Displacement Vector of Minute Hand: 8-8:20 & 8-9:00 a.m.

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SUMMARY

The displacement vector of the tip of a 2.0 cm minute hand from 8:00 to 8:20 a.m. is calculated using the formula r(t) = L (sin(2πT/12), cos(2πT/12)). For T = 1/3 (representing 20 minutes), the components are approximately r_x = 1.0 cm and r_y = 1.732 cm, resulting in a vector of (1.0, 1.732). From 8:00 to 9:00 a.m., T = 1, yielding r_x = 0 cm and r_y = -2.0 cm, resulting in a vector of (0, -2.0). These calculations provide the necessary displacement vectors for both time intervals.

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The minute hand on a watch is 2.0 cm in length. What is the displacement vector of the tip of the minute hand

From 8:00 to 8:20 a.m.? Express vector r in the form r_x, r_y, where the x and y components are separated by a comma.

and

From 8:00 to 9:00 a.m.? Express vector r in the form r_x, r_y, where the x and y components are separated by a comma.

I am no idea how to do these problems. could someone give some suggestions as to how to begin?
 
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[tex]\vec r(t) = \hat i L \sin \left( 2 \pi \frac {T}{12} \right) + \hat j L \cos \left( 2 \pi \frac {T}{12} \right)[/tex]

where L is the length of the minute hand and T is the time measured in hours.
 
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