Quinch Posted July 1, 2008 Share Posted July 1, 2008 Or... something like that. Originating from an accidentally misfiled thread, I'm trying to find the t in this equation; [math] s = {v_0}t + \frac{1}{2}at^2 [/math] Normally I'd try to shuffle the values around, adding, subtracting, dividing etc. until I ended up with what I wanted, but I've long since forgotten how to deal with multiple exponents and barely remembering the terminology {let alone in English} didn't help finding an answer either. Would someone be so kind as to give me a pointer or two how to deal with this? Link to comment Share on other sites More sharing options...
ajb Posted July 1, 2008 Share Posted July 1, 2008 Use The Formula. Link to comment Share on other sites More sharing options...
Klaynos Posted July 1, 2008 Share Posted July 1, 2008 Do what ajb said, you'll need to rearrange what you've got though. Link to comment Share on other sites More sharing options...
Air Posted July 20, 2008 Share Posted July 20, 2008 Or... something like that. Originating from an accidentally misfiled thread, I'm trying to find the t in this equation; [math] s = {v_0}t + \frac{1}{2}at^2 [/math] Normally I'd try to shuffle the values around, adding, subtracting, dividing etc. until I ended up with what I wanted, but I've long since forgotten how to deal with multiple exponents and barely remembering the terminology {let alone in English} didn't help finding an answer either. Would someone be so kind as to give me a pointer or two how to deal with this? When you rearrange the formula, you are able to get it into the form [math]ax^2 + bx + c =0[/math]. [math]s = {v_0}t + \frac{1}{2}at^2 \rightarrow \frac12 at^2 + {v_0}t - s = 0[/math]. This implies: [math]a = \left(\frac12 a\right), \ b={v_0}, \ c={-s}[/math]. For these type of formula, known as the quadratic equation (Highest power is 2), you can use the quadratic equation to find the value of [math]x[/math]. The quadratic formula is: [math]x= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/math] Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now