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Velocity vs displcement graph

by hav0c
Tags: displcement, graph, velocity
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hav0c
#1
Jul12-13, 11:34 AM
P: 59
i was wondering what the equation of the line/curve on a velocity vs displacement graph would be which would indicate constant acceleration.
I am totally stumped.
EDIT:please feel free to shift the thread to any other place if need be, but since it wasn't coursework this seemed to be the natural place.
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dipole
#2
Jul12-13, 12:00 PM
P: 433
Well you know that in general the equation of motion for a particle of constant acceleration is given by:

[itex] x(t) = \frac{1}{2}at^2 + v_0t + x_0 [/itex]

and

[itex] v(t) = at + v_0 [/itex]


So, solving the first equation for t:

[itex] \frac{1}{2}at^2 + v_0t + (x_0 -x) = 0[/itex]
[itex] t = \frac{-v_0 \pm \sqrt{v_0^2 - 2a(x_0-x)}}{a} [/itex]

Now plug that into the equation for v:

[itex] v(x) = \pm \sqrt{v_0^2 - 2a(x_0-x)} [/itex]

I skipped some steps, so I encourage you to work the math out yourself, it's quite simple. Do you understand why the velocity changes slower and slower as the displacement increases? Once you see that, then you'll understand why it had to have that shape even without doing the math.
hav0c
#3
Jul12-13, 12:29 PM
P: 59
Quote Quote by dipole View Post
[itex] v(x) = \pm \sqrt{v_0^2 - 2a(x_0-x)} [/itex]
isn't this just v^2-u^2=2ax?

Anyways, i see that with increase in x, a decreases
therefore the graph looks like this?
(see attachment)
Attached Thumbnails
Untitled.png  

dipole
#4
Jul12-13, 01:30 PM
P: 433
Velocity vs displcement graph

Yes it look like that, but the velocity doesn't decrease with x, the rate of change of the velocity decreases with x. Not that this is not acceleration, because that is the change of velocity with respect to time.


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