Learn How to Solve this Math Problem with Basic Knowledge | Need Help Now!

  • Thread starter Thread starter nyu
  • Start date Start date
AI Thread Summary
The discussion revolves around a math problem that the original poster struggled with after missing classes. They provided a specific equation involving variables t, T, v, and c, and sought clarification on how to solve it. A solution was proposed, detailing steps that included multiplying by the radical and squaring both sides to isolate v. The final expression derived for v was confirmed to be v = ± (c√(t-T))/t. The conversation highlights the importance of understanding algebraic manipulation and the use of LaTeX for clear mathematical communication.
nyu
Messages
1
Reaction score
0
This isn't homework, but I missed the last few classes so I'm in dire need of help. I was given a quiz today, amazingly - I was able to do all of it with just basic math knowledge...except one >_> And it's bothering me that I don't understand it >_>

Here is the problem:

[URL][PLAIN]http://www.fall3n.com/homework/solvev.gif[/PLAIN]

[PLAIN]http://www.fall3n.com/homework/solvev.gif[/PLAIN]

If anybody could explain it thoroughly, it'd be appreciated ;)
Thank you!
 
Last edited by a moderator:
Physics news on Phys.org
I assume that's a radical in the denominator.

Multiply both sides by the radical, square both sides then solve for v.
 
Ok um I am assuming the equation is t=T/(1-(V^2/c^2))

therefore

t = T*v^2
-------
v^2-c^2
therefore
v=-c sqroot(t)
-----------
sqroot(t-T)
 
Alright I kind of cheated and used the calculator last time
but I solved by hand and got the following

t=\frac{T}{{1 - v^2/c^2}}

t=\frac{Tc^2}{{c^2-v^2}}
c^2-v^2=\frac{Tc^2}{{t}}
-v^2=\frac{Tc^2}{{t}}-c^2
v^2=\frac{-Tc^2}{{t}}+\frac{c^2t}{{t}}
v^2=\frac{tc^2-Tc^2}{{t}}
v^2=\frac{c^2(t-T)}{{t}}
v=+ or - \frac{c\sqrt{(t-T)}}{{t}}
 
I am very tired though and could have made a mistake please check your answers with someone else or have someone confirm mine
 
Yes, Tom McCurdy's last response:

v=+ or - \frac{c\sqrt{(t-T)}}{{t}}

is correct.
 
YES... not only did i do some algerbra but I learned latex
 
Back
Top