CAPM Challenge 1

CAPM Challenge 1

a). 

The sketch above shows the setup for this experiment. We worked on a slanted table propped up with 2 thick textbooks. We aligned 2 meter sticks along the length of the table and used a metal cart with wheels as the object in which we would collect the data from. A piece of chalk was used to mark the position of the cart every half second which was timed out using a metronome. We performed a few test runs to make sure we were comfortable with where and when to put the marks as the cart rolled down the table. After adjusting, we ran the cart down the table 5 times so we would have 5 measurements originally in centimeters eventually converted to meters, for each half second that we would average together to have the most accurate results. 

b). 

Above is an image of our raw data table. The X column represents the time every half second starting at 1 second (we originally began at 1/2 second but soon realized that we skipped a half second before marking the position), and the Y column is the position which shows 5 measurements, and the final column represents the average of the 5 measurements. 

c).  

This graph was formed using the raw data. As can be seen, the line is not linear, therefore it does not allow us to make a prediction, and so the graph below represents the data after it was linearized, which was achieved by squaring the X values (time).

The x values are now squared in this graph making it linear, so we were able to find the equation of the line which will be explained more in depth in the next section. We used this equation to make our prediction for acceleration which will also be explained next. 

d). The equation we derived from the line is y=0.0785x+ 0.1135 or position=0.0785(time)^2+ 0.1135. We learned in a previous study that the slope of a velocity vs. time graph is the acceleration, but on a position vs. time^2, the slope is equal to half the slope (which is the acceleration) of the velocity time graph. In order to find the acceleration on a position vs. time^2 graph, we used the formula slope=1/2a. With the values plugged in, we had: .0785=1/2a. To isolate a we multiplied both sides by the reciprocal of 1/2 which was just 2, and we ended up with .157m/s^2.

e). RECAP TO FIND ACCELERATION AT 4 SECONDS: slope=1/2a

.0785=1/2a

(1/2a)(2)=(.0785)(2)

acceleration= .157 m/s^2

f). The velocity of the object at 4 seconds is equal to the acceleration just solved for in the previous step, therefore, the velocity is .157m/s. This is because we knew that the slope of a velocity time graph is equal to the acceleration of the object which was found in the previous step by multiplying the slope by 2. 

g). The formula for percent error is predicted value- actual value/actual value x 100. The actual value was .192m/s.  Strangely enough, in our physics class, all of the groups were a considerable amount off, and no one was within 10%, much unlike the other class. With our values plugged into the formula it would look like this: .192-.157/.192. x 100%= 18% error. Unfortunately, we were not within 10%, but we did complete the lab using all the proper steps and requirements.