By: Rebecca Harvey
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Summary: In this lab, we set up a cart on a slanted ramp resting on a box. We attached a measure of paper to the cart and strung it through a time ticker machine. We then started the time ticker machine and let the cart go, stopping it at the bottom of the ramp. The machine created dots on the measure of paper at the rate of 60 dots per second.We then took the paper and determined that six dots would have been created per every tenth of a second. We measured the distances of the intervals of every six dots and created the data table below.
Data Table:
Graph:
Verbal Model: As the time increases in seconds, the position in centimeters increases increasingly.
Math Model: Position = 39.625(cm/s^2)(time)^2 + 6.759(time) - 0.4664 cm
Explanation of Motion: the cart is moving at a positive consistently increasing rate.
Chart:
Graph:
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I knew that the amount of space in the intervals was the displacement, because we had defined the spaces in between the intervals on the tape as such. It wasn't a big leap to determine that if the area under the lines was the displacement, then the graph was a velocity vs. time graph. This graph tells me that the velocity is increasing constantly over a period of time.good
Tables:
Verbal model: as the time increases, the instantaneous velocity increases proportionally.
Math model: Instantaneous velocity = 78.786(cm/s/s)(time) + 6.3774 cm/s
Slope: the slope of the graph is the acceleration of the object, showing that the velocity in increasing constantly, because it is a straight increasing line. For every 1 second, the velocity increases by about 79 cm/s.
Y-intercept: our y-intercept should equal 0, but it's close enough to zero to be negligible. The y-intercept is equal to the initial velocity.
An acceleration v. time graph in this case would be a straight, horizontal line in the positive section of the graph. This is to show that the acceleration is positive and constant. The slope of the graph would be zero, because the acceleration does not change, and it would be the slope of the velocity v. time graph (in this case, about 79).good
how about how all slopes/area relate?
The two (and a half) new equations are:
Vfinal = at + Vinitial
Derived from the y = mx + b, used to state the equation of the velocity graph.
(Vf - Vi)/t = a OR (delta V)/t = a
Derived from the first equation, (we solved for a) to find the definition of acceleration.
displacement = 1/2(at)^2
This is not complete, but was derived from looking at the similarities between the velocity and position graphs, and how their slopes were related.
we did finish these the next day - all three from handwritten sheet
Acceleration -- the rate of change of velocity, how much the velocity changes over time
Average velocity -- how fast the object is travelling over a certain section of timedisplacement/time
Instantaneous velocity -- how fast the object is traveling at an exact point in time
No one should have the exact same results, because the length of the tapes were not exactly the same. Also, the slope of the ramp we let our carts slide down varied from group to group. Our ticker tape machines, while accurate, depended on where we stopped the cart and how long we let it decelerate. Therefore it was very difficult for everyone to get the same exact results. However, a lot of us got similar results, with equations just a couple digits away from being the same.
This lab was good, in that it was easy to execute, but I feel the way we worked on it was ineffective. Instead of letting us all work on it ourselves and having to explain it a bunch of different times, I think it might have been better to teach us all the basics first and then let us do the lab. I somewhat agree -we should have done the instantaneous velocity part together - that is clear to me in hindsight! You know what they say... :)This lab took too long because everyone lagged due to the fact that they didn't really know what was going on. Otherwise, I enjoyed the ticker tapes.
