Displacement/ Velocity/ Acceleration Clock Question

[popmop2]

A clock has a 20cm minute hand. From the 6PM mark to the 9PM 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 vector

 

[lewando]

Hello popmop2, welcome to PF! This is a great site and people will be happy to help you-- but per the rules, you have to show some effort. These questions are based on the definitions of the terms so give them a try.

 

[popmop2]

Sorry for not posting my work earlier.

So far I have...

a) On any given clock, from 6PM to 9PM, the minute hand is 20cm. So applying a^2+b^2=c^2 with a= 20cm and b=20cm...I get the displacement value of 28.2842cm

b) average velocity = Δd/Δt = 28.2842/(3-0) =9.4280 cm/h

c) average acceleration = vf-vi/ tf-ti = (0.1571cm/s) / (3hrs) = 0.0008cm/s^2

d) magnitude = 28.2842cm
direction is tan^-1 (-20/-20) = 45°

 

[lewando]

Questions a), b) and c) are looking for the answer to be a vector (magnitude and direction) in the unit vector notation. Are you familiar with that?

For a) you have found the magnitude.

For b) the average velocity vector is defined as the displacement vector divided by time. I think the question is implying time to be 15 minutes.

For c) the average acceleration vector formula is similar to what you have except that v_f and v_i are instantaneous velocity vectors. This will require you to find the speed of the arrow tip of the minute hand.

For d) you first need to find c).

 

[popmop2]

a) since i have found the magnitude, i will need to find the displacement
the displacement of the clock moving the minute hand (20cm) from 6pm to 9pm is equal to the parimeter of the circle from 6pm to 9pm
P=2∏r = 2∏(20) = 40∏= 125.60
then i will have to divide by 4 to get the perimeter of a 6pm to 9pm = 31.4cm

so for a) the answer would be 31.4 cm



b) the average velocity vector would be 31.4/(15-0) = 2.0933cm/min = 0.0348cm/second = 0.0003488 m/s

c) i am not sure how to find the speed of the arrow tip of the minute hand at time t=6pm and t=9pm

d) i cannot solve d since i am not sure on how to solve part c)

thanks a lot for your help, i really appreciate it

 

[haruspex]

a) since i have found the magnitude, i will need to find the displacement

No, you already found the correct magnitude (of the displacement). But the questions asks for a vector, so you also need to specify its direction. And it asks for this in 'unit vector notation'. Presumably you know what that means?
There's no point in looking at b c and d until that's clear.