uniform acceleration lab

jnimagine

We did a lab, where a cart was rolled down a ramp and a tickertape was attached to the cart as it rolled down. When the cart finished the course of rolling down, a line was drawn across the tape through every sixth dot to show that it is an interval of 0.10s.
Then we had to make 3 graphs: distance - time, average velocity - time, and acceleration - time.
When the displacement was graphed, the graph was a curve. The average velocity was calculated by subtracting the displacement at 0.10s from the displacement of 0.20s then dividing it by 0.10s.
When tangents were drawn on the distance time graph and slopes calculated to find out the instantaneous velocity, the slopes were slightly larger than all the average velocities. It said that the average and the instantaneous velocity must be the same. Why is that?? and what could explain the differences between the two in my result?
Also, what are some reasons why the average velocity - time graph does not pass through zero?
When the average velocity was graphed and a line of best fit was drawn, the slope was caculated to figure out the acceleration. This acceleration was only 6.5m/s^2 instead of 9.8m/s^2. What could be some errors in the lab (not human errors) that could have caused this problem??

If anyone could suggest some explanations to my questions, I would greatly appreciate it. Thank you. :)

qspeechc

What said that the average- and instantaneous- velocities should be the same? Maybe "it" meant they should be roughly the same.


Possible because you are using average velocity, and not instantaneous. Maybe you chose t=0 to be some point on the tape that was actually after you started the cart rolling. It could be innacurate drawing, or measuring.

The reason the acceleration came out as it did could be as a result of friction, wonky cart, maybe the tape was pulling on the cart and slowing it, bad drawing, measuring etc.