Vector Displacement Components for Minute Hand from 8:00am to 8:20am

[Lamnia]

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 8:00am to 8:20am. Express the vector in x and y components.
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I'm beginning to be a little frustrated with what I thought was a simple vector addition problem.

Here's my approach:

at 8 the y component is 2cm, and the x component is zero

as the minute hand moves in 5 minute intervals around the clock, it covers intervals of 30 degrees. Therefore at 8:20, the y component should be -(2cm*sin30) and the x component should be 2cm*cos30.

Upon addition of the components we would have x = sqrt 3 and y = 2-1 =1

However, I've submitted these answers, and they are incorrect.

Can someone please indicate where I went wrong so that I can recalculate and resubmit this homework problem? Thank you :)

 

[gendou2]

Let's be clear that:

(360 degrees) / (60 minutes) = 6 degrees/minute

Or, if you like:

(30 degrees) / (5 minutes) = 6 degrees/minute

It seems like at 8:20 the correct angle would be:

(6 degrees/minute) * (20 minutes) = 120 degrees

 

[Lamnia]

Thanks, so much... I'd even calculated 120 degrees and then just looked at the 30 degrees below the horizontal to determine the components :) I can finally move on to the next problem!