Question 4 At the starting gun, a runner accelerates at 1.7 m/s? for 4.8 s . The runner's acceleration is zero for the rest of the race. You may want to review Part A (Pages 29 - 33) . What is the speed of the runner at t = 2.0 s? Express your answer using two significant figures. ΑΣφ ? V = m/s Submit Request Answer Part B What is the speed of the runner at the end of the race? Express your answer using two significant figures. ? m/s Submit Request Answer

Question 4 At the starting gun, a runner accelerates at 1.7 m/s? for 4.8 s . The runner's acceleration is zero for the rest of the race. You may want to review Part A (Pages 29 - 33) . What is the speed of the runner at t = 2.0 s? Express your answer using two significant figures. ΑΣφ ? V = m/s Submit Request Answer Part B What is the speed of the runner at the end of the race? Express your answer using two significant figures. ? m/s Submit Request Answer

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Report 1. From t = 0 to t = 3.44 min, a man stands still, and from t = 3.44 min to t = 6.88 min, he walks briskly in a straight line at a constant speed of 1.94 m/s. What is his average velocity vavg in the time interval 1.00 min to 4.44 min? O a. 1.940 m/s O b. 0.000 m/s O c. 0.970 m/s O d. 0.564 m/s
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My question isn't how to solve the problem exactly. In fact, it's already been solved on this website. My question is about the acceleration. When I solve this problem myself, first I calculate the velocity by dividing 100m by 53s. I get 1.89m/s. Then I use that to find the acceleration using the equation vf = vi + at. That's 1.89/53 = 0.036m/s^2. That's not correct. The correct way to find the acceleration is to us the equation d = 1/2 at^2 and solve that way without taking the intermediate step of finding the velocity. Doing it that way, the acceleration is 0.0712m/s^2. My question is why you get a different result doing it the first way than you get doing it the second way.
Your answer is partially correct. A rock is dropped (from rest) from the top of a 73.9-m-tall building. How far above the ground is the rock 1.4 s before it reaches the ground? Number i eTextbook and Media Hint Units m
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