Graphing Distance and Time to find Acceleration

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The experiment involves rolling a cart down an inclined plane to measure acceleration by timing its travel over set distances. Participants must graph distance against time squared to create a linear relationship, where the slope represents half the acceleration. There is confusion about whether to take the natural logarithm of distance or time; the consensus is to apply ln to both sides of the equation to maintain equality. The correct approach is to graph distance versus time squared, simplifying the analysis. This discussion clarifies the method for determining acceleration through proper graphing techniques.
Terrence
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Homework Statement



I have to do an experiment tomorrow, in which we roll a cart down an inclined plane to find the acceleration of the cart. We are given set distances (20, 40, 60, and 80 cm) and will be told to time how long it takes the cart to go said distances. Once we have this information, we must create a graph to find the acceleration of the cart.

The catch is, we have to take the Natural Log (ln) of one, or both, parts of the data to create a straight line in our graph, and use the slope of that line to find the acceleration.

Homework Equations



D = .5*a*t^2

The Attempt at a Solution



I cannot figure out which side of the equation to take the ln of. I believe its the distance side, making ln(d) = .5*a*t^2, but I'm not sure.

Since I have not yet done the experiment I don't have any times to give (sorry :frown:) but I'm pretty sure that isn't required for what I'm asking.
 
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I have no idea why you would take ln of anything with this.

You're graphing d=mt^2...m=0.5a.

So you would want to graph d v. t^2 (d is y-axis, t^2 is x-axis) to give a straight line and take its slope.
 
Terrence said:
I cannot figure out which side of the equation to take the ln of.

Anything that you do to one side of the equation, you must also do to the other side. This is always the case. Otherwise the two sides will no longer be equal.

So, you have to take the natural log of both sides of the equation.
 
Apphysicist - Yeah that makes sense... I thought about it, but for some reason couldn't get ln out of my head. We've used it the most throughout this course. Thanks a lot.

cepheid - Thats what I was thinking too, which is why I was having so much trouble with this lol. I dunno, guess I just wasn't thinking straight. Thanks a lot guys.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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