I need help answering these questions using the graph. Please explain as simple as possible (Not honor class)(Not grading) Dota #2Graph #2TimeDistance(seconds)(meters)Distance (m) vs. Time (s)1.000751.010.50.040.25.090.00104.16Time (s).256.36.498.649..81101.03. 75%Normal textArial12| 1. I 2. | 3 I..4 | 5. | 6 .7a. Based on the data presented, describe the motion of the cart?(it is speeding up, it is slowing down, ifs speed remains constant)b. Describe the relationship between time and distance and write a general equationlinear-straightline, hyperbola-in verse curve, or parabola-power curve)for the curve.c. Write the general equation for the graph. ("y = kx, y=k/x or y= kx)d. Is the distance traveled greater between 0 and 1 second or between 7 and 8 seconds?e. Is the slope of the curve greater between seconds 0 and 2 or between seconds 8 and 10?

Answered: Image /qna-images/answer/c0a86f7b-c19e-4b91-a973-43412f0316e4.jpg

Answered: a. Based on the data presented,…

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter2: Representing Motion

Section: Chapter Questions

Problem 48A

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Similar questions

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According to Timothy Treadwell in 2001 "Now, the bears I live with average, the males are on average twelve hundred pounds. They're the largest bears in the world and they've been clocked at 41 [mph] and they've run a hundred meter dash in 5.85 seconds which human on stimulants doesn't even approach. a. Compute the speed of a grizzly bear using Mr. Treadwell's hundred-meter statement. b. Compute the momentum of a grizzly bear using the speed you calculated in part a. and the average mass stated by Mr. Treadwell. c. How fast would a 250 lb man have to run to have the same momentum you calculated in part b? d. How fast would a 4000 lb car have to drive to have the same momentum you calculated in part b?
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Solve the following operation(s) and express final answers to the correct number of significant figures. a. (5.72)(28.37)/(0.002231) b. (1.67 x 104)(0.15 x 10 5) c. 2.000 + 30.4 d. 19.26 - 7

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