dragon2309

Ok, all you physics whizzes out there, im usually quite good at physics, and im in the middle of something, its some crappy motion and acceleration work.

Starting with s=ut + 1/2 at^2

In this experiment im dropping the objuect, so obviously u = 0, s = displacement, which in this case ive got a range of values for. t = time, which again i have a range of values for., this leaves me with calculating a, all seems fine, but as a sidenote, the teacher has put, "a = g", so what is g

ive never heard of a g in this kind of topic. Its not gravity, cos im suposed to be working it out, lol, gravitational field strength is a constant at 9.81, so its not that.

and also, im getting rusty at this, does a correspond to finish speed or end speed, or is it something completely different like acceleration???

Thanks for the help, dragon2309

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Hairy_Lee

its 9.80665 m/s ^2..... 9.81 ms ^2 this is in fact accelleration due to gravity because you're dropping an object...... 9.81 ms ^2 will usually suffice for all calculations that you'll be doing

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dragon2309

ok, another physics question here, well actually it could be maths aswell, either, but here it is.

if something says plot "s" against "t", which one goes on the x axis and which one goes on the y axis.

Lol, i know ive asked questions like this before, it really gets me, i never know which to put where, if it helps t is time and s is displacemnt (or distance).

Thanks guys.

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Hairy_Lee

well the way i was taught physics meant that time always went along the bottom axis

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dragon2309

ok, thanks lee, youve been a bi help. i dont suppose you could lend a hand with this now could you...???



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

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Hairy_Lee

i would imagine that plotting the graph of s against t^2 would produce a straight line graph.... the gradient of this line should equal the value for (if i'm correct, i'm a little rusty at this) a.... now, seeing as this experiment involves objects falling (and all objects accellerate due to gravity at the same rate), you'll be looking at getting a value close to 9.8

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dragon2309

thanks for that info lee, can you help with the equation bit, i need to get a as the subject..... but im not sure what happens to t^2, do i square root it or not, and if i so then when...???

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

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Hairy_Lee

why do you need to make "a" the subject? if you plot the graph for the results you will get "a" from the gradient of the graph because the gradient of the graph is s/t^2 which is the same units for accelleration

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dragon2309

wow, i really had the wrong end of thwe stick there, so i just plot s against the squared value of "t", take the gradient of the straightline to find "a", lol, thats a lot more simple than what i was thinking, dammit

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jcnoernberg

You always put the independant variable on the x-axis. Things like time are independant bacause they are not affected by the other variable. time never changes.

put the dependant variable on the y-axis. this is always dependant on the other variable, like time.