The image at the right is the motion of the ball rolled up a ramp of 2 x 4's with a motion detector at the top of the ramp as shown below. If the animation has stopped, click the RELOAD or REFRESH button on your browser. The GENDIST program allows the user to set an origin and measure the distance toward the motion detector as a positive distance. The ball was rolled from below the origin toward the motion detector and allowed to return back down the ramp. The graph of the data set is shown at the left below. Three different mathematical models will be discussed with this activity.

The image at the right shows the same data set with positions marked on the data. The ball was held at the same position until being pushed up the ramp at point A. At point B, the ball reaches the position ramp arbitrarily determined to be the origin. At point C, the ball reaches its highest point on the ramp and begins to move back down the ramp, crossing the origin at point D on its way down. At point E, the ball is retrieved and held at the bottom of the ramp while the motion detector continues to collect data.

By tracing the graph, find the vertex or point at which the y-value is the maximum as shown in the image at the left. The vertex of the parabola occurs at (1.60, 0.49). Since the motion detector measured the distance in meters and the time in seconds, the object reached the point furthest from the origin (closest to the top of the ramp) at 1.60 sec after the motion detector began recording time. The distance from the origin is 0.49 meters. Knowing the values of the vertex (h, k), part of the vertex model can be completed.

y = a(x - 1.60)² + 0.48

To find the value of a in the model, students can use a trial and error method to understand the transformations involved in a vertical stretch. One may also use a second point on the graph to substitute for (x, y) and solve for a. The image at the right shows another point on the curve as (0.50, -.73). Substituting and solving for a, on gets -0.73 = a (0.50 - 1.60)² + 0.48 or

The units would be m divided by (sec)² or -1.00 m/sec². As discussed in the cart on the ramp lab, this a value is 1/2 of the acceleration of the ball.

The equation y = -1.00(x - 1.60)² + 0.48 graphed with the original data set is shown at the left. This appears to be a good fit for the data.

To get the mathematical model, we will replace the appropriate variables and use units to get D_(m) = -1.00 _m/sec^2 ( t_(sec) - 1.60_(sec))² + 0.48_(m).

The model demonstrates an acceleration of -2.00 m/sec^2 acceleration and a maximal distance above the origin on the ramp of 0.48 meters occuring 1.60 seconds after the motion detector was activated.

Finding the model y = a(x - r)(x - s) requires the individual to find the roots. The roots, x-intercepts, or zero's of the equation occur where the y-value is 0 or on the x-axis. Using the trace feature, trace to find the values closest to where y is 0. In some cases, averaging or rounding in a particular direction may be appropriate. In this case, we will round the x value down since the ball has already passed the origin. One of the roots is 0.89 seconds. The second root is shown in the image below and to the right. Since the ball has not yet reached the origin on its way back down the ramp, this value will be rounded up to 2.30 seconds. In some cases the points far enough from the origin that averaging the times where the ball is above and below the origin would be called for.

Substituting the two roots into the equation, we get y = a (x - 0.89)(x - 2.30). Just as was done in the vertex form, we will substitute a point and solve for the value of 'a'. Since the vertex (1.60, 0.48) was identified above, we'll use it for the substitution. 0.48 = a (1.60 - 0.89)(1.60 - 2.30). Solving for 'a', one gets
a = 0.48/( (1.60 - 0.89)(1.60 - 2.30) ) or -.97. Again the units are m / (sec * sec) or m/sec². As discussed in the cart on the ramp lab, this value is 1/2 of the acceleration of the ball. The acceleration would be -1.93 m/sec²

The graph of the equation y = -0.97 (x - 0.89)(x - 2.30) with the original data is shown at the left. Just as the vertex form fit the data, this graph also appears to fit the data. Changing the variables and inserting units, one gets a mathematical model of D(m) = -0.97 _(m/sec^2) ( t_(sec) - 0.89_(sec) ) ( t_(sec) - 2.30_(sec) ).

The meaning of this model is that the acceleration is -1.93 m/sec² and the ball crosses the position marked as the origin 0.89 seconds and 2.30 seconds after the motion detector was activated.

Using the CHOOSE or Select feature, the data for the parabolic portion of the data set is chosen and stored in lists 3 and 4 of the calculator. Now, the quadratic regression can be run on those lists. The quadratic model for this data is y = -0.98x² + 3.13x - 2.01. Once again, this model fits the data as shown in the image below and to the right. Replacing the variables to form a mathematical model, one gets D(m) = -0.98 _(m/sec^2) t_(sec) ² + 3.13 _(m/sec) t_(sec) - 2.01 _(m).

The explanation of the mathematical model as developed in the cart ramp lab, is that the value of 'a' in the quadratic model is 1/2 of the acceleration. The acceleration of the ball is -1.95 m/sec^2. The final two values indicate the velocity and position when time is 0 seconds IF the motion of the ball had been occurring continuously prior to the release. The initial velocity would have been 3.13 m/sec and the ball would have been 2.01 meters below the origin when the recording of data was initiated.

Extension:
How does this model compare to the factored form and vertex form. Can one find the vertex form and factored form from this model?