PDA

View Full Version : Simple algerba


mohdhm
Aug14-07, 02:48 PM
ok this is in an example in a first-year physics textbook, yet some how i spent an hour trying to solve it (identities, tricks, etc) and could not find a solution.

here is what the textbook states:
(1) v2f*cos\phi = 3.5*10^5 - v1f*cos(37)
(2) v2f*cos\phi = v1f*sin(37)

"now we sqaure these two equations and add them"

= 1.23*10^11 - 7*10^5*v1f*cos(37) + v1f^2*cos(37)^2 + v1f^2*sin(37)^2

(then it simplifies further)

I know that 1.23*10^11 is (3.5*10^5)^2 and that 7*10^5 is 2*3.5*10^5


any help in explaining how this equation is derived would be a great help (ESPECIALLY how 7*10^5 seems to pop out of no where)

neutrino
Aug14-07, 02:57 PM
Did you try squaring and adding the terms?

Extropy
Aug14-07, 03:06 PM
any help in explaining how this equation is derived would be a great help (ESPECIALLY how 7*10^5 seems to pop out of no where)

More importantly, where did that cos(37) come from?

mohdhm
Aug14-07, 03:41 PM
sorry the cos 35 is supposed to be cos 37. my mistake.

Yes i did try adding them and squaring them. your point? why the **** would i take the time to ****ing type it down clearly for your convenience.

neutrino
Aug14-07, 03:45 PM
Yes i did try adding them and squaring them. your point?
You obviously missed something doing that. Remeber that -2ab when expanding (a-b)^2? :grumpy:

mohdhm
Aug14-07, 03:55 PM
lol, you know what... i didn't forget that i should get -2ab... i made an even dumber mistake... i did (a^2 - b^2)

wow... thanks for pointing it out though, holy crap.