Find Displacement/ Velocity/ Acceleration of clock

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The discussion focuses on calculating the displacement, average velocity, and average acceleration of a clock's minute hand from 6 PM to 10 PM. The displacement vector has been calculated as -20cos(30)i + 30j, while the average velocity vector is derived from this displacement over the time interval of 1200 seconds. Participants express confusion about determining the average acceleration, emphasizing the need for initial and final velocities, which are critical for this calculation. The minute hand's motion is characterized as uniform circular motion, with the velocity vector being tangent to the circular path and perpendicular to the radius vector. Clarifications on the unit vectors of velocity at both time marks are sought to further understand the problem.
sakkid95
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A clock has a 20cm minute hand. From the 6PM mark to the 10PM mark, for the tip of the minute hand, calculate:

a) the displacement vector in unit-vector notation
b) the average velocity vector in unit-vector notation
c) the average acceleration vector in unit-vector notation
d) calculate the magnitude and direction of the average acceleration vectorSo I have solved for a is -20cos 30i + 30j
and b is (-20cos 30i+30j)/ 1200 seconds I believe

I really don't understand how you would find the avg acceleration for this problem. I know how to find the magnitude just a^2+b^2= c^2, and I'm not exactly sure how the direction is solved, i think it's tan-1 (Vi/Vf)?

So I just need some help and an explanation on C and how to find the direction. Thanks
 
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To find the average acceleration, you need to have the initial and the final velocities. What are they?
 
I have to calculate it, which is what is confusing me...
 
The tip of the minute hand is in uniform circular motion: what is the direction of its instantaneous velocity at any given hour mark? Its magnitude?
 
I think so, so let's say the minute hand at 6pm is r1 and 10 is r2, so I think v1 would be the vector tangent to r1 would be vi, and vector tangent to r2 is vf. So I believe v1 would be -20x2pi/1200 cm/s= -0.105 cm/s. I'm not sure about v2 though.
 
You have made an attempt to compute the magnitude of velocity at 6 PM. For that, you took the entire length of the minute circle and divided it by 20 minutes. I agree with the entire length, but 20 minutes is not compatible with the entire length. Does the magnitude of instantaneous velocity ever change for the minute hand's tip?

Then, the velocity vector in circular motion is not tangent to the radius vector. It is tangent to the circle of the motion, at the point where motion happens at any given time. It is also evident that the velocity vector is perpendicular to the radius vector.

What are the unit vectors of velocity at 6 PM and 10 PM?
 

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