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Physics question... halp??

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tigon
Posts: 131 - Karma: 10
2V1   +  3V2    = -5
2V1²  +  3V2²  = 35
solve for V1 and V2...

I can do it in ms paint if you can't read the equation...
thanks





Sep 5 2008 10:28PM
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morocco_mole
Posts: 635 - Karma: 17
(-5 - 3v2)/2=



Sep 5 2008 10:37PM
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MCIRONCOCK
Posts: 3094 - Karma: 30
I can't believe i passed physics on my first try
Sep 5 2008 10:39PM
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morocco_mole
Posts: 635 - Karma: 17
the square root of ((35 - 3v2squared)/2)

solve for v2
Sep 5 2008 10:40PM
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Aaaron.
Posts: 11817 - Karma: 903
you are now NS's resident nerd
Click for larger image
Sep 5 2008 10:41PM
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Aaaron.
Posts: 11817 - Karma: 903
oh
and i maxed out your karma =)
Sep 5 2008 10:41PM
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shabinka
Posts: 743 - Karma: 10
Let me try this out real quick.
Sep 5 2008 10:43PM
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tigon
Posts: 131 - Karma: 10
did anyone get an answer because I did it that way and I still couldn't ge t the right answer
Sep 5 2008 10:55PM
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shabinka
Posts: 743 - Karma: 10
Basically, I solved for V1 in terms of V2. Which V1= (-5-3V2)/2
Plug that into the second equation.
{2[(-5-35V2)^2/2]} + 3V2^2=35
Set the equation equal to zero, so your final equation is, 7.5V2^2+15V2-22.5=0
I guess you do the quadratic formula from there (-b+radical(b^2-4AC)/2a) but for some reason, that's not working but looking at the graph the 0's are x=1 and x=-3. I'm assuming you want 1. So plug 1 into the other equation, V1=-1.
Sep 5 2008 11:03PM
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tigon
Posts: 131 - Karma: 10
hm... my key says

V1=2 V2=-3    OR    V1=-4 V2=1

not sure if its a typo or what
Sep 5 2008 11:06PM
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shabinka
Posts: 743 - Karma: 10
OH! You are subtracting a negative when you are solving for V1 so, 2V1=-8, V1=-4. I can explain it better if you want to send me a message on AIM.
Sep 5 2008 11:08PM
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shabinka
Posts: 743 - Karma: 10
Basically, I solved for V1 in terms of V2. Which V1= (-5-3V2)/2
Plug that into the second equation.
{2[(-5-35V2)^2/2]} + 3V2^2=35
Set the equation equal to zero, so your final equation is, 7.5V2^2+15V2-22.5=0
(I can't believe I didn't notice this earlier)
Take out a 7.5. 7.5(V2^2+2V2-3)=0.
So, the V2^2+2V2-3 factors to (V2+3)(V2-1). Solve for V2, which gives you -3 and 1. Ok, so that was simple.
Then just plug one of those numbers into the first eqation for V2 (2V1+3V2=-5) to solve for V1.
The V2^2 is V2 squared incase you haven't noticed, and the weird brackets are just how I grouped things so I could understand them.
Sep 5 2008 11:12PM
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tigon
Posts: 131 - Karma: 10
alright I got it... thanks mate
Sep 5 2008 11:20PM
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shabinka
Posts: 743 - Karma: 10
Yeah man, just drop me a message if you need any more physics help.
Sep 5 2008 11:21PM
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morocco_mole
Posts: 635 - Karma: 17
that sounds right,   one equation is a strait line and the other is a curved line  so they can intersect twice
like this


  \                                                 /
   \                                              /
----x-------------------------------x---------
          \                                   /
                \                      /
                       _____


i think
           

Sep 5 2008 11:23PM
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tBatt
Posts: 9632 - Karma: 580
i noticed there was a lack of an icon in this thread, so i decided to ruin the trend
too bad this has nothing to do with physics.
Sep 6 2008 11:44AM
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morocco_mole
Posts: 635 - Karma: 17
thats interesting
Sep 6 2008 10:32PM
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B-VUR
Posts: 1237 - Karma: 167
I noticed that as well
Sep 6 2008 11:10PM
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