Rearranging the kinematic equations

  • Thread starter BogMonkey
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  • #1
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The equations for linear motion I've memorized are:
1.) v = u + at
2.) v^2 = u^2 + 2as
3.) s = ut + 1/2(at^2)
v = final velocity, u = initial velocity, a = acceleration, t = time, s = displacement

In an example my physics teacher uses the equation x = x0 + (v^2-u^2)/2a

I'm assuming he derived that from the second equation I listed but I can't figure out why (v^2-u^2) is divided by 2a. When I rearrange that equation I get s = v^2 - y^2 - 2a
Am I just using incorrect algebra or is that a different equation altogether?
 

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  • #2
Doc Al
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I'm assuming he derived that from the second equation I listed but I can't figure out why (v^2-u^2) is divided by 2a. When I rearrange that equation I get s = v^2 - y^2 - 2a
Show the steps in your derivation.
Am I just using incorrect algebra or is that a different equation altogether?
You are making an algebra error. Also, realize that what is called 's' in one equation is called 'x - x0' in the other.
 
  • #3
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The equations for linear motion I've memorized are:
1.) v = u + at
2.) v^2 = u^2 + 2as
3.) s = ut + 1/2(at^2)
v = final velocity, u = initial velocity, a = acceleration, t = time, s = displacement

In an example my physics teacher uses the equation x = x0 + (v^2-u^2)/2a

I'm assuming he derived that from the second equation I listed but I can't figure out why (v^2-u^2) is divided by 2a. When I rearrange that equation I get s = v^2 - y^2 - 2a
Am I just using incorrect algebra or is that a different equation altogether?

He is. But what you have as s your teacher has apparently called the change in position or delta x, which is really just the final position on a line we might call the x- axis, subtracted by the initial position. He has written final x position as x, and x0 to symbolize initial x position. Knowing this does the algebra now make sense?

oops you apparently realized this now that I look back. you are mixing dividing and multiplying with subtracting and adding in your algebra. This is gonna create big problems.
 
  • #4
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Alright heres the steps
v^2 = u^2 + 2as
1.) Moved the v^2 over and got 0 = u^2 - u^2 + 2as
2.) Moved the s and got -s = u^2 - v^2 + 2a
3.) Changed all the signs and got s = v^2 - u^2 - 2a

I never fully learned algebra which seems to be the source of 90% of the problems I have with physics. Like you said pgardn I mix up the addition/subtraction with the multiplication/division when moving things from one side of an equation to the other. Can you recommend a method of knowing exactly how to rearrange equations when you have combinations of addition/subtraction/multiplication/division?
 
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  • #5
Doc Al
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Alright heres the steps
v^2 = u^2 + 2as
1.) Moved the v^2 over and got 0 = u^2 - u^2 + 2as
OK. (But better to just move the u^2 to the left.)
2.) Moved the s and got -s = u^2 - v^2 + 2a
No good. You must do the same operation to both sides. On the left, you subtracted 's'; on the right you picked one term and divided it by 's'.

Move the entire '2as' term to the left side.
 
  • #6
656
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Alright heres the steps
v^2 = u^2 + 2as
1.) Moved the v^2 over and got 0 = u^2 - u^2 + 2as
2.) Moved the s and got -s = u^2 - v^2 + 2a
3.) Changed all the signs and got s = v^2 - u^2 - 2a

I never fully learned algebra which seems to be the source of 90% of the problems I have with physics. Like you said pgardn I mix up the addition/subtraction with the multiplication/division when moving things from one side of an equation to the other. Can you recommend a method of knowing exactly how to rearrange equations when you have combinations of addition/subtraction/multiplication/division?

When you write moved you really have to know whether you are adding/subtracting v. multiplying/dividing in order to isolate some variable. You cannot move the s over like you did(subtracting/multiplying). It has to move with the 2a as well...the whole 2as has to be added to each side...

sorry again, already posted by Dr. Al.

I dont mean to be a downer, but you are really going to have to work on your algebra or this is going to be very tough (as you noted) if your physics class is anything but purely conceptual.
 
  • #7
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Thanks a lot. So it appears its better to look at it like doing the same operation on both sides than moving something from one side of the equation to the other.

Yeah its gotten very tough already. Its basically just this rearranging equations part I have trouble with but that seems to be one of the most essential maths skills to have when it comes to physics so I suppose I better master it.
 
  • #8
656
2
Thanks a lot. So it appears its better to look at it like doing the same operation on both sides than moving something from one side of the equation to the other.

Yeah its gotten very tough already. Its basically just this rearranging equations part I have trouble with but that seems to be one of the most essential maths skills to have when it comes to physics so I suppose I better master it.

At least you understand what you do not understand.

Yes you are manipulating both sides in exactly the same way to isolate one of the variables. Some people have done this so thoroughly that they can perform this very quickly. For those who have not had the practice, practice is in order.

A small sample of what you did...

A = B - CD Read A is equal to B, when B has CxD subtracted from it. So you wanted to get the D as the variable by itself. First you "move" the B to the other side by subtracting B from both sides. Now you have A - B = -(CD) . Second you get the C "moved" by dividing both sides by C. Now you will have A - B all divided by C is equal to - D. Multiply each side by -1... This of course is a long process, but necessary if you have not had a lot of practice.

It is kind of confusing for people just coming back to algebra again with the notation and the rules. Sorry if I oversimplified, but this is basically the same thing you had with your kinematic equation.
 

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