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(Rofl)
No, you don't have to do anything with vectors or anything "advanced". Just look at the angles made with the vertical.
Originally Posted by jackiemoon
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.