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March 3, 2015

March 3, 2015

Posted by **Kyle** on Tuesday, August 16, 2011 at 7:38pm.

For t_delta = 0.5 s calculate the magnitude of the average acceleration a_av.

I have no idea how to start this problem because I'm trying to use this equation:

a_av = v^2/R

But the answer isn't correct.

I'm always trying to use the equation a_av = (vf - vi)/(t2 - t1)

But I'm not sure how to find vf and vi to subtract the two vectors.

- Physics -
**Damon**, Tuesday, August 16, 2011 at 7:42pmThe centripetal acceleration is constant:

v^2/r = 5.80^2/2.3 = 14.6 m/s^2

- Physics -
**Kyle**, Tuesday, August 16, 2011 at 7:44pmBut is centripetal acceleration the same as average acceleration?

What's the point of the problem giving information about time if it's not even used in the process?

- Physics -
**Damon**, Tuesday, August 16, 2011 at 8:06pmNow if you wanted to approximate that in x y coordinates you could do the following:

w = angular frequency = 2pi/period

period = 2 pi r/v

so w = v/r = 5.8/2.3 = 2.52 radians/second

Vx = -5.8 sin wt = 0 at t = 0

Vy = 5.8 cos w t = 5.8 at t = 0

for going in a circle starting at (5.8 , 0)

after .5 s

Vx = 5.8sin1.26radians = -5.52

Vy = 5.8cos1.26 = 1.77

change in Vx = -5.52

change in Vy = 1.77 - 5.8 = -4.03

for acceleration A divide by .5 sec

Ax = -11.04

Ay = -8.06

magnitude of A = sqrt(121+64) = 13.6 etc

but what a waste of time :)

- Physics -
**Damon**, Tuesday, August 16, 2011 at 8:10pmI suspect the 1/2 second and average was just given to confuse you but you could do it the way I did calculating the change in velocity over he half second and dividing by the time. I suspect that they did not expect you to do all that.

The point is that although the centripetal acceleration is constant, you can approximate it by using

average Acceleration = change in velocity / change in time

- Physics -
**Damon**, Tuesday, August 16, 2011 at 8:13pmBy the way that averaging came out a lot closer to the real answer than I thought it would. After all we went almost 1/4 of the way around the circle during that half second.

- Physics -
**Kyle**, Tuesday, August 16, 2011 at 8:25pmOh okay. I think this makes sense. Thanks for the explanation!

For the first Vx, how come the 5.8 is negative?

- Physics -
**Kyle**, Tuesday, August 16, 2011 at 8:35pmOkay I understand the problem now and I got the answer right doing it your way.

Thanks a lot for the explanation. It really helped!

- Physics -
**Jordan**, Sunday, December 1, 2013 at 3:07pmI had a question actually... why is your Vx related to the sin? and your Vy related to the cos? isn't it usually the other way around?

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