Rearranging Kinematic Equations. Help.

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Homework Help Overview

The original poster seeks assistance with rearranging kinematic equations, specifically focusing on the equation Δd = ½a(Δt)^2 to isolate the variable for acceleration. The context involves introductory physics concepts, particularly kinematics.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the general approach to isolating variables in equations, emphasizing operations that can be performed on both sides. There are questions regarding the interpretation of the squared term in the equation, specifically whether it applies only to Δt or to a larger expression.

Discussion Status

Some participants have provided guidance on how to manipulate the equation to isolate acceleration, while others have engaged in clarifying the notation used in the equation. Multiple interpretations of the notation are being explored, but there is no explicit consensus on the overall approach.

Contextual Notes

The original poster mentions missing school, indicating potential gaps in understanding foundational concepts. There is a request for explanations in plain language, suggesting a need for accessible communication in the discussion.

KayAr
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Homework Statement


Homework Equations


The Attempt at a Solution



The template doesn't really apply to what I need help with. I've missed some school and I need help rearranging Kinematic equations. I am new to physics so please use plain english.

Here's the specific one I am having problems with, but I need a general rule in order to rearrange them:

Δd = ½a(Δt)^2 (trying to get equation for acceleration.)
 
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Glad to see you on the physics forums. Welcome!

The general idea is to do the same thing to both sides: add or subtract something, multiply or divide by something. Try to gradually get the variable you are looking for all by itself. Sometimes it is very helpful to ask what has been done to that variable when the formula was created. In this case, your acceleration has been multiplied by 1/2 and by (Δt)^2. To undo these, you divide both sides by 1/2 (or multiply by 2) and divide both sides by (Δt)^2.

Δd = ½a(Δt)^2
2Δd = a(Δt)^2 after multiplying by 2
2Δd/(Δt)^2 = a after dividing by (Δt)^2
 
Delphi51 said:
Glad to see you on the physics forums. Welcome!

The general idea is to do the same thing to both sides: add or subtract something, multiply or divide by something. Try to gradually get the variable you are looking for all by itself. Sometimes it is very helpful to ask what has been done to that variable when the formula was created. In this case, your acceleration has been multiplied by 1/2 and by (Δt)^2. To undo these, you divide both sides by 1/2 (or multiply by 2) and divide both sides by (Δt)^2.

Δd = ½a(Δt)^2
2Δd = a(Δt)^2 after multiplying by 2
2Δd/(Δt)^2 = a after dividing by (Δt)^2

Wow, after your explanation it all came back to me.

Thank you :smile:
 
Most welcome!
 
is Δt the only thing squared or is it the entire term that it is a part of? if it is only Δt then why is Δt in parentheses?
 
Only delta t is squared. The brackets make it clear that it is delta t that is squared, not just t.
When you write (x*y*z)^2 it means all of x*y*z is squared as in
x*y*z*x*y*z.
 
Delphi51 said:
Only delta t is squared. The brackets make it clear that it is delta t that is squared, not just t.
When you write (x*y*z)^2 it means all of x*y*z is squared as in
x*y*z*x*y*z.

Thanks again
 

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