Recent content by FUR10N

K, I just scanned in my work: http://imgur.com/nTmiO.jpg The equation at the bottom is the correct final form.

The TI-89 has a ridiculously crazy-awesome Algebra System built in. All I did was use the "solve()" function on the equation you had posted. It took about 5 seconds to calculate and gave me that monstrous equation. As for how to accomplish that by hand: I have absolutely no idea. On another...

B=-90*(asin((cos(a)*d + sin(a)*c)*h/(\sqrt{c^2+d^2}*k))*pi/90-atan(c/d)*pi/90+(sign(d)-2*(2*i+1))*pi)/(pi) where sign(d) = the sign of d and i = any positive integer p.s. there is also a negative version of this equation, just multiply the right side by -1 After solving that I get 690...

Will do. I've been in the library drinking coffee working on this for the past 6 hours though, so I'm going to get some sleep. Once I get a presentable result I'll post it.

I think I understand it all now. So, I came up with some simple numbers just to see: P1 = (0,4) V1 = <1,0> P2 = (3,0) |V2| = 2 c = -3 d = -4 k = 2 h = 1 B = ? A = 0 After solving your final equation I ended up with -3sinB + -4cosB = (-3sin0 + 4cos0)/2 simplified to: -3sinB + -4cosB = 2 I...

I was playing around with equation (2), the one you got by solving the position equation for t; and I saw that t = some vector divided by another vector. So there is not really anything we can do with that right? Also, when you substitute (2) into (3), both of the those equation were derived...

Homework Statement There is one particle traveling at a constant velocity. Its velocity vector and initial position are known. A second particle needs to collide with the first particle. We know the initial position of this particle, and the speed (magnitude of velocity). Find the direction...