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

  • Thread starter Thread starter Infamous
  • Start date Start date
AI Thread Summary
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.
Infamous
Messages
3
Reaction score
0
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.
 
Physics news on Phys.org
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top