Physics Lab - Position, Velocity, Acceleration

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1 Position, Velocity and Acceleration Lab Online Purpose The purpose of this activity is to s tudy some of the basic behaviors of a mass that is being uniformly accelerated ; that means, experiencing a constant acceleration. Theory Displacement is defined to be the straight line length measured from what is taken to be the initial position, and the final position. ∆𝑥 = 𝑥 − 𝑥 ௢ The average velocity is defined to be the time rate of change of position , therefore it is the displacement divided by the time interval the displacement took place over . 𝑣 ௔௩௚ = ∆𝑥 𝑡 = 𝑥 − 𝑥 ௢ 𝑡 The average acceleration is defined as the time rate of change of velocity , therefore it is the average velocity divided by the time interval the change in velocity took place over . 𝑎 ௔௩௚ = ∆𝑣 ௔௩௚ 𝑡 = 𝑣 − 𝑣 ௢ 𝑡 For a mass to be accelerated uniformly, the value of the acceleration must be constant, meaning it always has the same value, and therefore in such a case we can drop the ‘avg’ subscript for the acceleration in the above equation. 𝑎 = 𝑣 − 𝑣 ௢ 𝑡 When the acceleration a mass is experiencing is constant we say that that mass is being uniformly accelerated . Since acceleration is the time rate of change of velocity, then for a mass being uniformly accelerated, its velocity will be changing at a constant rate such that its average velocity will also be given by the following equation. 𝑣 ௔௩௚ = 𝑣 + 𝑣 ௢ 2 This equation tells us that for a mass being uniformly accelerated its average velocity is just the average value between its initial and final velocities. Keep in mind that this equation is only true for masses experiencing uniform acceleration!

2 Just using these definitions, and with the assumption of uniform acceleration, we can easily construct the Linear Kinematic Equations of Motion. Some of the equations we can construct are the following Uniformly Accelerated Motion 𝑎 = 𝑎 ௢ (constant value) 𝑣 = 𝑣 ௢ + 𝑎𝑡 𝑥 = 𝑥 ௢ + 𝑣 ௢ 𝑡 + ଵ ଶ 𝑎𝑡 ଶ Plotting these three equations out as functions of time will wield graphs similar to the following; A concept that is rarely discussed in freshman physics classes is the jerk. The average jerk is defined to be the time rate change of the acceleration; therefore, it is the change in acceleration divided by the time interval the change in acceleration took place over. 𝐽 ௔௩௚ = 𝑎 − 𝑎 ௢ 𝑡 In this exercise we will be ignoring whatever little jerk there might be.

3 Setup 1. Go to the following website: https://www.walter-fendt.de/html5/phen/acceleration_en.htm 2. You should now see the following: Procedure 1. Move the little green bar that is near the top center of the large yellow area as far to the right as you can. 2. In the green area on the right side of the screen make the following settings: a. Initial Position: 0.00 m b. Initial Velocity: 0.00 m/s c. Acceleration: 0.75 m/s 2 3. Note that the instantaneous values for position, velocity, and acceleration can be seen at the bottom of the large yellow area. a. We will be recording the instantaneous values for position, velocity, and acceleration as well as the value for time in our table during the experiment. b. The time of the experiment can be read off from the three different experiment clocks in the large yellow area. 4. Start the experiment by clicking on the yellow start button located in the green section on the right of your screen. a. Once you click the start button it will turn into a pause button.

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4 b. At approximately 1 second into the experiment, pause the experiment, and record the time of the experiment and the instantaneous values for position, velocity, and acceleration in Table 1. c. Restart the experiment only to pause it again at approximately 2 seconds to record the time and instantaneous values in Table 1. d. Repeat this process for approximately 10 seconds.

5 Analysis of Position, Velocity and Acceleration Lab Online Name Melissa Fernandez Ayala Course/Section PHY 1951 011 Instructor Christopher Dunn, TA: Amilcar Torres Quijano Table 1 (10 points) t(s) x(m) v(m/s) a(m/s 2 ) 0.00 0.00 0.00 0.75 1.001 s 0.38 m 0.75 m/s 0.75 m/s 2 2.001 s 1.50 m 1.50 m/s 0.75 m/s 2 3.005 s 3.39 m 2.25 m/s 0.75 m/s 2 4.001 s 6.00 m 3.00 m/s 0.75 m/s 2 5.008 s 9.41 m 3.76 m/s 0.75 m/s 2 6.042 s 13.69 m 4.53 m/s 0.75 m/s 2 7.023 s 18.50 m 5.27 m/s 0.75 m/s 2 8.034 s 24.20 m 6.03 m/s 0.75 m/s 2 9.014 s 30.47 m 6.76 m/s 0.75 m/s 2 10.078 s 38.09 m 7.56 m/s 0.75 m/s 2 (I know its three significant figures but I wanted to be more accurate with the seconds!) 1. From the data in Table 1, and using Excel or some other graphing software, make the 3 following graphs: a. Position vs. time b. Velocity vs. time c. Acceleration vs. time For the Position vs. time graph have software display BOTH the linear fit, and the quadratic fit on the graph. For the Velocity vs. time graph have the software display the linear fit on the graph. For the acceleration vs. time graph have the software display the linear fit on the graph. Make sure to turn these graphs in with the lab worksheet. (20 points)

6 Figure 1. Position vs. Time Graph ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Figure 2. Velocity vs. Time Graph ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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7 Figure 3. Acceleration vs. Time Graph Table 2 (From the ‘fits’ displayed on your graphs fill in Table 2) (10 points) Position vs Time Value Linear Fit m 3.7507 B -5.6554 y = mx + b y = 3.7507x – 5.6554 Quadratic Fit A 0.3818 B -0.0902 C 0.1126 y = Ax 2 + Bx + C 𝑦 = 0.3818𝑥 ଶ − 0.0902𝑥 + 0.1126 (x 1 , y 1 ) first point (0, 0) (x 2 , y 2 ) last point (10.078, 38.09) Slope 38.09 - 0/10.078 - 0 = 3.78 m/s OR: if finding the slope of the secant and the two points where the graphs intersect, it

8 would be 3.75 (1.87,1.357) and (7.954, 24.178) Velocity vs. Time Value M 0.7503 B -0.0012 y = mx + b y = 0.7503x – 0.0012 (x 1 , y 1 ) first point (0, 0) (x 2 , y 2 ) last point (10.078, 7.56) v avg 3.78 v avg time 5.04 Vavg = total distance /total time taken  38.09/10.078 = 3.78 Vavg Input 3.78 = y into 3.78 = 0.7503(x) - 0.0012  5.04 Vavg time Acceleration vs. Time Value M -5E-17 B 0.75 y = mx + b Y = -5E-17x - 0.75 I’m assuming since the slope is the “jerk” that there is a little bit of air resistance, hence the negative value for the slope (despite it being such a tiny number). If not then the graph was just giving me a weird number, though I am well aware that the constant acceleration does remain 0.75 m/s 2 . 1. What are the appropriate units for the slope of the: (a) Position vs Time graph? (2 points) The appropriate units for the slope of the Position vs. Time graph is (m/s) - velocity (b) Velocity vs Time graph? (2 points) The appropriate units for the slope of the Velocity vs. Time graph is (m/s 2 ) - acceleration (c) Acceleration vs Time graph? (2 points) 𝑣 𝑎𝑣𝑔 = 𝑣 + 𝑣 𝑜 2

9 The appropriate units for the slope of the Acceleration vs. Time graph is (m/s 3 ) – the “jerk” 2. For Position vs Time data: (a) Did your quadratic fit of this graph provide initial position? If yes, what is its value? (4 points) Yes, the quadratic fit of this graph did provide an initial position, located at 0 m (b) Did your quadratic fit of this graph provide initial velocity? If yes, what is its value? (4 points) Yes, the quadratic fit of this graph provided an initial velocity (slope is velocity), this value is 0 m/s (c) Did your quadratic fit of this graph provide acceleration? If yes, what is its value? (4 points) While the graph itself does not provide a value for acceleration, you may find this number by taking the second derivative of the quadratic equation generated. When doing so, 0.3818 x 2 = 0.7636 m/s 2 as the acceleration. While it is a little larger than 0.75 m/s 2 this is evidently due to error in the times, as they were not paused at the exact second. (d) What specific physical quantity does the slope of the two middle points from the Position vs. Time graph represent? (4 points) The slope of the two middle points from the position vs time graph represents the velocity of the object in motion (if we want to get really technical the slope of the secant line generated with these two points gives the average velocity). 3. For Velocity vs Time data: (a) Did your linear fit of this graph provide initial position? If yes, what is its value? (4 points) While the linear fit of the graph didn’t explicitly provide a value for the initial position, to determine the initial position from a velocity vs. time graph you must

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10 find the function of the graph and take the integral to figure out the displacement, which is the initial position. Plus, from the data we already know that the initial position is 0 m, but again, since neither axis is labeled strictly for the position, it isn’t labeled directly on there. (b) Did your linear fit of this graph provide initial velocity? If yes, what is its value? (4 points) Yes, the linear fit of the graph provided an initial velocity of 0 m/s (c) Did your linear fit of this graph provide acceleration? If yes, what is its value? (4 points) Yes, the linear fit of the graph provided an acceleration (the slope) of 0.75 m/s 2 4. For Acceleration vs Time data: (a) Did your linear fit of this graph provide initial position? If yes, what is its value? (4 points) No, the linear fit of this graph does not have a provided initial position explicitly written on it. However, it is possible to find the initial position when given an acceleration vs. time graph considering acceleration is the change in velocity over the change of time, and velocity is the change in position over time, meaning if we work backwards, it is possible to determine the initial position if only given the acceleration. (b) Did your linear fit of this graph provide initial velocity? If yes, what is its value? (4 points) Yes, the initial velocity given by the graph is 0 m/s. Acceleration is the change in velocity over the change of time (m/s/s), by rearranging the equation, one can determine the initial value of velocity. (c) Did your linear fit of this graph provide acceleration? If yes, what is its value? (4 points) Yes, the linear fit of the graph provided an acceleration of 0.75 m/s 2

11 (d) What are the SI units of the Jerk? (4 points) (change in acceleration over time) The SI units of the jerk are m/s 3 7. What is the general shape of each graph and why does each have that shape? (10 points) Position vs. Time – quadratic (parabola) with a linear trendline . The graph holds a quadratic shape because of the velocity (slope) changing at a constant rate, which is supported by the data gathered in the experiment. In addition, when the acceleration is constant, the curvature will continue, which again is supported by the table as the acceleration constantly displays “0.75”. In fact, when comparing the values on our data table, it seems that the velocity changes approximately by 0.75 every time it increases, further proving this reasoning. Velocity vs. Time – linear, positive slope. The acceleration is positive (acceleration has already been determined to be the slope), given the m value being 0.75 (the value of the acceleration) and moving at a constant rate, hence the linear nature and positive slope of the graph. Acceleration vs. Time – linear, straight line that is constant and parallel to the x-axis. Given our data, there was no change in the acceleration (remained a constant 0.75 m/s 2 ), thereby, the graph depicted the expected outcome of no slope aside from the constant 0.75 m/s 2 .

Physics Lab - Position, Velocity, Acceleration

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