It is a really easy question, and I was doing a bunch of them yesterday but im having a brain fart.
A speed skater moving across frictionless ice at 8.10m/s hits a 4.70-m-wide patch of rough ice. She slows steadily, then continues on at 5.30m/s
what is her acceleration on the rough ice

This post was edited by Elements_Fury on Sep 9 2013 04:09pm

Use: V²_f = V²_i + 2aΔx

let _f and _i be subscripts for final and initial.

(5.3)² = (8.1)² + 2*a*4.7

solve for a.

these doesn't seem to be right, after i get my solution it says it is still wrong

and what answer did you get, and units? magnitude?

it's right though.

Vi = 8.1
Vf = 5.3
s = 4.7

use formula...
Vf = Vi + 2a (s)

5.3 = 8.1 + 2a (4.7)

a = -0.297872

This post was edited by labatymo on Sep 10 2013 08:32am

which isn't the correct formula

what units is that? (bananas lengths)/(time to tie your shoes)^2?

in m/s^2 it is a ridiculous answer.



this^^^

This post was edited by Azrad on Sep 10 2013 12:06pm

Then what is the right solution? =p im confused

This post was edited by Elements_Fury on Sep 10 2013 12:22pm

I got a = -3.991489... m/s^2 ≈ -3.99 m/s^2

This post was edited by Azrad on Sep 10 2013 01:02pm