Question

Can someone explain to me why the slope on a distance vs. time^2 graph represents 1/2 the acceleration of that object?

And also how to figure out the spring constant given a force vs. displacement graph?

Thanks

Answer 1

1) Maybe this gives you a clue:
\[\frac{dv}{dt^2}=\frac{dv}{2tdt}=\frac{a}{2t}\]
2) Find the slope,

Answer 2

explain like i'm five? ^^"

Answer 3

y = 1/2 * gt^2
the slope of the line is
y = ax^2

so my thinking is x^2 corresponds to t^2
while a corresponds to 1/2*gt

but i don't know how to express that algebraically

Answer 4

The slope of distance vs time^2:
\[\frac{dy}{dt^2}= \frac{dy}{2tdt}=\frac{v}{2t}=\frac{a}{2} \]
Is this clear enough?

Answer 5

\(\Large\frac{d^2s}{dt^2}=a\) This is acceleration by definition.
\(\Large\frac{ds}{dt}=at+C_1=v\) This is velocity by definition.
\(\Large s=\frac{1}{2}at^2+C_1t+C_2\) This is distance. Now we figure out what the constants are.
C1 here is the initial velocity, while C2 here is the starting distance. Thus we have
\(\Large\frac{ds}{dt}=at+v_0=v\)
\(\Large s=\frac{1}{2}at^2+v_0t+s_0\)

Answer 6

So, where'd the 1/2 come from? well take the derivative of at^2. You'll get 2at, which is twice off of what we want, which is at. In fact, we fix this issue by dividing at^2 by 2. Try the derivative of (1/2)at^2, you should get at.

Answer 7

so I have slope y = 4.8711x

Answer 8

how do you find a

Answer 9

Can you like give me what your problem that you have to solve is? :S

Answer 10

x here t^2?
If yes, then a=1/2*4.8711
:D

Answer 11

it's to compute the experimental value of g (gravity) from the slope value which is y = 4.8711x

Answer 12

You're giving me an equation, but I have no idea what this equation means. Can you clarify?

Answer 13

imron, how'd you know to substitute that way though?

Answer 14

so i have this graph where the slope of the distance vs time^2 is y=4.8711x

Answer 15

you could just draw the graph, but nvm I got it now.

Answer 16

i did

Answer 17

If this graph is what you get?

Answer 18

Kk. you have
\(\huge s(x)=\frac{1}{2}at^2\)
s(x) is the distance, a is acceleration, and t is time.

Answer 19

yezzir

Answer 20

From the derivation before, you can see that the slope is 1/2*a. a=2*4.8711
(I was mistaken back then :) )

Answer 21

Well you have a 1/2 in that equation that comes from integrating that I just explained earlier. a is assumed to be constant (gravity), and there you go it's the equation.

Answer 22

assume we don't know a

Answer 23

or g rather

Answer 24

We don't know a, but it's a constant.

Answer 25

4.8711x = (1/2)at^2

Answer 26

do the x and t^2 cancel out somehow?

Answer 27

since x represents t^2 in the graph?

Answer 28

They are the same thing.

Answer 29

So there you go. Cancel them both out, multiply both sides by 2, and here you have a.

Answer 30

hmm

Answer 31

So, where did the magical formula come from? Well I explained it way back there.
\(\Huge s(x)=\frac{1}{2}at^2+v_0t+s_0\)
But, we know that the initial velocity and the initial distance are both 0, we we effectively get
\(\Huge s(x)=\frac{1}{2}at^2\)

Answer 32

wtf who are you and tracks?

Answer 33

sorry. proceed

Answer 34

I finished already. ;)