Uniform acceleration kinematics lab

[Trevormbarker]

Homework Statement

This is more of a general question but it is part of my physics course so i decided to post here, I am sure many of you have done this lab but what it is is a cart with tickertape attached to it going down a ramp, once we plot a d-t graph there are lots of application questions. One of them says plot a d-t^2 graph and a v-t graph, it says to calculate the slope of each graph and compare them.

Homework Equations

None really just slope, (y2-y1)/(x2-x1)

The Attempt at a Solution

By looking at the units i thought that the slope of each graph should both represent acceleration as cm/s^2 and (cm/s)/s are the same thing, since they came from the same data set they should have the same slope but I get two different slopes, they are exactly off by a factor or 2. Am I making an arithmatic error or is it supposed to be like this, ( I get a slope of 61 for the d-t^2 graph and 120 for the v-t

 

[PhanthomJay]

What is the equation that relates velocity with acceleration and time? What is the equation that relates distance with acceleration and time?

 

[Trevormbarker]

What is the equation that relates velocity with acceleration and time? What is the equation that relates distance with acceleration and time?

v2^2=v1^2 + 2aD?
D= v1T + 1/2aT^2?

im sorry but I am not sure what exactly you are getting at? why are my accelerations different if the units worked out correctly and they were both taken from the same data set

 

[PhanthomJay]

Your 2nd equation relates distance with acceleration and time,

D = 1/2at^2

you didn't express the equation which relates velocity with acceleration and time,

v = at

Now when you plot v vs t, the slope of the line is a , correct?

Now plot D vs. t^2. The slope of the line is ?

 

[Trevormbarker]

D vs t^2 should also be acceleration no! m=rise/run = D/t^2= cm/s^2
Or are you implying I use your D=1/2at^2 and t^2 for it and do (1/2at^2)/(t^2) which leaves me with 1/2 a. Which makes sense slightly but what is this the acceleration for? And howcome it doesn't work when i just use the units on the axis, only when i use an equivelant relationship to D

 

[PhanthomJay]

When you plot D vs. t^2, you should get more or less a straight line sloping up and to the right. Since distance = 1/2 at^2 (d = mt^2, where m is the slope)then the slope of the line is ?. Be sure to plot distance on the y axis, and time squared (not time) on the x axis.

 

[Trevormbarker]

1/2a. Thanks for all the help I just don't understand what this acceleration represents. Is it just the graph is set up like the kinematic equation so "solving for the slope" is really just isolating the 1/2a, being 1/2 of the acceleration of the cart. Is doing D over t^2 times 2 just the graphical equivelent of the kinematic equation. I think that's all been cleared up thanks alot. What is the the actual graph, ie what does it really mean, when i read it what information can i extract without differntiating or integrating it.