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Question Number 17449 by ajfour last updated on 06/Jul/17
Between 2:00 and 2:15, what time  is it exactly when the hour,  minute, and second′s hand of a  clock occupy the same angular  position.
$$\mathrm{Between}\:\mathrm{2}:\mathrm{00}\:\mathrm{and}\:\mathrm{2}:\mathrm{15},\:\mathrm{what}\:\mathrm{time} \\ $$$$\mathrm{is}\:\mathrm{it}\:\mathrm{exactly}\:\mathrm{when}\:\mathrm{the}\:\mathrm{hour}, \\ $$$$\mathrm{minute},\:\mathrm{and}\:\mathrm{second}'\mathrm{s}\:\mathrm{hand}\:\mathrm{of}\:\mathrm{a} \\ $$$$\mathrm{clock}\:\mathrm{occupy}\:\mathrm{the}\:\mathrm{same}\:\mathrm{angular} \\ $$$$\mathrm{position}. \\ $$$$ \\ $$
Commented by mrW1 last updated on 06/Jul/17
in the interval [2:00−2:12) the hour  hand remains at 2:00≡position 10  since the hour hand moves a step  every 12 minutes.    in the interval [2:10−2:11) the minute  hand remains at position 10, since  the minute′s hand moves a step  every 60 seconds.    at 2:10:10 the second hand is at   position 10.
$$\mathrm{in}\:\mathrm{the}\:\mathrm{interval}\:\left[\mathrm{2}:\mathrm{00}−\mathrm{2}:\mathrm{12}\right)\:\mathrm{the}\:\mathrm{hour} \\ $$$$\mathrm{hand}\:\mathrm{remains}\:\mathrm{at}\:\mathrm{2}:\mathrm{00}\equiv\mathrm{position}\:\mathrm{10} \\ $$$$\mathrm{since}\:\mathrm{the}\:\mathrm{hour}\:\mathrm{hand}\:\mathrm{moves}\:\mathrm{a}\:\mathrm{step} \\ $$$$\mathrm{every}\:\mathrm{12}\:\mathrm{minutes}. \\ $$$$ \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{interval}\:\left[\mathrm{2}:\mathrm{10}−\mathrm{2}:\mathrm{11}\right)\:\mathrm{the}\:\mathrm{minute} \\ $$$$\mathrm{hand}\:\mathrm{remains}\:\mathrm{at}\:\mathrm{position}\:\mathrm{10},\:\mathrm{since} \\ $$$$\mathrm{the}\:\mathrm{minute}'\mathrm{s}\:\mathrm{hand}\:\mathrm{moves}\:\mathrm{a}\:\mathrm{step} \\ $$$$\mathrm{every}\:\mathrm{60}\:\mathrm{seconds}. \\ $$$$ \\ $$$$\mathrm{at}\:\mathrm{2}:\mathrm{10}:\mathrm{10}\:\mathrm{the}\:\mathrm{second}\:\mathrm{hand}\:\mathrm{is}\:\mathrm{at}\: \\ $$$$\mathrm{position}\:\mathrm{10}. \\ $$
Commented by ajfour last updated on 06/Jul/17
yes sir, i understood your answer  (with floor function) only a little  later after i commented.
$$\mathrm{yes}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{understood}\:\mathrm{your}\:\mathrm{answer} \\ $$$$\left(\mathrm{with}\:\mathrm{floor}\:\mathrm{function}\right)\:\mathrm{only}\:\mathrm{a}\:\mathrm{little} \\ $$$$\mathrm{later}\:\mathrm{after}\:\mathrm{i}\:\mathrm{commented}. \\ $$
Commented by mrW1 last updated on 07/Jul/17
I think, if the hands move  continiously, they stay all in the same  position only at 12:0:0.
$$\mathrm{I}\:\mathrm{think},\:\mathrm{if}\:\mathrm{the}\:\mathrm{hands}\:\mathrm{move} \\ $$$$\mathrm{continiously},\:\mathrm{they}\:\mathrm{stay}\:\mathrm{all}\:\mathrm{in}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{position}\:\mathrm{only}\:\mathrm{at}\:\mathrm{12}:\mathrm{0}:\mathrm{0}. \\ $$
Commented by ajfour last updated on 07/Jul/17
complete research, thank you sir.
$$\mathrm{complete}\:\mathrm{research},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by mrW1 last updated on 06/Jul/17
the hands on a clock don′t move  continiously.  there are totally 60 possible positions  for the three hands on a clock:  p=0,1,2,...59.    let′s say x hours after 2 o′clock.  position of hour′s hand is p_h = 10+⌊5x⌋  position of minute′s hand is p_m = ⌊60x⌋  position of second′s hand is p_s = ⌊((3600x)/(60))⌋  at 2:10:10  x=0h10m10s=0.1694... hour  p_h =10+⌊5×0.1694⌋=10  p_m =⌊60×0.1694]=10  p_s =⌊((3600×0.1694)/(60))⌋=10  ⇒they are at the same position.
$$\mathrm{the}\:\mathrm{hands}\:\mathrm{on}\:\mathrm{a}\:\mathrm{clock}\:\mathrm{don}'\mathrm{t}\:\mathrm{move} \\ $$$$\mathrm{continiously}. \\ $$$$\mathrm{there}\:\mathrm{are}\:\mathrm{totally}\:\mathrm{60}\:\mathrm{possible}\:\mathrm{positions} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{three}\:\mathrm{hands}\:\mathrm{on}\:\mathrm{a}\:\mathrm{clock}: \\ $$$$\mathrm{p}=\mathrm{0},\mathrm{1},\mathrm{2},…\mathrm{59}. \\ $$$$ \\ $$$$\mathrm{let}'\mathrm{s}\:\mathrm{say}\:\mathrm{x}\:\mathrm{hours}\:\mathrm{after}\:\mathrm{2}\:\mathrm{o}'\mathrm{clock}. \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{hour}'\mathrm{s}\:\mathrm{hand}\:\mathrm{is}\:\mathrm{p}_{\mathrm{h}} =\:\mathrm{10}+\lfloor\mathrm{5x}\rfloor \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{minute}'\mathrm{s}\:\mathrm{hand}\:\mathrm{is}\:\mathrm{p}_{\mathrm{m}} =\:\lfloor\mathrm{60x}\rfloor \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{second}'\mathrm{s}\:\mathrm{hand}\:\mathrm{is}\:\mathrm{p}_{\mathrm{s}} =\:\lfloor\frac{\mathrm{3600x}}{\mathrm{60}}\rfloor \\ $$$$\mathrm{at}\:\mathrm{2}:\mathrm{10}:\mathrm{10} \\ $$$$\mathrm{x}=\mathrm{0h10m10s}=\mathrm{0}.\mathrm{1694}…\:\mathrm{hour} \\ $$$$\mathrm{p}_{\mathrm{h}} =\mathrm{10}+\lfloor\mathrm{5}×\mathrm{0}.\mathrm{1694}\rfloor=\mathrm{10} \\ $$$$\left.\mathrm{p}_{\mathrm{m}} =\lfloor\mathrm{60}×\mathrm{0}.\mathrm{1694}\right]=\mathrm{10} \\ $$$$\mathrm{p}_{\mathrm{s}} =\lfloor\frac{\mathrm{3600}×\mathrm{0}.\mathrm{1694}}{\mathrm{60}}\rfloor=\mathrm{10} \\ $$$$\Rightarrow\mathrm{they}\:\mathrm{are}\:\mathrm{at}\:\mathrm{the}\:\mathrm{same}\:\mathrm{position}. \\ $$
Commented by mrW1 last updated on 06/Jul/17
other times when the three hands of a clock  are at the same position:  12:0:0  1:5:5  2:10:10  3:16:16  4:21:21  5:27:27  6:32:32  7:38:38  8:43:43  9:49:49  10:54:54  11:59:59
$$\mathrm{other}\:\mathrm{times}\:\mathrm{when}\:\mathrm{the}\:\mathrm{three}\:\mathrm{hands}\:\mathrm{of}\:\mathrm{a}\:\mathrm{clock} \\ $$$$\mathrm{are}\:\mathrm{at}\:\mathrm{the}\:\mathrm{same}\:\mathrm{position}: \\ $$$$\mathrm{12}:\mathrm{0}:\mathrm{0} \\ $$$$\mathrm{1}:\mathrm{5}:\mathrm{5} \\ $$$$\mathrm{2}:\mathrm{10}:\mathrm{10} \\ $$$$\mathrm{3}:\mathrm{16}:\mathrm{16} \\ $$$$\mathrm{4}:\mathrm{21}:\mathrm{21} \\ $$$$\mathrm{5}:\mathrm{27}:\mathrm{27} \\ $$$$\mathrm{6}:\mathrm{32}:\mathrm{32} \\ $$$$\mathrm{7}:\mathrm{38}:\mathrm{38} \\ $$$$\mathrm{8}:\mathrm{43}:\mathrm{43} \\ $$$$\mathrm{9}:\mathrm{49}:\mathrm{49} \\ $$$$\mathrm{10}:\mathrm{54}:\mathrm{54} \\ $$$$\mathrm{11}:\mathrm{59}:\mathrm{59} \\ $$

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