Using photogates to find acceleration

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The discussion focuses on using photogates to measure the acceleration of a cart on an inclined track. Participants are tasked with timing the cart's travel over specific distances and graphing the results to determine acceleration. A key point of confusion is the appropriate graphing method; instead of plotting distance versus time, participants should plot distance against the square of time to achieve a linear relationship. This approach allows the slope of the graph to represent half of the cart's acceleration. Understanding this relationship clarifies how to derive acceleration from the collected data.
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Perform the following experiment, using the dynamic cart and track, XPLorer
photogates meter stick and blocks. The track is set to an angle of about 5 degree of incline.
Determine the acceleration of the cart from the time it takes to travel the following distances: 20,
40, 60, 80 cm. Use the photogates and XPLorer for timing . Do only one good run per distance.
Place this data in a well-formed and well-labeled table. Use table and grid on the back.
Graph this data in such a way that the data should be a straight line and that the slope of
this line is proportional to the acceleration.
Determine the slope of this line. On the graph identify the run and rise used.
Use this slope to determine the acceleration of the cart. SHOW YOUR WORK!

Homework Equations


The Attempt at a Solution


I am just confused about the graphing part. Since I am measuring the time it takes for a cart to travel a certain distance, and i graph a distance vs. time graph, then the slope of this would be the velocity. I do not see how I can get the acceleration. Would I first have to divide the distance by the time it takes to travel that distance, to get velocity, and then graph velocity vs. time? Any help would be appreciated..thanks.
 
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pradeepk said:
Graph this data in such a way that the data should be a straight line and that the slope of
this line is proportional to the acceleration.

Homework Equations





The Attempt at a Solution


I am just confused about the graphing part. Since I am measuring the time it takes for a cart to travel a certain distance, and i graph a distance vs. time graph, then the slope of this would be the velocity. I do not see how I can get the acceleration. Would I first have to divide the distance by the time it takes to travel that distance, to get velocity, and then graph velocity vs. time? Any help would be appreciated..thanks.

You should not plot distance vs. time. You should devise a plot as indicated by the question. I assume your cart starts from rest. What equation do you know that relates distance, time and acceleration in this case?
 
kuruman said:
You should not plot distance vs. time. You should devise a plot as indicated by the question. I assume your cart starts from rest. What equation do you know that relates distance, time and acceleration in this case?

Well the equation that relates distance, time and acceleration is x=xo + vo(t) + 0.5at2. And since the cart starts at rest, it would be D=0.5at2. I am not sure how I can make a graph from this equation..it says the graph is supposed to be linear, but this looks quadratic.
 
pradeepk said:
Well the equation that relates distance, time and acceleration is x=xo + vo(t) + 0.5at2. And since the cart starts at rest, it would be D=0.5at2. I am not sure how I can make a graph from this equation..it says the graph is supposed to be linear, but this looks quadratic.
Yes it looks quadratic if your y-axis is D and your x-axis is t. Suppose you did not put t on the x-axis, but you put t2 instead. What would the plot look like and what does the slope give you?
 
kuruman said:
Yes it looks quadratic if your y-axis is D and your x-axis is t. Suppose you did not put t on the x-axis, but you put t2 instead. What would the plot look like and what does the slope give you?

ahh i see now, you would have a linear plot and the slope would give you half the acceleration. I never knew thought about relating these kinematic equations to the problem..thank you so much!
 
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