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?

Use The Formula.

Do what ajb said, you'll need to rearrange what you've got though.

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]