Fall22 Postion, Velocity, and Acceleration Lab Online EDITED 8

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Jun 18, 2024

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1 Analysis of Position, Velocity and Acceleration Lab Name: Nishat Shama Course/Section: PHY-1611-02F-202430 Instructor: Tyler Rucas Table 1 (10 points) t(s) x(m) v(m/s) a(m/s 2 ) 0.00 0.00 0.00 0.75 1.005 .38 .75 .75 2.042 1.56 1.53 .75 3.006 3.39 2.25 .75 4.173 6.53 3.13 .75 5.127 9.86 3.85 .75 6.012 13.55 4.51 .75 7.097 18.89 5.32 .75 8.128 24.77 6.10 .75 9.005 30.41 6.75 .75 10.087 38.16 7.57 .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)
2 y = 3.7727x - 5.6883 y = 0.3751x 2 - 0.0008x + 0.0007 -10 -5 0 5 10 15 20 25 30 35 40 45 0 2 4 6 8 10 12 Position (m) Time (s) Position vs. Time y = 0.7504x - 0.0022 -1 0 1 2 3 4 5 6 7 8 0 2 4 6 8 10 12 Velocity (m/s) Time (s) Velocity vs. Time
3 Table 2 (From the ‘fits’ displayed on your graphs fill in Table 2) (10 points) Position vs Time Value Linear Fit m 3.7727x B - 5.6883 y = mx + b y = 3.7727x - 5.6883 Quadratic Fit A 0.3751x 2 B - 0.0008x C 0.0007 y = Ax 2 + Bx + C y = 0.3751x 2 - 0.0008x + 0.0007 (x 1 , y 1 ) first point (0, 0) (x 2 , y 2 ) last point (10.087,38.16) Slope 3.783 Velocity vs. Time Value M 0.7504x B - 0.0022 y = mx + b y = 0.7504x - 0.0022 (x 1 , y 1 ) first point (0, 0) (x 2 , y 2 ) last point (10.087,7.57) v avg .7504 v avg time 3.7727 y = -5E-17x + 0.75 0.7497 0.751275 0 2 4 6 8 10 12 Acceleration ( m/s2) Time (s) Acceleration vs. Time
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