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At what time exactly does the minute hand and the hour hand in a clock becomes straight between 7O'clock & 8O'clock. Iam getting the time is in between 7:05 and 7:10 but what after that. can it be solved using graph theory ?

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    $\begingroup$ You should calculate the speeds of hour and minute hand. I thinhk physics would be more usefull then graph theory:) $\endgroup$ – Antoine Jul 13 '13 at 13:13
  • $\begingroup$ When I was in high school — but that was in an earlier geological epoch — this was a typical question for 11th-graders taking algebra. $\endgroup$ – Lubin Jul 13 '13 at 16:47
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Try writing expressions for the angle of the minute hand and the angle of the hour hand with respect to time during that interval. Then set the expressions equal and solve.

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  • $\begingroup$ Wouldn't the expression for the minute-hand angle have to have a $\mod 360$ in it? $\endgroup$ – DJohnM Jul 14 '13 at 0:02
  • $\begingroup$ No, because the time interval is only one hour. If you measure angle from the 12 on the clock face, it's just a linear expression. $\endgroup$ – augurar Jul 14 '13 at 7:46
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A visual proof. In the blue points the hands are overlapping, in the green points they are straight, int the red points they are orthogonal. enter image description here

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So if you look at a clock, I'd hope you can figure out that the hour hand covers 60 minutes in one hour. The minute hand covers 5 minutes in one hour. Now assume that both hands move at a constant rate (though not the same constant rate for each hand). Could you figure it out from here?

Alright, if you haven't tried to figure it out, the hour hand moves at 12 times the rate of the minute hand since (60/5)=12. At 7 o'clock the minute hand has a 35 minute head start on the hour hand. At 7:35 the hour hand will reach the 7:35 mark, and the minute hand will have traveled (5*(35/60))=(35/12) of the way to 7:40. So, the minute hand has almost reached 7:38. Once the hour hand reaches 7:38, the minute hand will have moved past 7:38 since the minute will have reached (5*(38/60))=(19/6) of the way to 7:40. But, once the hour hand moves one more minute, the minute hand won't have reached that spot also. Consequently, the time where they meet lies between 7:38 and 7:39 on the clock.

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Hour hand goes $360$ degrees in $12$ hours, or $30$ degrees/hour, or $0.5$ degrees per minute.

Minute hand goes $360$ degrees in $60$ minutes. or $6$ degrees/minute.

So the minute hand moves $5.5$ degrees/minute faster then the hour hand.

The minute hand catches the hour hand when the minute hand travels $360$ degrees more than the hour hand.

So the minute hand will catch the hour hand for the seventh time after $\frac{7 \times 360}{5.5}$ minutes, or $458.1818...$ minutes.

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I have written an equation that models the total angle between the second, minute, and hour hands on the clock:

|6(t mod 60)-(1/10)(t mod 3600)|+|6(t mod 60)-(1/120)(t mod 43200)|+|(1/10)(t mod 3600)-(1/120)(t mod 43200)| = ∆hº

I plugged it in to Desmos where you can see the equation and the graph. Finding the zeros of the graph will yield the the times when all three hands overlap (accurate to the second). The graph turns out to be pretty interesting: https://www.desmos.com/calculator/i7z7yfuk2w

The zero between 7 and 8 hours turns out to be 7:38:38.

As a note on the graph, all times are represented in seconds, i.e. 1:05 = 3900 seconds.

Hope this helps

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At the instant that the hands are in line, let the number of degrees after the hour be 'n' for the hour hand. At this instant it follows that the number of degrees after the hour for the minute hand will therefore be 'n + 180'. Since the minute hand travels 12 times faster than the hour hand, it follows that at this instant: 12n - n = 180 So: 11n = 180. So: n = 16.36 (recurring) degrees. Converting this back to minutes of time, n = 16.36 x (60/360) = 2.7272 minutes = 2 minute 44 seconds. Thus: The two hands of a clock will be aligned in a straight line at 32 minutes 44 seconds after each hour. I'd welcome the pointing out of any flaws in this argument!

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  • $\begingroup$ The main flaw is that the only time the hands are aligned at the top of the hour is at $12$. Indeed the first time after $12$ when the hands are at $180$ degrees is at $12$:$32$:$43.6363\ldots$. As pointed out in the question, however, the first $180$-degree alignment after $7$ is sometime between $7$:$05$ and $7$:$10$. In particular, it's exactly $32.7272\ldots$ minutes after the previous alignment of the hands, which occurred a few minutes after $6$:$30$. $\endgroup$ – David K May 22 '16 at 16:28

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