Linearizing Distance vs. Time^2 Graphs

Recently a great deal has been written regarding the Modeling Workshops use of linearizing data from a distance vs time^2 graph. Of particular concern has been the error which results from some of the linearizations. We have encountered errors in lineariztion while piloting a combined precalculus/physics course at Lincoln High School, Lincoln, NE. Let me attempt to detail what we have learned from a mathematical perspective using a TI-8? calculator and data gathered using a CBL with a sonic range finder. If you have questions or would like to have the dropball program mentioned below, please e-mail jwelker@lps.org

Setup

The following lab uses a dropball.8?p program to record the height of a ball(m) as a function of time (sec). The dropball program is written so that the distance measured is the height of the ball above the floor. The height is plotted on the y-axis while the time is on the x-axis. Image 1 shows the results of the initial collection of data from the CBL.

Note the flat portion at the start of the graph. During this phase, the ball was being held below the motion detector but had not been dropped. Image 2 shows the use of the trace key to determine the first point which will be used in the actual fitting of a curve to the data. Image 3 shows the grapher displaying the final point prior to the ball hitting the floor beneath the motion detector.

Texas Instruments provides a program called CHOOSE which will select any portion of a curve and store that portion in the calculator lists L3 and L4. After running the CHOOSE program, the curve is shown in Image 4. The data from the graph is displayed in Table 1.

A Mathematical Model

The data shown in Image 4 appears to be in the form y = k*x^2 or in this case d = k*t^2. By calculating the regression equation of the quadratic function y = ax^2 + bx + c, the equation in Image 5 is given. Recall that the physics formula for vertical displacement is d = (1/2)*a*t^2 + v_i*t + d_i where "a" is the acceleration, "v_i" is the initial velocity and "d_i" is the initial height of the object. The regression equation is d = -4.77t^2 + 2.15t + 1.39. Interpreting the data, we find 1/2 of the acceleration is -4.77 m/s^2. To find the acceleration, multiply this value by 2 and the acceleration of the ball is -9.54 m/s^2 (theoretical value -9.8 m/s^2, % error = 2.6%). The initial velocity is 2.15 m/s with an inital height of 1.39 m. Since the initial velocity is not zero, the first point in the data set is after the ball was dropped.

The exact point at which the ball was dropped will be the vertex of the parabola which can be found mathematically from a quadratic equation by a procedure known as completing the square (CTS). CTS converts the equation to parabolic form d = a(t-h)^2 + k. The vertex is given as (h, k) where h is the time at which the maximum height occurs. The value of k is the maximum height. In our experiment, the maximum height is the height from which the ball is dropped and h is the time from starting the CBL until the ball was dropped. Using the values from the quadratic regression equation, the vertex of the parabola is (0.23 s, 1.63 m) as illustrated in Image 6. Using this particular regression method indicates the ball was dropped 0.23 seconds after the timer was triggered. Image 7 shows the graph of the regression equation plotted on the same graph with the experimental data.

Using a Linearization Method

The modeling workshop method of linearizing the data would be to graph the distance vs the square of the time. If the data fits the d vs. t^2 model, the data should be a straight line. The data is displayed in Table 2.

Graphing the linearized data from Table 2 yields the linear graph in Image 8. When performing a linear regression on the data in the graph of d vs. t^2, the result is y = -2.82 x + 1.94 (Image 9). The graph of the linear equation with the linearized data is shown in Image 10.

To utilize the variables given, the equation would best be put as d = -2.82 t^2 + 1.95. The value of -2.82 m/s^2 is far from the expected value of -4.9 m/s^2 (42.4% error). Graphing this equation with the data (Image 11), one will note that the curve does not fit the data as well as the quadratic regression (Image 7). One possible explanation for this error is the assumption that the parabola resulting from linearization begins when t=0. The linearized form is also in the parabolic form d = -2.82(t-0)^2 + 1.95. The vertex is at (0 s, 1.95 m) and assumes that the ball was released at time zero from a height of 1.95 m. However, Image 1 clearly shows that the ball was not released when t=0. We chose the first data point well after the ball had already started to drop. Hence, the error in the linearization method. To test this theory, we will shift the data to make the time the ball was dropped equal zero. Using the values from the vertex in the quadratic regression, make the appropriate transformations, and check to see if the linearization method is closer to the expected value.

Transforming the data

To improve the accuracy of the linearization method, one must shift the data so that the vertex of the parabola occurs when t=0. This can be accomplished by mathematically subtracting the time prior to release of the ball. The best value for this time is the vertex of the parabola determined from the quadratic regression equation. Therefore, subtract 0.23 seconds from the time (see Table 3). The end result will be to shift the graph 0.23 seconds to the left and place the vertex of the parabola at time t = 0 (y-axis of the graph).

Linearizing a second time

Now, we must linearize the data again. We want the graph of d vs. (t - 0.23)^2, so me must square the time and find a new graph.

The graph of the linearized data in Table 4 is shown in Image 12. The regression equation for the line is y = -4.77x + 1.63 or d = -4.77 t^2 + 1.63 (Image 13). The -4.77 m/s^2 is much closer (2.7% error) to the expected value of -4.9 m/s^2 with the data shifted so that the vertex is at t=0 seconds. The graph of the linear regression equation with the linearized data is shown in Image 14.

Image 15 displays the graphs of the two linearization methods. The first method without the transformation of time is the top line with the flatter slope (m=-2.82 m/s^2) while the second method with the transformation for time is shown by the lower line with the steeper slope (m=-4.77 m/s^2).

Finally, to check this equation against the original data in Table 1, we will graph the parabolic equation found by linearizing and shifting the data. Recall that d = a(t-h)^2+k is the standard form of a parabola and that the time at which the vertex occurred was 0.23 sec, the value for h is 0.23. When substituting 0.23 into the equation for h, we get d= -4.77(t - 0.23)^2 + 1.63. Image 16 shows this equation with the original data from Table 1. Note that the linearized equation which was shifted horizontally fits the data much better than the previously linearized equation shown in Image 11. To further demonstrate the difference between the two linearized equations, Image 17 shows the equations d = -2.82 t^2 + 1.95 and d = -4.77(t - 0.23)^2 + 1.63 on the same graph with the original data from Table 1. Note that the graph of the first equation has its vertex on the origin (upper parabola) as compared to the second equation (lower parabola) which has its vertex at approximately 0.23 seconds. Which one is a better fit? Which one has a lower percent error?

Conclusion

It appears that the linearization method does not work well when the time at the start is not zero and the graph is in the form d = k*t^2. If data is collected prior to the motion starting, the use of the linearization method results in significant error. To see what happens if the time t=0 at the start but there is initial velocity in the Drop the Picket Fence lab.

As time permits, I will attempt to display a page where the linearization method works well.

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