
Distance vs. Time  Modeling Acceleration and the Quadratic Equation
The distance (m) as a function of time (sec) graph shown at the right was collected by dropping 5 cm a picket fence vertically through a photogate. 

In the chart below, distance as a function of time is converted to velocity as a function of time. Since velocity is distance divded by time, we will calculate the velocity between each pair of points. First, find the difference in the distance and difference in the time between each point. Second, find the average time between each pair of points. Finally, find the velocity (d/t)
Time
(sec) 
Distance
(m) 

Time
(sec) 
Distance
(m) 

Avg Time
(sec) 
Velocity
(m/sec) 
0 
0.00 

0.020 
0.05 

0.010 
2.523 
0.020 
0.05 

0.019 
0.05 

0.030 
2.707 
0.038 
0.10 

0.017 
0.05 

0.047 
2.885 
0.056 
0.15 

0.016 
0.05 

0.064 
3.049 
0.072 
0.20 

0.016 
0.05 

0.080 
3.205 
0.088 
0.25 

0.015 
0.05 

0.095 
3.392 
0.102 
0.30 

0.014 
0.05 

0.110 
3.494 
0.116 
0.35 

0.014 
0.05 

0.124 
3.647 
0.130 
0.40 

0.013 
0.05 

0.137 
3.771 
0.144 
0.45 

0.013 
0.05 

0.150 
3.897 
0.156 
0.50 

0.013 
0.05 

0.163 
4.013 
0.169 
0.55 

0.012 
0.05 

0.175 
4.146 
0.181 
0.60 

0.012 
0.05 

0.187 
4.255 
0.193 
0.65 






Download Excel Spreadsheet with the calculations shown above.

Graphing the Calculated Velocity vs Avg Time data obtained from the chart above results in the graph at the right. The linear data results in a regession equation of V = 9.81 m/sec^2 + 2.43 m/sec. The slope of the line is v / t which is acceleration. Therefore, the calculated acceleration is 9.81 m/sec^2. The yintercept is 2.43 m/sec which is the velocity when t=0 sec or the velocity when the picket fence breaks the photogate. 
Finding Displacement from Velocity vs Time graph.
The area under the curve of a velocity vs time graph is displacement. To begin, we will look at the rectangular portion of the graph. The area of the rectangle is the base times the height of the rectangle. The base of the rectangle is the change in time and the height of the rectangle is the yintercept of the graph which is the initial velocity.


Combining the area of the rectangle and the area of the triangle is the displacement of the object the resulting equation is:
Displacement = (1/2)*a*t^2 + v_{i} *t.
If data collection begins when t=0, the elapsed time t is the same as t and the resulting equation is
Displacement = (1/2)*a*t^2 + v_{i}*t.
If the object starts at an initial distance from the origin, the displacement is added to the initial position and the model is:
Position = (1/2)*a*t^2 + v_{i}*t + d_{i}. Note this equation is in quadratic form, ax^2 + bx + c.
The Quadratic Regression and Interpretation
The graph at the right is the original distance vs.time graph with a quadratic regression fit to the data. The regression is y = 4.91x^2 + 2.43x + 0.00. The model implied by the data is D(m) = 4.91 (m/sec^2) * t^2 + 2.43 (m/sec) * t + 0.00 (m) where t is the time in seconds.
Combining the information gained from graphing the velocity vs time data, the quadratic model suggests that 1/2 of the acceleration is 4.91 m/sec^2 or the acceleration is 9.82 m/sec^2. The "b" term is the velocity when time is zero. The quadratic model suggests 2.43 m/sec which is the same as the yintercept of the velocity vs. time graph. Finally, the initial position of the object was 0.00 m. The fence was dropped through the photogate and the distance was measured from the photogate as the fence passed through it. 

Vertex Form
Using algebraic manipulations, it can be shown that the value of the vertex form of the parabola y = ax^2 + bx + c is
or the vertex can be found by graphing and finding the maximum or minimum value of the vertex.
The vertex of the parabola y = 4.91x^2 + 2.43x + 0.00 is
y = 4.91(x +0.25)^2  0.30.
The model would be
D(m) = 4.91 m/sec^2 (t + 0.25 sec)^2  0.30 m
The vertex (0.25 sec, .30 m) is special because the slope of the distance vs. time graph is velocity. At the vertex, the slope of the tangent line is zero. The vertex is the time and position where the object had zero velocity. The model predicts that the object was released or had a velocity of zero 0.25 seconds prior to breaking the the photogate. Since the distance was measured as positive when the object dropped through the photogate, positive distance is measured in the downward direction. The negative distance would be distance above the photogate. The vertex predicts the velocity was zero when the object was 0.30 m above the photogate. 


