LAB 01 Linear Kinematics

09/14/2023 PHY 121 L69 TA: Jake Hassan Linear Kinematics

Introduction: Linear kinematics is the study of an object’s motion with respect to time. The study of an object’s motion is demonstrated by the relationships between displacement, velocity, and acceleration. These quantities are related. This experiment will demonstrate these relationships by pushing a cart back and forth. The resulting motion can be explained with the following equations, v x = dx dt , which means velocity ( v ) over a certain period of time is the instantaneous change in position ( dx ) over the instantaneous change in time ( dt ), a x = d v x dt means acceleration ( a ) over a certain period of time equals the instantaneous change in velocity ( d v x ¿ over the instantaneous change in time ( dt ). x − x 0 = ∫ 0 t v x dt , states that the displacement ( x − x 0 ) or change in position is equal to the area under velocity vs. time graph. v − v 0 = ∫ o t a x dt , states that the change in velocity ( v − v 0 ¿ is equal to the area under the acceleration vs. time graph. Equivalently, velocity is the slope of the change in position and change in time, and acceleration is the slope of the change in velocity and the change in time. Methods/Procedure: 1. Set up iO Lab Device. 2. Open iO Lab application and calibrate device. 3. Place the device on wheels down on a table and press record. 4. Give the cart a push and allow it to stop on its own. 5. Zoom into data graphs vertically and horizontally until the data fills the chart. 6. Enter “analysis mode” and highlight where the velocity is linearly decreasing toward zero.

9. Find the starting and ending position of the wheel at that section. 10. Find the area under one acceleration spike. Find the average velocity before and after the spike. Calculate the difference. 11. Compare the slope of the displacement curve to the corresponding average velocity value. 12. Compare the change in position to the area under the velocity curve for the same time interval. 13. Compare the change in the velocity to the area under the acceleration curve. Part II 1. Turn on the Accelerometer sensor. 2. Set the device on a table with the positive z direction facing upwards. 3. Record 10 seconds while the device is on the table then pause. 4. Continue to record data as the device is dropped onto a pillow with the positive z direction facing upwards. Results: According to Figure 2, the velocity vs. time graph has a slope of 0.18 m/s 2 . The average acceleration of the acceleration vs time graph is 0.16 m/s 2 . This data verifies the equation a x = d v x dt which states that the change in velocity over the change in time (slope) is equal to the acceleration. The equation v − v 0 = ∫ o t a x dt is supported by the acceleration vs. time graph. The area under the acceleration vs. time graph is equal to the change in velocity. The area under the acceleration vs. time graph is 0.418m/s. Figure 2. Linearly decreasing velocity. The position, velocity, and acceleration data with respect to time are shown for a cart pushed and allowed to come to a stop on its own.

An acceleration spike can be seen on the acceleration vs. time graph between 1.07 seconds and 1.22 seconds. The area of this acceleration spike is -0.41 m/s. The average velocity before the spike is 0.16 m/s and the average velocity after the spike is -0.28 m/s. The difference between these acceleration spikes is -0.44 m/s. Figure 3. The displacement, velocity, and acceleration data with respect to the time when the cart is pushed and the average velocity is positive. Figure 4. The table displays data collected from the displacement, velocity, and acceleration graphs. Figure 5. The graph displays the x, y, and z acceleration for the iOLab device that was dropped on a pillow for the Accelerometer experiment.

For the negative slope -0.22 ± 0.02 m/s = -0.08 m / 0.39 s is true. And -0.08 m = -0.22 ± 0.02 m/s * 0.39 s is also true. During the accelerometer experiment, when the iOLab device was sitting on the table, there was no activity in the acceleration vs. time graph. That is because there was no movement at all. When the device was in free fall, the acceleration graph spiked. This happened because of the gravitational acceleration of -9.81m/s 2 . Proof