A squirrel has a- and y- coordinates (1.0 m, 3.5 m) at time t₁ = 0 and coordinates (4.9 m, -0.50 m) at time t2 = 3.4 s. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Calculating average and instantaneous velocity. Find the magnitude of the average velocity. Express your answer in meters per second. V= [ΠΙ ΑΣΦ Submit Part D Request Answer Find the direction of the average velocity. Express your answer in degrees. m/s

A squirrel has a- and y- coordinates (1.0 m, 3.5 m) at time t₁ = 0 and coordinates (4.9 m, -0.50 m) at time t2 = 3.4 s. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Calculating average and instantaneous velocity. Find the magnitude of the average velocity. Express your answer in meters per second. V= [ΠΙ ΑΣΦ Submit Part D Request Answer Find the direction of the average velocity. Express your answer in degrees. m/s

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Description: Needs Complete typed solution with 100 % accuracy. A squirrel has x- and y-coordinates (1.0 m, 3.5 m) at time t₁ = 0and coordinates (4.9 m, -0.50 m) at time t2 = 3.4 s.For related problem-solving tips and strategies, you may want to viewa Video Tutor

Physics for Scientists and Engineers with Modern Physics

10th Edition

ISBN:9781337553292

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter3: Vectors

Section: Chapter Questions

Problem 35AP: A person going for a walk follows the path shown in Figure P3.35. The total trip consists of four...

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I took 2.28km and converted it to 2280 meters then divided by 3200 and ended up with .71m/s but it keeps saying that its incorrect. Im not sure where I keep messing up. John walks 1.80 kmkm north, then turns right and walks 1.40 kmkm east. His speed is 1.60 m/sm/sduring the entire stroll. What is the magnitude of his displacement, from beginning to end? (2.28) If Jane starts at the same time and place as John, but walks in a straight line to the endpoint of John's stroll, at what speed should she walk if she wants to arrive at the endpoint just when John does?
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.
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