What Is the Skater's Acceleration on the Rough Ice?

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Homework Help Overview

The problem involves a skater transitioning from frictionless ice to a rough patch, where she experiences a change in velocity. The subject area pertains to kinematics, specifically acceleration and motion under varying conditions of friction.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the kinematic equation and the variables involved, including initial and final velocities, as well as distances. There are attempts to clarify the algebraic manipulation required to isolate acceleration.

Discussion Status

The discussion is ongoing, with participants expressing uncertainty about algebraic transposition and the correct application of the kinematic equation. Some guidance has been offered regarding the need to follow standard algebraic rules, but no consensus has been reached on the correct approach.

Contextual Notes

There appears to be confusion regarding the algebraic manipulation of the equation, with participants questioning assumptions about variable placement and the implications of their changes. The original poster has indicated difficulty in setting up the formula correctly.

Ashkon
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Homework Statement


A skater moving on a frictionless ice at 8.0 m/s hits a 5.0-m-wide patch of rough ice. She slows steadly, the continues at 6.0 m/s. What is her acceleration on the rough Ice

Homework Equations


v2 = vo 2 + 2a (d-do)

The Attempt at a Solution


I have tried to set up the formula but I'm not sure how to do it properly.

a = vo 2 + 2v2 (d-do)
 
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v and v0 are the final and initial velocities.
d and d0 are the distances from the skater's starting point to the end and beginning of the rough ice
 
Ashkon said:

Homework Statement


A skater moving on a frictionless ice at 8.0 m/s hits a 5.0-m-wide patch of rough ice. She slows steadly, the continues at 6.0 m/s. What is her acceleration on the rough Ice

Homework Equations


v2 = vo 2 + 2a (d-do)

The Attempt at a Solution


I have tried to set up the formula but I'm not sure how to do it properly.

a = vo 2 + 2v2 (d-do)
Your algebra in transposing the Equation in 2.) to the Equation in 3 is a little dodgy.

You can't just switch the positions of a and v2 like you show. :frown:
 
andrewkirk said:
v and v0 are the final and initial velocities.
d and d0 are the distances from the skater's starting point to the end and beginning of the rough ice

I understand that, I just have problems with setting up the formula.
 
Ashkon said:
I understand that, I just have problems with setting up the formula.
Well, we can't do your algebra for you. You'll have to take the Equation in Section 2 and solve for a using standard algebraic rules.
 
Deleted
 
Ashkon said:
I'm not very good at transposing. I think that moving the v2 will make it a fraction an it would be on the bottom? Or will it just stay the same but v2 goes in another place?
Do you or do you not understand algebra?

You've already tried switching a and v2, and we told you that wasn't correct.
 

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