find out how much time passes between the times when the clock coincide.
find out how much time passes between times when the hands of the clock are perpendicular.
It would be great if some one could help me by telling me how to do it/ give me hints--don't give exact answer without explaining, please!!!
Here are a few similar problems from which you will acquire the knowledge needed to solve your particular problem.
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<< As a man leaves to go out to lunch, he notices the exact positions of the hands on the clock. When he returns, he notices that the hands have simple exchanged places. How long was he gone for lunch? >>
Consider the hour hand to be h degrees past noon and the minute hand to be m degrees past noon, m greater than h.
The hour hand moves at the rate of 30º/hr while the minute hand moves at the rate of 360º/hr.
The time for the hour hand to move to the minute hand position is T = (m-h)/30.
The time for the minute hand to move to the hour hand position is T = [360 - (m-h)]/360.
Equating both and simplifying yields 13m - 13h = 360 or (m-h) = 27.6923º.
The angle between the two hands at T1 is given by 360T1 - 30T1 = 27.6923 making T1 = .0391 hours or 5.0349 minutes after noon.
Thus, m = 5.0349(360)/60 = 30.2094º.
Similarly, h = 5.0349(30)/60 = 2.51745º.
m - h = 27.69195º.
Therefore, the elapsed time for the two hands to exchange positions is (360 - 27.6923)/360 = .9230 hours = 55.384 minutes.
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<< The minute and hour hands of a clock are exactly aligned at 12:00 Noon. What is the next time that they are EXACTLY aligned? >>
The hour hand will move 360º in 12 hours or .5º/min.
The minute hand will move 360º in 60 min. or 6º/min.
For the minute hand to catch up with the hour hand again, it will have to move 360º plus some interval from 12 past 1.
The angle x through which the hour hand moves in this time period is .5N, where N is the total number of minutes from 12N to the time being sought.
The angle 360 + x, through which the minute hand moves in this time period is 6N.
Solving for N and equating, we get x/.5 = (360 + x)/6 or 5.5x = 180 from which x = 32.727272 degrees.
Since the minutes hand moves 6º/min. or (1/6)min/deg, 32.727272deg = 32.727272(1/6) = 5.454545 min = 5 min.-27.27sec.
Thus, the hour and minute hands will again be coincident at 1:05:27.27 PM.
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<< The minute hand moves faster than the hour hand. It is currently 4:00 PM at what time will the minute hand overtake the hour hand? >>
They will meet at an angle µ from the 12. µ(h) for the hour hand is 120 + .0083333º/sec.(x). µ(m) for the minute hand is .10º/sec.(x), x being the number of seconds to their meeting. Since µ(h) must equal µ(m), we have 120 + .0083333x = .10x from which x = 1309.09 seconds = 21 min. - 49.09 sec.
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<< At what times of the day do the hands of the clock form right angles? >>
Start out at 12 midnight. The hour hand moves 30º in 60 minutes for a rate of .5º/m. The minute hand moves 360º in 60 minutes for a rate of 6º/m.
In proceeding from 12 midnight to 1 AM, the hour hand will rotate Hº = .5t degrees in time t. The minute hand will rotate Mº = 6t degrees in time t. We are looking for the point in time when the minute hand is 90º away from the hour hand or where Mº = Hº + 90 or 6t = .5t + 90 or 5.5t = 90 making t = 16.3636 minutes after midnight.
Similarly, when the minute hand has moved around the dial toward 1 AM, we seek the point where 6t = .5t + 270 making t = 49.0909 minutes after midnight.
In going from 1 AM to 2 AM, we experience 6t = .5t + 30 + 90 = .5t + 120 making t = 21.8181 minutes after 1 AM. Some 30 odd minutes later we have 6t = .5t + 30 + 270 = .5t + 300 making t = 54.5454 minutes after 1 AM.
In going from 2 AM to 3 AM, we experience 6t = .5t + 60 + 90 = .5t + 150 making t = 27.2727 minutes after 2 AM. Clearly, the next event occurs when 6t = .5t + 60 + 270 = .5t + 330 making t = 60 minutes or 3 AM.
See if you can figure out how many other times the event occurs and what the other times are.
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>This problem is hard and I have no way of going about starting it, i realy
>need some help.
>
>PART 1
>
>Thyme Daily was the only person to witness Ima Ford drive off the bridge into the turbulent water of the River Pow. When the police questioned her, all she could remember was that when she glanced at her watch it was between 2 and 3 o'clock and that the hands of her watche were precisely overlapping each other. The detective, wanting to be precise, but not relly sharp in math, is in need of help to find out the time. What time did Ima fall into
the river?
>
>PART 2
>
>Ms. Ford had left her home at 2 o'clock sharp. So tell me now how many
>degrees did the minute hand of Ms. Daily's watch move untill the moment of
>the accident. [This time you must expess your answer to the nearest 10th of a
>degree.]
>
>I am supposto find an equation to solve both of these but have no clue what
>to do. Any help I could be given would be great. Thank You. Ryan
The minute hand moves faster than the hour hand. Starting at 2:00, at what time will the minute hand overtake the hour hand, at which time they will overlap one another?
The numbers on the clock face are 30º apart. The hands will meet at an angle µ from the 12. µ(h), the angle for the hour hand, is [60 + (.0083333º/sec)x]. µ(m) for the minute hand is x[.10º/sec.], x being the number of seconds from 2:00 to their meeting. Since µ(h) must equal µ(m), we have 60 + .0083333x = .10x from which x = 654.545 seconds = 10min.-54.5sec. after 2:00.
The angle of the minute hand from 2:00 is therefore 654.545(.10) = 65.45º.
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<< At what time after 4:00 will the minute hand and the hour hand of a clock first be in the same position? >>
1--The hour hand moves 1 minute every .50º or 1M/.5º.
2--The minute hand moves 1 minute every 6º or 1M/6º.
3--When the two hands are coincident, the hour hand has moved Xº and the minute hand has moved (120 + X)º.
4--So, in T minutes, the two movements are (120 + X)1/6 = X/.5.
5--This leads to 60 + .5X = 6X or X = 10.909º.
6--Therefore, the hour hand has moved 10.909º, the minute hand 130.909º, and the minute hand has moved 20 + 10.909090(1/6) = 21.818 minutes past 4 o'clock.
Alternatively:
1--Let the speed (rate) of the hour hand be "r" and the speed (rate) of the minute hand be "12r" in minutes per minute.
2--Let the distance the hands move be measured by the minutes of the hour.
3--We are looking for the distance (in minutes) that the minute hand must move to overtake, and be coincident with, the hour hand.
4--Let this distance (or time in minutes) be "x."
5--Let the distance the hour hand must move in the same time period be (x - 20), the hour hand already being located at the "20 minutes after the hour" point.
6--As time is equal to distance divided by speed, the time for the minute hand to reach the point of coincidence is given by x/12r.
7--Similarly, the time for the hour hand to reach the same point is given by (x - 20)/r.
8--SInce the two hands each move the same amount of time, we can write x/12r = (x -20)/r or x = (12/11)20 = 21 9/11 = 21.8181 minutes.
9--Therefore, the two hands coincide at 04:21:49.09 or 4h-21m-49.09s after 12.
10--It is worth noting that in the expression x = (12/11)20, the quantity 20 represents the number of minutes that the hour hand is ahead of the minute hand to start out with.
11--Therefore, the value of 12/11 represents the number of minutes required for the minute hand to overtake the hour hand, per minute of head start.
12--For example, if we were seeking the time after 9 o'clock that the two hands overlapped, we need only multiply (12/11)45 = 49.0909 to derive the time of 09:49.0909 or 09h-49m-5.45s after 12.
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Shortly after 12 o'clock, the hour and minute hands of a clock forms an angle of 55 degrees. A person observes this then leaves for a short time, returning before one o'clock. He is astounded to observe that the clock hands now also form a 55 degree angle. How much time (exactly) elapsed while the person was absent? I have no clue how to do this and I guessed 15 mintues, but I know
>that's not right.
>
The hour hand moves at the rate of 30/60 = .5º per minute.
The minute hand moves at the rate of 360/60 = 6º per minutes.
The hour positions on the clock face are 30º apart.
Shortly after 12 o'clock, the hour hand is slightly past the 12 and the minute hand is 55º away from the hour hand.
Therefore, letting x represent the number of minutes elapsed from 12 o'clock, the minute hand has moved (6x)º and the hour hand has moved (.5x)º or 6x - .5x = 55. Solving, x = 10 meaning that the two hands are 55º apart at exactly 12:10 o'clock.
As for the return time, we can write 360 - 6x + .5x = 55 making 5.5x = 305 or x = 55.4545 = 54m-27.27s or 12:55:27.27 o'clock.
Thus, the elapsed time between the two events is 12:55:27.27 - 12:10:00 = 45:27.27 minutes.
