Calculus I 2nd Partial Project Applications of Motion: Position, Velocity, and Acceleration Fernando Jair Balcázar A01570174 René Alejandro Ponce A01570252 Alberto Frese A01570191 María Renée Gamboa A01570128
Introduction The meaning of physics is not completely defined since there are always discoveries that lead not only to more discoveries and findings but also to speculations. Despite this, one of the most defined concepts and uses of physics is Kinetics, which studies energy and motion. This one manages the position, velocity and acceleration of an object in a given time. This three are related since whenever we get the function of position according to the time, its velocity is its derivative as well as acceleration is the derivative of the velocity of the same object. This information is useful and makes it easier to understand and analyze some functions. To show this, we will work with the following data and derivate the functions that complete the scatter plot in order to get the velocity and acceleration functions in relation to time. Data t f(t) g(t) h(t) F(t) G(t) H(t) -5-4.5-4 -3.5 3.00673 8 2-4.44225-8.5-2 3.01110 9 2.875-4.35721-7.75-1.5 3.01831 6 3.5-4.25992-7 -1 3.03019 7 3.875-4.14471-6.25-0.5 0.14285 7 0.15384 6 0.16666 7 0.18181 8-3 3.04978 7 4-4 -5.5 0 0.2-2.5 3.08208 5 3.875-3.7937-4.75 0.5 0.22222 2-2 3.13533 5 3.5-3 -4 1 0.25-1.5 3.22313 2.875-2.2063-3.25 1.5 0.28571 4-1 3.36787 9 2-2 -2.5 2 0.33333 3-0.5 3.60653 1 0.875-1.85529-1.75 1.5 0.4
0 4-0.5-1.74008-1 1 0.5 0.5 4.64872 1-2.125-1.64279-0.25 0.5 0.66666 7 1 1.5 2 5.71828 2-4 -1.55775 0.5 0 1 7.48168 9-6.125-1.48171 1.25-0.5 2 10.3890 6-8.5-1.4126 2-1 N.P. 2.5 15.1824 9-11.12 5-1.34904 2.75-1.5-2 3 23.0855 4-14 -1.29002 3.5-2 -1 3.5 36.1154 5-17.12 5-1.23483 4.25-2.5-0.6666 7 4 57.5981 5-20.5-1.18288 5-3 -0.5 4.5 93.0171 3-24.12 5-1.13374 5.75-3.5-0.4 5 151.413 2-28 -1.08707 6.5-4 -0.3333 3
Analysis of the data According to the behavior of the points given, we found out that this followed the function 7.9511e0.363x. Because of the variable in the exponent, the graph curves exponentially and follows the points. However, it is not shown completely in this picture, so the curve cannot be appreciated completely because of the domain and range chosen for the function. To get the function of velocity, we must get the derivative of it. First, we need to know where the x is in the function and then apply the corresponding rule. Since x is located as an exponent of e, then we follow the next rule f (x) = u e u. By following this, we get: f (t) = 0.363 79511e 0.363x. After multiplying, we get that the function of velocity in relation to time is f (t) = 28862.49e 0.363x. To get the acceleration we get the derivative of the previous function. Since the x is located in the same spot, we follow the same rule: f (x) = u e u. We get that f (t) = 10477.08e0.363x.
The graph that adequates the most to the points given in the original table is a parabola, formed by a quadratic function. We can see that, different to the previous graph, this one passes through all the points given. However, similar to the last graph, the complete graph is not shown complete so it may not seem as a parabola. This is also because of the domain and range of the graph. As we said in the previous graph, to get the derivative, we must first locate x. However, we ll now use a more vertical way to display the process in order to make it easier to read. Since it is multiplying a constant, we can use power rule. g (t) =. 5x 2 3x. 5 g (t) = 2(. 5x) 1 (3) g (t) = x 3 Then, the derivative of this function will give us the acceleration function. g (t) = 1( 1 ) g (t) = 1
Since the software we used in the first two graphs (Excel) couldn t work really well with this one, we decided to use another one, which ended up being GeoGebra. In this software, we had more freedom to manipulate the degree of the function in order to make the graph more exact and close to the points. In this case, the degree of the function ended up being 6, as it is its highest power and can be appreciated by the form of the graph. Velocity: h (t) = 5(.12x 4 ) 4(.05x 3 ) 3(.52x 2 ) + 2 (.39x) +. 28 h (t) =. 6x 4. 2x 3 1.56x 2 +. 78x +. 28 Acceleration: h (t) = 4(.6x 3 ) 3(.2x 2 ) 2 (1.56x) +. 78 h (t) = 2.4x 3. 6x 2 3.12x +. 78 The scatter plot in this case formed a clear line that was easy to cover with a linear function. There were only two things that had to be analyzed and they were the slope and the y-intercept. As we can see, the graph passes through all the points of the scatter plot, so it is a satisfactory function. Like some of the previous examples, the complete graph is not needed, so it is shortened and cut. Velocity: F (t) = 1.5x 1 F (t) = 1.5(1)x 1 1 1 (0) F (t) = 1.5 Acceleration: F (t) = 1.5(0) F (t) = 0 Since velocity is constant, meaning that it s slope doesn t change, it means that there s no acceleration.
For this equation, we used GeoGebra to scatter plot all the data and determine the function. This one acts almost as an absolute value with the data we re given but we can see some variations that make it difficult to adapt it so we chose the function: G (t) = 0.01x⁷ + 0.02x⁶ 0.06x⁵ 0.09x⁴ + 0.35x³ 0.09x² 1.22x + 1.05 Velocity: G (t) = 0.01x⁷ + 0.02x⁶ 0.06x⁵ 0.09x⁴ + 0.35x³ 0.09x² 1.22x + 1.05 G (t) = 7(.01)x 6 + 6(.02)x 5 5(.06)x 4 4(.09)x 3 + 3(.35)x 2 2(.09)x 1.22 G (t) =. 07x 6 +. 12x 5. 3x 4. 36x 3 + 1.05x 2. 18x 1.22 Acceleration: G (t) =. 07x 6 +. 12x 5. 3x 4. 36x 3 + 1.05x 2. 18x 1.22 G (t) = 6(.07)x 5 + 5(.12)x 4 4(.3)x 3 3(.36)x 2 + 2 (1.05)x. 18 G (t) =. 42x 5 +.6x 4 1.2x 3 1.08x 2 + 2.1x. 18
To solve the scatter plot of this function, we used Excel and the function that the program gave us was an exponential one with degree 4. The base form of this function looks like a flat parabola and this tendency can be clearly seen, but the graph is cut in half so this shape is not shown completely. Similar to previous cases, the domain and range are more limited than in the base form of the function, because the information needed can be shown by the section chosen. Velocity: H (t) = 0.0035x 4 0.0037x 3 0.126x 2 + 0.4395x 1.7147 H (t) = 0.0035(4)x 4 1 0.0037(3)x 3 1 0.126(2)x 2 1 + 0.4395(1)x 1 1 1.7147(0) H (t) = 0.014x 3 0.0111x 2 0.252x + 0.4395 Acceleration: H (t) = 0.014x 3 0.0111x 2 0.252x + 0.4395 H (t) = 0.014(3)x 3 1 0.0111(2)x 2 1 0.252(1)x 1 1 + 0.4395(0) H (t) = 0.042x 2 0.0222x 0.252 Conclusions René Ponce: The use of derivatives are essential to calculate real life events such as to get the acceleration and velocity of an object. This project has helped me get more understanding of this. Fernando Jair: This project not only helped to understand the importance of derivatives in different functions, but it also helped me review the scatter plot subjects and how different kind
of functions behave, showing me also the difference that small details such as the degree of the function can make in the graph. Renée Gamboa: This project has aided me to difference between velocity and acceleration in terms of derivatives, it explained why when velocity is constant it has no acceleration, and this was due to it having a non-changing slope. Alberto Frese: This project really helped me to imagine how some of the derivatives translate into things that I would initially have never thought could be obtained by getting the derivative of a function. Bibliography - Khan Academy. (n.d.). What is Physics?. Retrieved in October 19th, 2017 from: https://www.khanacademy.org/science/physics/one-dimensional-motion/introduction-to-p hysics-tutorial/a/what-is-physics - Fisicanet.(n.d.). Conceptos Básicos de Cinemática. Retrieved in October 19th, 2017 from: https://www.fisicanet.com.ar/fisica/cinematica/ap01_cinematica.php