926CF758-F3FF-448E-B0F4-D40AA24A3049

1 Position, Velocity and Acceleration Lab Online Purpose The purpose of this activity is to study 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 𝑜 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 𝑎𝑣𝑔 = ∆𝑥 𝑡 = 𝑥 − 𝑥 𝑜 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 𝑎𝑣𝑔 = ∆𝑣 𝑎𝑣𝑔 𝑡 = 𝑣 − 𝑣 𝑜 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 𝑜 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 𝑎𝑣𝑔 = 𝑣 + 𝑣 𝑜 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 𝑜 (constant value) 𝑣 = 𝑣 𝑜 + 𝑎𝑡 𝑥 = 𝑥 𝑜 + 𝑣 𝑜 𝑡 + 1 2 𝑎𝑡 2 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 𝑎𝑣𝑔 = 𝑎 − 𝑎 𝑜 𝑡 In this exercise we will be ignoring whatever little jerk there might be.

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 Name Marian Garcia Course/Section PHY 1611-011 Instructor Professor John Harrison Table 1 (10 points) t(s) x(m) v(m/s) a(m/s 2 ) 0.00 0.00 0.00 0.75 1.2 0.51 0.87 0.75 2.1 1.56 1.53 0.75 3.0 3.38 2.25 0.75 4.1 6.37 3.09 0.75 5.0 9.53 3.78 0.75 6.1 14.05 4.59 0.75 7.2 19.66 5.43 0.75 8.3 25.47 6.18 0.75 9.1 31.19 6.84 0.75 10.1 38.41 7.59 0.75 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)

7 Table 2 (From the ‘fits’ displayed on your graphs fill in Table 2) (10 points) Position vs Time Value Linear Fit m 3.84 B -5.971 y = mx + b y = 3.84x - 5.971 Quadratic Fit A 0.3735 B 0.0249 C - 0.0396 y = Ax 2 + Bx + C y = 0.3735x 2 + 0.0249x - 0.0396 (x 1 , y 1 ) first point (0, 0) (x 2 , y 2 ) last point (10.1, 38.41) Slope 2.67 Velocity vs. Time Value M 0.7527 B - 0.0138 y = mx + b y = 0.7527x - 0.0138 (x 1 , y 1 ) first point (0, 0) (x 2 , y 2 ) last point (10.1, 7.59) v avg 3.8318 v avg time 19.577 Acceleration vs. Time Value M 0 B 0.75 y = mx + b Y= 0.75

8 1. What are the appropriate units for the slope of the: (a) Position vs Time graph? (2 points) (m/s) (b) Velocity vs Time graph? (2 points) (m/s 2 ) (c) Acceleration vs Time graph? (2 points) (m/s 3 ) 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 initial position was 0 m. (b) Did your quadratic fit of this graph provide initial velocity? If yes, what is its value? (4 points) Yes, the initial velocity was 0 m/s. (c) Did your quadratic fit of this graph provide acceleration? If yes, what is its value? (4 points) No (d) What specific physical quantity does the slope of the two middle points from the Position vs. Time graph represent? (4 points) It represents 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) Yes, the initial position is 0 m.