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nyu
Sep27-04, 11:35 PM
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:

http://www.fall3n.com/homework/solvev.gif

http://www.fall3n.com/homework/solvev.gif

If anybody could explain it thoroughly, it'd be appreciated ;)
Thank you!

Tide
Sep28-04, 12:23 AM
I assume that's a radical in the denominator.

Multiply both sides by the radical, square both sides then solve for v.

Tom McCurdy
Sep28-04, 12:27 AM
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)

Tom McCurdy
Sep28-04, 12:47 AM
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}}

Tom McCurdy
Sep28-04, 12:48 AM
I am very tired though and could have made a mistake please check your answers with someone else or have someone confirm mine

HallsofIvy
Sep28-04, 08:41 AM
Yes, Tom McCurdy's last response:

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

is correct.

Tom McCurdy
Sep28-04, 03:39 PM
YES... not only did i do some algerbra but I learned latex