Constant Acceleration Motion: Experimental Analysis Guide

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

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4 Run 1a Run 1b

5 Run 2 Run 3

6 Run 4 PART 2: Free Fall Table 6. Position vs. Time graph Curve Fit Parameters A B C gravitational acceleration ( meters per second per second ) − 4.763 18.45 − 10.78 − 9.526 m/s 2 Table 7. Velocity vs. Time graph Linear Fit Parameters Slope Y-intercept gravitational acceleration ( units ) − 9.667 18.78 -9.667

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7 DATA ANALYSIS ( 10 points ): the section includes sample calculations and error analysis. Be sure to include equations! PART 1: Uniformly accelerated motion along the dynamic track. Show how the acceleration of the cart was calculated using the coefficients of quadratic fit. (Refer to Table 2). 1 2 𝑎𝑎𝑡𝑡 2 + 𝑣𝑣 0 𝑡𝑡 + 𝑥𝑥 0 = 𝐴𝐴𝑥𝑥 2 + 𝐵𝐵𝑥𝑥 + 𝐶𝐶 thus 1 2 𝑎𝑎 = 2 ⋅ 𝐴𝐴 Write equations of the motion (position as a function of time) for the cart for each of the run, using the parameters in Table 2. 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

8 Run 4: 𝑥𝑥 ( 𝑡𝑡 ) = 0.2137 𝑥𝑥 2 − 1.475 𝑥𝑥 + 2.687 𝑣𝑣 ( 𝑡𝑡 ) = 0.4270 𝑥𝑥 − 1.476 𝑎𝑎 ( 𝑡𝑡 ) = 0.4270 Write equations of the velocity for the cart for each of the run, using the parameters in Table 4. See Above (v(t)) PART 2: Free Fall

9 Run A: Calculate the percent discrepancy between accepted value of gravitational acceleration and your experimental result: Percent Error = � | Actual − Experimental | | Experiemental | � × 100 = |9.80 − 9.53| |9.53| × 100 ≈ 2.8 % Run B: From the average (mean) time of fall, calculate the experimental gravitational acceleration. Show equations and calculations. average time to fall 𝑡𝑡 avg = (1.33 + 1.21 + 1.24) 3 = 1.26 seconds 𝐻𝐻 = 1 2 𝑔𝑔𝑡𝑡 2 7.42 = 1 2 ( 𝑔𝑔 )(1.26) 2 solving for g we get: 𝑔𝑔 = 9.35 m/s 2 Knowing that the uncertainty in height measurement is ∆H = 0.02 m and the average human response time is ∆t = 0.2 s, estimate the uncertainty ( ∆𝑔𝑔 ) in your experimental g. Show equations and calculations.

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10 � ∆𝑔𝑔 𝑔𝑔 � 2 = � ∆𝐻𝐻 𝐻𝐻 � 2 + � 2 ∆𝑡𝑡 𝑡𝑡 ̅ � 2 � Δ𝑔𝑔 9.35 � 2 = � 0.02 7.42 � 2 + � 2 × 0.2 1.26 � 2 � Δ𝑔𝑔 9.35 � 2 = 0.1007883184 Δ𝑔𝑔 9.35 = √ 0.1007883184 Δ𝑔𝑔 = ±2.968360956 Δ𝑔𝑔 ≈ 2.97 Calculate the final velocity (eq. 2) of the ball at the bottom of the fall and its uncertainty. (Hint: � ∆𝑉𝑉 𝑓𝑓 𝑉𝑉 𝑓𝑓 � 2 = � ∆𝑔𝑔 𝑔𝑔 � 2 + � ∆𝑡𝑡 𝑡𝑡 ̅ � 2 ) � Δ𝑉𝑉 𝑓𝑓 𝑉𝑉 𝑓𝑓 � 2 = � 2.97 9.35 � 2 + � 0.2 1.26 � 2 Δ𝑉𝑉 𝑓𝑓 2 𝑉𝑉 𝑓𝑓 2 = 2.97 2 + 0.2 2 9.35 2 + 1.26 2 Δ𝑉𝑉 𝑓𝑓 𝑉𝑉 𝑓𝑓 = √ 8.8609 √ 89.0101 Δ𝑉𝑉 𝑓𝑓 𝑉𝑉 𝑓𝑓 = 2.976726 9.434516 Δ𝑉𝑉 𝑓𝑓 ≈ 2.98 and 𝑉𝑉 𝑓𝑓 ≈ 9.44 Δ𝑉𝑉 𝑓𝑓 is 2.98 m/s 𝑉𝑉 𝑓𝑓 is 9.44 m/s

11 RESULTS (3 POINTS) Note: Read the rules of the number significant figures to report in the final results. (pg. 6 Lab Manual). PART 1: Acceleration (m/s 2 ) Run # Position vs. Time graph Velocity vs. Time graph Write equation of motion for each run 1a 0.34 0.34 𝑥𝑥 ( 𝑡𝑡 ) = 0.1698 𝑥𝑥 2 − 0.2701 𝑥𝑥 + 0.1804 1b 1.2 1.2 𝑥𝑥 ( 𝑡𝑡 ) = 0.5952 𝑥𝑥 2 − 1.186 𝑥𝑥 + 0.6651 2 − 0.34 − 0.34 𝑥𝑥 ( 𝑡𝑡 ) = − 0.1688 𝑥𝑥 2 + 0.2222 𝑥𝑥 + 1.858 3 − 0.35 -0.35 𝑥𝑥 ( 𝑡𝑡 ) = − 0.1721 𝑥𝑥 2 + 1.463 𝑥𝑥 − 1.487 4 0.43 0.43 𝑥𝑥 ( 𝑡𝑡 ) = 0.2137 𝑥𝑥 2 − 1.475 𝑥𝑥 + 2.687 PART 2: Free Fall Part: (gravitational acceleration ± uncertainty) (m/s 2 ) Final velocity ± uncertainty) (m/s) 2A 9.6m/s 2 ± 2.8 N/A 2B 9.5 m/s 2 ± 3.0 9.4 m/s ± 3.0

12 DISCUSSION & CONCLUSION ( 10 points ): The purpose of this experiment was to further solidify the model for constant acceleration, by demonstrating how a cart might move with constant acceleration, calculating it, and experimentally testing it. The main concept used was the constant acceleration model learned, where the shape of the position function is parabolic, and the velocity function is linear. We saw how to model equations based on the direction they are heading, with motion heading away from the origin having a positive sign, and towards the origin having a negative sign. We calculated equations for each run’s position and velocity. In part two, we calculated the uncertainty in an experimental versus theoretical g value, being 9.80 m/s 2 and our value being 9.53 m/s 2 , and the percent error being 2.8 %. Then for part B, we calculated the final velocity and uncertainty for that, being 9.4 m/s ± 3.0 . To tell if the cart was not moving with constant velocity, it would not be a parabolic shape, that is one way to tell, another way to tell is if you plot the acceleration and obtain it’s derivative, it is not linear, as the velocity needs to be linear for acceleration to be constant, with its magnitude telling us the slope and whether or not it is traveling towards the origin or not. When m is positive, it is travelling away from the origin, and when m is negative it is traveling towards the origin. This only applies to the velocity function, if acceleration is positive or negative, it does not mean the object is going a certain direction. In the position vs. time graph, our coefficient A tells us half of the acceleration, so for constant acceleration, the A coefficient tells half the magnitude of acceleration and if it’s positive or negative. This is due to the power rule in calculus, since A is always paired with 𝐴𝐴𝑥𝑥 2 , so if the derivative of the position function is velocity, the slope is going to be (2 ⋅ 𝐴𝐴 )x , and the slope of the velocity function IS the acceleration. Thus, we obtain 1 2 𝑎𝑎 = 𝐴𝐴 In run 1, the calculated value using the position vs time graph compares to the velocity vs time graph are almost identical, with 2 significant figures being the same number. Graphs for 1a and 1b show very similar graphs, with a similar parabolic shape, positive velocity function and positive acceleration. This makes sense as these are both travelling away from the sensor (origin), just at different accelerations, with 1b having a stepper velocity slope than 1a. In run 2, it differs from run 1 by being negative, meaning the cart in run 1 is moving away from the cart, but if we look at the calculated acceleration values, they are only different in sign, meaning if they were travelling in the same direction, they will run parallel to each other. In run 3, the acceleration is negative while the velocities are positive because the velocity being positive means it is travelling away from the sensor (origin), but this velocity is decreasing over the time period until the cart reaches the top of its path, since the velocity

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13 is DECREASING at a constant rate, this constant rate is negative, and it’s also the slope, and as we know the slope of the velocity is the acceleration, thus acceleration is negative. The cart moved in such a way because the angle we had, upon reaching the top of its path, it would begin to return to the origin (having a parabolic shape). In run 4 the acceleration is positive even though the cart is slowing down with time, because it’s like run 3, the position graph has a negative slope, while the velocity graph has a positive slope (but some of the values are negative since the b-intercept is negative), the slope of the velocity then results in a positive acceleration. We know when the cart is heading to the sensor, the velocity is positive, and this makes sense because the position is decreasing from the origin as the cart is moving up its path. As we have seen the direction of motion depends on knowing the signs of position and velocity, so just knowing the acceleration does not tell us much about which direction an object is going, same with if it is slowing down or speeding up. The graphs of part 2 and 1 are related by the fact of having constant acceleration, and we see that in the graph shape, with the position function in part 1 being parabolic, and the y- position being parabolic as well, the main difference is that in part 1, we are looking at the motion of the whole object, while in part two we are looking at only the y-coordinate motion of the ball, not it’s x-coordinate as that is constant. Free fall motion is constant acceleration and can be modeled as such. In the formula for uncertainty the numerator has a dominating effect as mathematically as the numerator increases so does the ratio between the two, so for bigger values of Δ𝑔𝑔 , we will obtain larger uncertainty values. While the opposite is true for increase 𝑔𝑔 , for bigger values of 𝑔𝑔 we obtain smaller values. The theoretical value of 𝑔𝑔 does fall within the range of my reported result, one source of error could have been the timing interval of the simulation, with each measurement in time being 0.033 seconds, decreasing the interval of time per measurement, we would obtain more graph points and thus could estimate the value of 𝑔𝑔 much better. Another one that was systemic in nature could have been how computers represent floating point numbers (decimal numbers), as they are not really precise and can mess up doing mathematical operations with them.

Lab 2

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