Lab 2

1 Constant Acceleration Motion in One Dimension (along a straight line). Devin Pridgeon 188357 Harish Hemming Week 2

2 OBJECTIVE ( 3 points ): Constant acceleration can be modeled and calculated. We will see how experimental values compare with those calculated. EXPERIMENTAL DATA ( 3 points ): Run 1 a : 𝑥𝑥 ( 𝑡𝑡 ) = 0.1698 𝑥𝑥 2 − 0.2701 𝑥𝑥 + 0.1804 𝑣𝑣 ( 𝑡𝑡 ) = 0.3404 𝑥𝑥 − 0.2726 𝑎𝑎 ( 𝑡𝑡 ) = 0.3396 Run 1b: 𝑥𝑥 ( 𝑡𝑡 ) = 0.5952 𝑥𝑥 2 − 1.186 𝑥𝑥 + 0.6651 𝑣𝑣 ( 𝑡𝑡 ) = 1.192 𝑥𝑥 − 1.188 𝑎𝑎 ( 𝑡𝑡 ) = 1.192 Run 2: 𝑥𝑥 ( 𝑡𝑡 ) = − 0.1688 𝑥𝑥 2 + 0.2222 𝑥𝑥 + 1.858 𝑣𝑣 ( 𝑡𝑡 ) = − 0.3402 𝑥𝑥 + 0.2299 𝑎𝑎 ( 𝑡𝑡 ) = − 0.3402 Run 3: 𝑥𝑥 ( 𝑡𝑡 ) = − 0.1721 𝑥𝑥 2 + 1.463 𝑥𝑥 − 1.487 𝑣𝑣 ( 𝑡𝑡 ) = − 3.449 𝑥𝑥 + 1.465 𝑎𝑎 ( 𝑡𝑡 ) = − 3.449 Run 4: 𝑥𝑥 ( 𝑡𝑡 ) = 0.2137 𝑥𝑥 2 − 1.475 𝑥𝑥 + 2.687 𝑣𝑣 ( 𝑡𝑡 ) = 0.4270 𝑥𝑥 − 1.476 𝑎𝑎 ( 𝑡𝑡 ) = 0.4270 PART 1: Uniform Accelerated Motion on a Dynamic Track Table 1. Run 1a Time interval ( seconds ) Coordinates ( seconds, meters ) Distance ( meters ) Δ𝑡𝑡 = 4 − 3 (3,0.8986) & (4,1.8159) 1.357 Δ𝑡𝑡 = 2.125 - 2 (2, 0.3191) & (2.125, 0.3729) 0.1361 Table 2. Position vs. Time Curve Fit Coefficients

3 Run # A B C Acceleration ( units ) 1a 0.1698 − 0.2701 0.1804 0.3396 1b 0.5952 − 1.186 0.6651 1.1904 2 − 0.1688 0.2222 1.858 − 0.3402 3 − 0.1721 1.463 − 1.487 − 0.3449 4 0.2137 − 1.475 2.687 0.4270 Table 3. Position vs. Time Curve Fit Parameter Definitions Coefficients Name of Physics quantity (i.e. position, distance, velocity, etc.) A 1 2 𝑎𝑎 = one half of the acceleration B initial velocity = 𝑣𝑣 0 C 𝑥𝑥 − intercept or initial position Table 4. Velocity vs. Time Linear Fit Parameters Run # Slope Y-intercept Acceleration ( units ) 1a 0.3404 -0.2726 0.3404 1b 1.192 − 1.188 1.192 2 − 0.3402 0.2299 − 0.3402 3 − 0.3449 1.465 − 0.3449 4 0.4270 − 1.476 0.4270 Table 5. Velocity vs. Time Linear Fit Parameter Definitions Coefficients Name of Physics quantity (i.e. position, distance, velocity, etc.) m (slope) acceleration b (y-intercept) initial position

4 Run 1a Run 1b