '19 at 14:48
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    $\begingroup$ My counternitpicking is that they can't "meet" if they are already together! $\endgroup$ – Stefano Nov 28 '19 at 18:27
  • $\begingroup$ @eagle275 Sorry if this solution offends you. I only posted it because it neatly avoids making assumptions about how the second hand should move on a clock (or the minute hand, for that matter – in my conception of an ideal clock, the minute hand should jump by 0.1 degrees each second instead of moving continuously). $\endgroup$ – wrtlprnft Nov 28 '19 at 20:26
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If we assume that only seconds hand goes backwards, then

It will take 59 and 1/120 seconds. Because when seconds hand reaches 1 second, the minute hand will have 1/60 of the segment left to go. Obviously, they will meet in the middle of the remainder.

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  • $\begingroup$ Will it really be in the middle of the remainder though? One is moving at 60 times the speed of the other (if both move continuously). Of course, if the second hand jumps in one second increments (like many clocks) the solution becomes more complex still... $\endgroup$ – Stiv Nov 27 '19 at 12:08
  • $\begingroup$ They will not meet in the half. They have different speed. $\endgroup$ – user64076 Nov 27 '19 at 12:23
  • $\begingroup$ @user64076 Yeah, I got it. $\endgroup$ – Ver Nick says Reinstate Monica Nov 27 '19 at 15:32
  • $\begingroup$ @Stiv Yeah, it's too complicated ) $\endgroup$ – Ver Nick says Reinstate Monica Nov 27 '19 at 15:32
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    $\begingroup$ You're assuming that for that last 1/120 of a second, the two hands have at the same average speed. This seems unlikely to me. Almost all the time, the second hand either travels considerably faster than the minute hand, or is stationary. $\endgroup$ – Dawood says reinstate Monica Nov 28 '19 at 2:13
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Say we're on the unit circle, oriented upwards though, and that the second hand starts at $2\pi$, while the minute hand starts at $0$. They spin in oposite directions.

The second hand spins with angular velocity $\omega_1 = -2\pi/60$, while the minute hand spins with angular velocity $\omega_2 = 2\pi/3600$. Equation will be $$2\pi - \omega_1 t = \omega_2 t \; \Rightarrow \; 1 = (1/60 + 1/3600) t$$ They meet after $\frac{3600}{61}$ seconds from starting time.

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