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Math Help - Clock problem

  1. #1
    Junior Member
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    Clock problem

    Hi,

    Can anybody help with this problem please:

    At 3pm the big hand of a clock is pointing at 12, and the small hand pointing at 3 (at right angles, obviously). The question is what time will it be when the hands are at right angles again for the first time?

    This problem is driving me nuts. I know there's probably a very simple way to do it but it's eluding me at the moment. I have worked out position vectors and velocity vectors and i'm thinking that it's something to do with dot product equal to 0 and solving for t or else using the distance between the 2 position vectors being root 2 (assuming the clock radius is 1 unit).

    Any suggestions? Apart from switching to a digital clock

    Thanks
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  2. #2
    MHF Contributor

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    Quote Originally Posted by jackiemoon View Post
    Hi,

    Can anybody help with this problem please:

    At 3pm the big hand of a clock is pointing at 12, and the small hand pointing at 3 (at right angles, obviously). The question is what time will it be when the hands are at right angles again for the first time?

    This problem is driving me nuts. I know there's probably a very simple way to do it but it's eluding me at the moment. I have worked out position vectors and velocity vectors and i'm thinking that it's something to do with dot product equal to 0 and solving for t or else using the distance between the 2 position vectors being root 2 (assuming the clock radius is 1 unit).

    Any suggestions? Apart from switching to a digital clock

    Thanks
    No, you don't have to do anything with vectors or anything "advanced". Just look at the angles made with the vertical.

    As the hour hand goes from 3 to 4 the minute hand will travel all the way around the clock so there must be a time between 3 and 4 when the two hands are at right angles. x minutes after 3, the minute hand will make an angle of (x/60)(360)= 6x degrees with the vertical. During that same time the hour hand will have moved a fraction x/60 of the angle between 3 and 4. Since there are 12 hour steps in 360 degrees the angle between 3 and 4 is 360/12= 30 degrees. The hour hand will have moved (x/60)(30)= x/2 degrees and so will be at 90+ x/2 degrees. The angle between the minute hand and the hour hand, x minutes after 3, is 6x- 90- x/2= (11/2)x- 90= 90. Solve that for x.
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