# How to get x vs t and v vs t equations?

1. Oct 17, 2009

### CChubbs

1. The problem statement, all variables and given/known data

I was out for the lab we did a few weeks ago. What we did was measure the position and time data for a cart rolling down a ramp. Simple for everyone else, difficult for me because I wasn't there, plus I'm in about my first week of physics in school. This was the data my partner collected:

time(sec) ... velocity (m/s)
.15... .015
.40... .65
.65... .119
.85... .158
1.10... .212
1.3... .252
1.5... .285
1.75... .32

2. Relevant equations

x = Ax^2 + Bx + C

v = m(t) + b

3. The attempt at a solution

My partner told me the position versus time equation is x = .1015t^2 - .065t + .0119, but I don't think that's right.

I got v = (.188m/s)t - .002m/s, but I'm not sure if that's correct either.

2. Oct 17, 2009

### dotman

Hello!

Yeah, that stuff you have there is, as you suspected, all wrong. You can tell by looking at the units. You're looking for an equation for velocity, but your equation gives units of meters - meters/s, which doesn't make any sense. Similarly, your partners equation also doesn't make sense (units of sec^2 + sec?).

Now, that said, its unclear what your data is there. It's listed as time and velocity, but I'm guessing its really time and position? Because that's typically what you would measure in a lab like this.

The equations of motion, in one dimension, are these:

x = x0 + v0*t + 1/2*a*t^2
v = v0 + a*t

Of course x0 = 0, by definition, and v0 = 0, since you started from rest.

Hope this helps, what does the lab say to do?

3. Oct 17, 2009

### CChubbs

Never mind, I figured it out! Thanks much for your help!!