Fall22 Postion, Velocity, and Acceleration Lab Online EDITED 8.16.22-1

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! 1

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. Setup 1. Go to the following website: https://www.walter-fendt.de/html5/phen/acceleration_en.htm 2. You should now see the following: 2

Analysis of Position, Velocity and Acceleration Lab Name: Course/Section: Phy-1611-014 Instructor: Table 1 (10 points) 4

t(s) x(m) v(m/s) a(m/s 2 ) 0.00 0.00 0.00 0.75 1.010 0.38 0.76 0.75 2.001 1.50 1.50 0.75 3.010 3.40 2.26 0.75 4.040 6.13 3.03 0.75 5.141 9.92 3.86 0.75 6.010 13.54 4.51 0.75 7.003 18.38 5.25 0.75 8.010 24.06 6.01 0.75 9.002 30.40 6.75 0.75 10.013 37.60 7.51 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 5

Position vs Time Value Linear Fit m 3.7495 B -5.6191 y = mx + b y = 3.7495x - 5.6191 Quadratic Fit A 0.375 B 0.0001 C 0 y = Ax 2 + Bx + C y = 0.375x 2 + 0.0001x + 0 (x 1 , y 1 ) first point (0, 0) (x 2 , y 2 ) last point (10.013, 37.60) Slope 3.76 Velocity vs. Time Value M 0.7499 B 0.0014 y = mx + b y = 0.7499x + 0.0014 (x 1 , y 1 ) first point (0, 0) (x 2 , y 2 ) last point (10.013, 7.51) v avg 3.77 v avg time 5.03 Acceleration vs. Time Value M 0 B 0.75 y = mx + b Y= 0x + 0.75 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 Position vs Time would be m/s. b. Velocity vs Time graph? (2 points) 7

The appropriate units for the slope of Velocity vs Time would be m/s 2 . c. Acceleration vs Time graph? (2 points) The appropriate units for the slope of Acceleration vs Time would be 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) The quadratic fit of this graph did provide the initial position, the initial position being 0m at time 0s since the car is at rest. b. Did your quadratic fit of this graph provide initial velocity? If yes, what is its value? (4 points) The quadratic fit of this graph did provide the initial velocity, that of which being 0m/s at time 0s since the car is at rest. c. Did your quadratic fit of this graph provide acceleration? If yes, what is its value? (4 points) Yes, the value of acceleration of Position vs Time is 0.7499m/s 2 . d. What specific physical quantity does the slope of the two middle points from the Position vs. Time graph represent? (4 points) The specific physical quantity the slope of the two middle points from the Position vs. Time graph represents is 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) The linear fit of this graph did not provide the initial position, as it is only giving the velocity of the car. b. Did your linear fit of this graph provide initial velocity? If yes, what is its value? (4 points) The linear fit of this graph did provide the initial velocity, with the initial value being 0m/s at time 0, as the car is at rest. c. Did your linear fit of this graph provide acceleration? If yes, what is its value? (4 points) The linear fit of this graph did provide acceleration, the value being 0.7499m/s, 8