Solve for a in "S = ut + 1/2 at^2" - Dragon2309

[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

 

[Hootenanny]

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

 

[dragon2309]

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

 

[Hootenanny]

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

 

[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?

 

[dragon2309]

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?

Yes i do, i have a set of values for s and t, I am getting really confused, i just don't know what I am supposed to be doing now, and tryig to think about it just threw up more questions.

 

[Hootenanny]

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 $s$.

 

[Hootenanny]

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