Alissa Castano- PHYS 1P91_ Lab 3

TABLES: UNITS VALUE UNCERTAINTY Track distance m 0.70 0.05 UNIT VALUE Maximum y value (positive peak) m 0.608 Minimum y value (negative peak) m 0.089 UNIT VALUE Data Collection Time s 10.0 UNIT VALUE Average Velocity *(from equation 3) m/s 0.07 Average Acceleration *(from equation 4) m/s^2 0 UNIT VALUE UNCERTAINTY Average Velocity *(iolab online) m/s 0.07210 N/A Average Velocity x scan time m 0.721 N/A UNIT VALUE UNCERTAINTY Average Acceleration *(iolab online) m/s 2 0.0000 N/A Change in velocity (equation 4) m/s 0 N/A UNIT VALUE UNCERTAINTY Average Velocity *(iolab online) m/s 0.07110 N/A Average Velocity x m 0.711 N/A

scan time UNIT VALUE UNCERTAINTY Average Acceleration *(iolab online) m/s 2 0.0000 N/A Change in velocity (equation 4) m/s 0 N/A FIGURES: Figure 1: Track and iOlab set up. Figure 2: Position time graph of iOlab moving along the track.

Figure 5: Trial 1 acceleration-time graph of iOlab moving back and forward along the track. Figure 6: Trial 2 velocity-time graph of iOlab moving erratically back and forward along the track.

Figure 7: Trial 2 acceleration-time graph of iOlab moving erratically back and forward along the track. DISCUSSION: How centered was your center marker? Explain how you were able to determine this. If you are way off (>5cm) from what you expect, think about how you moved the iOLab, what might be the cause (think of perspective and what part of the iOLab moved where). The piece of tape placed to mark the estimated center of the track was far from the true center of the track. As seen in figure 2, the two humps based on the movement of the iOlab are very uneven, indicating that the iOlab was not stopped at the actual center of the track. This is also proven by the maximum y value and the minimum y value obtained, +y= 0.608 and -y= 0.089, respectively. The difference between the positive peak and the negative peak is over 5 cm. This could have happened for multiple reasons. Firstly, the marker was not placed at the accurate center of the track. Secondly, the position of the pauses taken at the +y, -y, and estimated center mark may not have been consistent, the positions were not precise. Compare your position- and velocity-time graphs. Your position graph has 2 humps (one positive and one negative) while your velocity has 2 humps for every hump in the position-time graph. Explain why this happens physically using your knowledge of how you moved the iOLab. For the position-time graph, it shows the position of the iOlab with respect to the time. The iOlab was moved from the center starting point in the +y direction, then back to the center point, leading to a positive hump. Then the iOlab was moved from the center point in the -y direction, creating a negative hump. The positive and negative humps correspond to the way the iOlab was moved back and forward. On the other hand, the velocity-time graph

Did your average acceleration and average velocity change depending on how cRaZy you moved it? Should it? No, the average acceleration and velocity did not change even when the iOlab was moved in a more erratic way. The acceleration value did not change because there was still no change in velocity as it is still a position-time graph, meaning the velocity is constant all the way through. As for the velocity, it did not change because the length of the track (total displacement) was still 0.70 m and the scan time was still 10 s. The average velocity is found by dividing the change in position by the change in time. No matter how sporadically the iOlab was moved, the change in position and time were the same, leading to the same value as a more tame movement. How would your average acceleration change if the iOLab was still moving when it stopped collecting data? If the iOlab was not stopped when collecting data then the final velocity would not be zero, neither would the change in velocity. This means that when calculating the acceleration, there would be a change in velocity, therefore an acceleration value would be obtained. How would you expect your average velocity to change if we doubled the length of your track, but kept everything else (including scan time) the same? Would average acceleration change too? If the length of the track was doubled, then only the average velocity would change. The average velocity is dependent on the displacement of an object, if the displacement is double, then so is the velocity. For example, if the track length in this experiment was doubled then so would the velocity: Normal length: 0.70m average velocity = ∆𝑥 ∆? = 0.70𝑚 10? = 0. 07 𝑚/? Doubled length 0.70 m * 2 = 1.40 m average velocity = ∆𝑥 ∆? = 1.40𝑚 10? = 0. 14 𝑚/? However, the acceleration would not change, because the velocity is still constant, the length of the track would not create a change in velocity.