S = ut + 1/2 at^2

1. Feb 22, 2006

dragon2309

hi there, i need a little help:

s = ut + 1/2 at^2

0.3 = 0 x 0.02 + 0.5 x a x 0.02^2

0.3= 0.5 x a x 0.02^2

Thats wher i got a bit confused, what do i do to get a the subject, and what happens to the t^2, bearing in ind that the graph i have to plot at the end involves t^2, and not t

Thanks, dragon2309

2. Feb 22, 2006

Hootenanny

Staff Emeritus
What are you plotting $t^2$ against what?

3. Feb 22, 2006

dragon2309

im plotting t^2 against s, which is displacement (or distance)

4. Feb 22, 2006

Hootenanny

Staff Emeritus
And is $t^2$ on the $y$ axis? If that is the case you need to re-arrange until you get something like $t^2 = k.s + c$.

5. Feb 22, 2006

assyrian_77

You are plotting $t^2$ against s? But in your equation above, the only unknown is a. Do you have a set of values for t and s and have to determine the acceleration from them?

6. Feb 22, 2006

dragon2309

Yes i do, i have a set of values for s and t, im getting really confused, i just dont know what im supposed to be doing now, and tryig to think about it just threw up more questions.

7. Feb 22, 2006

Hootenanny

Staff Emeritus
I'll help you through it, ignore the numbers for the moment. Like I said before you trying to get something that looks like $t^2 = k.s + c$. Start with $s = ut + \frac{1}{2} a t^2$ and see how you can manipulate it. It would be easier however to plot $t$ against $s$.

Last edited: Feb 22, 2006
8. Feb 22, 2006

Hootenanny

Staff Emeritus
HINT: You can cancel the $ut$ because $u =0$ $\Rightarrow s = \frac{1}{2} a t^2$. Nevermind, it seems $t^2$ is easier to plot. oops

Last edited: Feb 22, 2006