Ch 02 HW

3/7/23, 11:31 PM Ch 02 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460266 1/58 Ch 02 HW Due: 10:59pm on Tuesday, March 7, 2023 You will receive no credit for items you complete after the assignment is due. Grading Policy PhET Tutorial: Moving Man Learning Goal: To understand the relationships between position, velocity, and acceleration. NOTE: These activities use Java, and are therefore not screen-reader accessible and may not work on a mobile device. If the browser you're using no longer supports Java, try a different browser and download the Java plugin for this content. For this tutorial, use the PhET simulation The Moving Man . This simulation allows you to drag a person back and forth and look at the resulting position, velocity, and acceleration. You can also enter a position as a function of time mathematically and look at the resulting motion. Start the simulation. When you click the simulation link, you may be asked whether to run, keep, or save the file. If no prompt appears, you can also right-click the link and select "Copy link address", then manually paste this URL into a new tab (or window). If you are still unable to open this PhET in your browser, select the following link to open a separate compatible version of this simulation. This version may not perform as well, therefore it is offered as a secondary option: Moving Man (backup link) . Under the Charts tab you can click and drag the person left and right, or enter a numeric value in the boxes on the left panel to see plots for the person’s position, velocity, and acceleration as a function of time. Click the Play button to start a simulation and the Pause button to stop a simulation. You can also watch a playback by selecting the Playback radio button instead of the default Record radio button. You can click Clear to remove the current plot while maintaining your settings for position, velocity and acceleration or click Reset All to start over. In the Playback mode, the grey bar can be dragged over the plot to any value in time, and the digital readouts will show the corresponding values of the position, velocity, and acceleration. Under the Special Features menu, the Expression Evaluator option produces a second window in which you can mathematically type in any function for the position as a function of time, x ( t ) . After typing in a function, click the Play button to start the simulation. To zoom in vertically, click any of the three + buttons to the top right of each plot. To zoom in horizontally, click the + button to the bottom right of the acceleration plot. Feel free to play around with the simulation. When you are done, click Reset All on the Charts tab before beginning Part A. Part A First, you will focus on the relationship between velocity and position. Recall that velocity is the rate of change of position. This means that the velocity is equal to the slope of the Position vs. Time graph.

3/7/23, 11:31 PM Ch 02 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460266 2/58 Move the person to the position x = − 6 m or enter –6.00 in the position box. If you dragged the person to position, click the Pause button and then the Clear button. Next, drag the person to the right to roughly x = 6 m and reverse his direction, returning him to the original position, at x = − 6 m . Move the person relatively quickly, about a few seconds for the round trip. Your plots should look something like those shown below. Look at the Position vs. Time and Velocity vs. Time plots. What is the person's velocity when his position is at its maximum value (around 6 m )? ANSWER: Correct When the person’s position is a maximum, the slope of the position with respect to time is zero, which means the velocity is also zero. However, due to the person’s acceleration, the velocity does not remain zero; he eventually moves to the left. Part B Similarly, acceleration is the rate of change of the velocity, so it is the slope of the Velocity vs. Time graph. Because it is difficult to drag the person in a consistent and reproducible way, use the Expression Evaluator under the Special Features menu for this question. Click Reset All and type in the function x ( t ) = 8 t − 2 t t in the Expression Evaluator . Click the Play button and let the simulation run roughly 5 simulation seconds before pressing the Pause button. Use the zoom buttons to adjust the plots so they fit in the screen. You should see a plot similar to what you got in the previous question, but much smoother. Look at the Position vs. Time, Velocity vs. Time, and Acceleration vs. Time plots. Hint 1. How to approach the problem The person's velocity is zero. positive. negative.

3/7/23, 11:31 PM Ch 02 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460266 3/58 In Playback mode, use the grey vertical bar. Slide the bar until the value x = 8 m is displayed in the position box on the left panel. What are the values of velocity and acceleration when x = 8 m ? ANSWER: Correct At x = 8 m , the person turns to go back in the opposite direction. His velocity is zero, but his acceleration is negative since the velocity is decreasing with time. This is similar to throwing a ball straight up into the air; at its highest point, the velocity is zero but the acceleration is still directed downward. Part C Keep the function x ( t ) = 8 t − 2 t t in the Expression Evaluator. What is the value of the person’s acceleration a x at t = 2 s ? Hint 1. How to approach the problem Use the grey vertical bar. Slide the bar until it coincides with t = 2 s on the horizontal axis. What is the value displayed in the acceleration box on the left panel? ANSWER: Correct This is an example of one-dimensional motion with constant acceleration. The position of an object undergoing this type of motion obeys the kinematic equation x ( t ) = x 0 + v x , 0 t + 1/2 a x t 2 . In this case, the initial velocity is v x , 0 = 8 m/s and the acceleration is a x = − 4 m/s 2 (since 1/2 a x = − 2 m/s 2 ). Part D In the previous question, the person had an initial velocity of 8 m/s and a constant acceleration of −4 m/s 2 . How would the maximum distance he travels to the right of the origin change if instead his initial velocity were doubled ( v x , 0 = 16 m/s )? When the person is 8 m to the right of the origin, both the velocity and the acceleration are zero. the velocity is zero but the acceleration is negative. both the velocity and the acceleration are nonzero. the velocity is zero but the acceleration is positive. a x = 4 m/s 2 −2 m/s 2 0 −4 m/s 2

3/7/23, 11:31 PM Ch 02 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460266 4/58 Hint 1. How to approach the problem Go to the Introduction tab to run the simulation using the new initial velocity v x , 0 = 16 m/s and the same acceleration of −4 m/s 2 , and read the value for position when the velocity equals zero. Remember to remove the walls from the simulation by clicking on the red close button on the walls. In Playback mode the simulation can be run slowly and paused when the velocity is zero. ANSWER: Correct Because it takes twice as much time to momentarily stop, and because his average velocity will be twice as fast, the distance he travels will be four times greater. Using the kinematic equation, x (4 s) = (16 m/s)(4 s) − (1/2)(4 m/s 2 )(4 s) 2 = 32 m . Alternatively, since his final velocity (i.e. the point he turns around) is zero, you can use the other kinematic equation v 2 x , f = v 2 x , 0 + 2 a x ( x f − x 0 ) , or x f = − v 2 x , 0 2 a x to directly observe that doubling the initial velocity results in the distance increasing by a factor of 4. Now, assume going forward that the position is given by the equation x ( t ) = 0.2 t 3 − 2.4 t 2 + 7.2 t − 5 . Enter this function in the Expression Evaluator as x ( t ) = 0.2 t t t − 2.4 t t + 7.2 t − 5 and run the simulation by clicking the Play button in the Record mode, waiting until the person collides with the wall to select the Pause button. Consider the person's motion from the beginning until the moment just before colliding with the brick wall. Part E Which of the following statements is true of the position at t = 1.0 s ? ANSWER: Correct The slope of the Position vs. Time graph is positive (but decreasing in magnitude) at this time. Part F Which of the following statements is true of the velocity at t = 1.0 s ? The maximum distance would not change. The maximum distance would double. The maximum distance would increase by a factor of four. The position is decreasing at a rate increasing in magnitude. The position is decreasing at a rate decreasing in magnitude. The position is increasing at a rate increasing in magnitude. The position is increasing at a rate decreasing in magnitude.

3/7/23, 11:31 PM Ch 02 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460266 5/58 ANSWER: Correct Since the slope of the Position vs. Time graph is positive, the velocity is positive (i.e. the person is moving to the right), although since this slope is becoming smaller in magnitude, this implies the magnitude of the person's velocity, or speed, is decreasing (i.e. the person is slowing down). Part G How many times are the person's position, velocity, and acceleration equal to zero? Sort each item into the bin corresponding to the number of times it is equal to zero. ANSWER: Correct Notice that since the position is given by x = 4 t 3 , when the time is t = 1 s , the position is x = 4(1) 3 m = 4 m. Part H At approximately what times does the person change direction? Select all that apply. Hint 1. How to approach the problem The velocity is positive and the person is speeding up. The velocity is negative and the person is speeding up. The velocity is positive and the person is slowing down. The velocity is negative and the person is slowing down. Reset Help acceleration velocity position never 1 2 3 4