Calculate Magnitude & Angle of Clock's Vector Displacement

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The discussion focuses on calculating the magnitude and angle of the displacement vector of a clock's minute hand, which is 12 cm long, over three specific time intervals. For the first interval, from a quarter past the hour to half past, the magnitude is 12 cm, and the angle is 90 degrees counterclockwise from the +x direction. The second interval, from half past to the next quarter past, also has a magnitude of 12 cm but an angle of 180 degrees. The final interval, from a quarter past to the next hour, results in a magnitude of 12 cm and an angle of 270 degrees. The confusion arises from the assumption that the displacement vectors for different intervals are the same.

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The minute hand of a wall clock measures 12 cm from its tip to the axis about which it rotates. The magnitude and angle of the displacement vector of the tip are to be determined for three time intervals. What are the (a) magnitude and (b) angle from a quarter after the hour to half past, the (c) magnitude and (d) angle for the next half hour, and the (e) magnitude and (f) angle for the hour after that? Give all angles as positive values measured counterclockwise from the +x direction (to the right, or 3 o'clock).

I thought that the answer from (a) should be the same for (c) as (a) would be the square root of .12^2+.12^2 . Apparently its not. Any help would be appreciated. Doing vectors on clocks confuses me.
 
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Are you drawing the vectors out?

Part (c) is asking for the next half hour. Part (a) involved a quarter hour. If the minute hand is at the six, and then goes to the 12, what is the displacement? It won't be the same as (a).
 

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