PHY 111Spring 2021In-Class Activity #1Name: Peony MoradiaPart I: This graph is a simplified (and slightly approximated) version of the position versus time graph from the experiment where I pushed a cart, initially near the middle of a level track, away from the motion sensor (the origin). The cart struck a barrier and reversed its direction of travel.a.What is the position of the cart at the times t = 2.0 s and t = 8.0 s? How did you determine the values?b. Is the cart at rest at t = 0? How do you know?c. During what time interval is the cart moving away from the origin? How did you determine your answer?T= 0d. During what time interval is the cart moving toward the origin? How did you determine your answer?e. What is the cart’s displacement between t = 1.0 s and t = 3.0 s? Interpret your result.1.1 mf. What is the cart’s displacement between t = 0 s and t = 10.0 s? Interpret your result.-.74 mg. Use your answers to parts e and f to determine the average velocity for the two different time intervals.0.55 m/s-0.074 m/sh. What is the cart’s velocity at t = 1.0 s and at t = 3.0 s? How did you determine the values?t=1 is 1.1 m/st=3 is 0.9 m/si. What is the cart’s velocity at t = 5.0 s and at t = 8.0 s? How did you determine the values?-0.55 m/s-0.16 m/sj. How can you use the given position versus time graph to figure out the total distance traveled by the cart during the 10 s interval shown in the graph? Is this equal to the magnitude of the displacement of the cart? (See your answer to part f.) Why do you think the results are different?Distance= 5.25 mDisplacement= -0.74 mThe distance does not equal the magnitude of the displacement because the cart traveled in reverse meaning the final position is negative, but distance is the total distance of the path taken to reach said final position k. What is the average speed of the cart during the interval t = 0 to t = 10 s?0.525 m/sl. Draw a velocity versus time graph for the cart. Use your previous answers to construct and label the graph. m. Write a verbal description of the velocity versus time graph. The velocity would have positive slope from t=0 to t=4 and a negative slope from t=4 to t=10Part II: This graph is a simplified (and slightly approximated) version of the position versus time graph from the experiment where I pushed a cart up an inclined track, away from the motion sensor (the origin). The cart does not contact the barrier at the top of the incline.a.During the motion depicted here, what is the total displacement of the cart? Use your result to determine if the cart ends up at a point closer to, or farther from the origin. What about the total displacement tells you the answer?Displacement: -0.3 mCloser to the origin because the cart bounced back and traveled in reverse closer to the origin b. Is the cart in motion at t = 0? How did you determine your answer?Yes, because the motion graph shows a slope meaning that it’s in motionc. Is the cart in motion at the moment when the cart is at its greatest distance from the origin? How did you determine your answer?Yes because there is no plateau in the graph to signify a standstill position d. During what time interval is the cart moving away the origin? Is the cart speeding up, slowing down, or traveling with constant velocity during this time interval? How did you determine your answers?From t=0 to t=5 the cart is moving away from the origin and the cart is slowing down because it travels at a faster rate in the beginning and slows down directly afterward e. During what time interval is the cart moving toward the origin? Is the cart speeding up, slowing down, or traveling with constant velocity during this time interval? How did you determine your answers?From t=0 to t=5 the cart is moving towards the origin and is speeding up because the cart picks up speed as it covers more area in less time.f. This one will be harder to put (good) numbers on at this time so just sketch the graph of velocity versus time. Try to label important features.g. Write a verbal description of the velocity versus time