How Do You Rearrange v²=u²+2as to Solve for 's'?

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To rearrange the equation v² = u² + 2as for 's', isolate '2as' by moving it to one side, resulting in s = (v² - u²) / 2a. The discussion emphasizes the importance of correctly manipulating the equation and using fractions. Participants also mention the use of LaTeX for better formatting in forum posts. Understanding this rearrangement simplifies further calculations for users. Overall, the thread provides clarity on solving for displacement in kinematic equations.
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I have been trying but i can't quite figure out how to rearrange this so that 's' (displacement) is the subject. Please help me figure this out.

Sorry if this is in the wrong place.
 
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What have you gotten so far?

Start by isolating "2as" to one side.
 
Mostly what i had tried involved me immediately trying move 's' to the other side (so eg. v²/s=u²+2a), which hadnt been working for me. So, by isolating 2as, would this be correct as what i end up with:

v²-u²
_____ = s ?

2a

(Ordinarily 's' would be on the other side, but i couldn't work out how to do fractions on this forum).

So, is that pretty much right?
 
Yep, you got it. :biggrin:
Infamous said:
(Ordinarily 's' would be on the other side, but i couldn't work out how to do fractions on this forum)
You can use Latex for it on the advanced posting screen, the button(\Sigma) is on the upper right:

s = \frac{v^2-u^2}{2a}
 
Last edited:
Awesome, now that I understand this the rest of my work is easy. Thanks!

mplayer said:
You can use Latex for it on the advanced posting screen, the button(\Sigma) is on the upper right:

s = \frac{v^2-u^2}{2a}

Thanks for telling me about this, many forums i visit I've needed something like this, but never known how to use it.
 
Infamous said:
Awesome, now that I understand this the rest of my work is easy. Thanks!



Thanks for telling me about this, many forums i visit I've needed something like this, but never known how to use it.

No problem, just take a little bit to get used to. I'm still trying to figure it out too.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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