# Clock, calculate displacement, avg. velocity, avg. acceleration.

1. May 5, 2014

### timnswede

1. The problem statement, all variables and given/known data
A clock has a 20 cm minute hand. From the 6 PM mark to the 9 PM 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.

2. The attempt at a solution
a. I named my 6 PM vector B and my 9 PM vector A. A=-20i and B=-20j. A-B=C(the displacement vector) which equals -20i+20j

b. V=(-20i+20j)/15min= -4/3i+4/3j cm/min

c. Vi=Vf=40∏/60=2∏/3
Vi=-2∏/3i
Vf= 2∏/3j
Aave= (2∏/3i+2∏/3j)/15min = 2∏/45i+2∏/45j cm/min^2

d. √((2∏/45)^2)+(2∏/45)^2) = .197 cm
tanΘ=2∏/45/2∏/45 = 45°

I am not sure if I did b, c, or d right.

Last edited: May 5, 2014
2. May 5, 2014

### paisiello2

c. Vi and Vf are wrong directions.

d. Wrong units. And technically tanθ ≠45°.

Last edited: May 5, 2014
3. May 5, 2014

### timnswede

Oh shoot, I didn't mean to put pi in for part b, I'll edit that. But for part c I used it to find the velocity of the minute hand, the circumference divided by the time it takes the minute hand to do one rotation.

c. Why are Vi and Vf wrong directions? Wouldn't the velocity at 6 PM be to the left so it would be negative and at 9 PM it would be straight up so it would be positive?

d. And I am not sure what the units are supposed to be then, but the angle is wrong because of what I did wrong in part b I am assuming.

Last edited: May 5, 2014
4. May 5, 2014

### paisiello2

c. But you defined 6PM to be your A vector = -20i. Your velocity can't be in the same direction.

d. What are the units for acceleration? I think you got the right answer for θ, you just stated it wrong as tanθ.

5. May 5, 2014

### timnswede

Ah, it was supposed to be the other way, my A vector is 9 PM, and my 6 PM is my B vector. Sorry, i need to work on organization apparently.
Oh, I was thinking of magnitude as a unit of length for some reason, but it would be cm/min^2 then.
Is my average acceleration vector right then?

6. May 5, 2014

### paisiello2

I didn't check your numbers in detail but your answer says the average acceleration is 45 degrees towards the 2 o'clock/3 o'clock position. Intuitively that seems right to me.