(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

"The relation between the distanse S of the trolley as it moves on air track and the time t is given by S = Vt+(1/2)at^2, where V is the initial velocity of the trolley. The following set of data were taken for the distance S of a trolley on an air track apparatus, whose angle of inclination is 5 degrees, as a function of time t."

Given data:

S: 0 - 0.15 -0.388 - 0.600 - 0.930 - 1.37 - 2.45

t: 0 - 0.975 - 1.48 - 2.04 - 2.46 - 3.03 - 3.95

The questions:

a-) Calculate t^2 for each case.

b-) Plot a graph of S (y-axis) against t^2 (x-axis)

c-) From the graph determine the magnitude of the acceleration of the trolley.

d-) From the graph, determine the initial velocity of the trolley.

From the above results, determine the experimental value of the acceleration due to gravity.

2. Relevant equations

s = vi*t + 0.5at^2, where s is the distance, vi is the initial velocity, t is the time, and a is the acceleration.

3. The attempt at a solution

Parts a and b are easy, as you just square t and plot the graph. As for part c, as I recall the slope of a distance against a time squared graph is equal to half the acceleration, and as such the acceleration is equal to 2 times the slope.

As for d, the initial velocity can be determined by the formula given. According to my prof, any value of vi is OK (given the given data is used, obviously).

What I can't get right is the last part, determining the experimental value of acceleration due to gravity. I tried a couple of times, but I keep getting a = 0.30 from the graph, which, when divided by sin of the angle (5 degrees), gives the value 3.44 (a = g*sin (theta)).

What am I doing wrong? Is the acceleration determined in a different way? Does the experimental value really have that big of an error percentage?

Thanks.

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# Determining the value of acceleration due to gravity

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