A bike travels 6.53 x 103 m at 67.9 degrees North of East in a time of 2.56 x 103 seconds. The biker then has a drink for 1.23 x 103 seconds. She then rode a distance 6.05 x 103 m South in a time of 2.93 x 103 m seconds. She paused for 1.88 x 103 seconds, and it takes her back to her starting point. What was the magnitude of the average velocity of their entire ride?

Name: A bike travels 6.53 x 103 m at 67.9 degrees North of East in a time of
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Description: A bike travels 6.53 x 103 m at 67.9 degrees North of East in a time of 2.56 x 103 seconds. The biker then has a drink for 1.23 x 103 seconds. She then rode a distance 6.05 x 103 m South in a time of 2.93 x 103 m seconds. She paused for 1.88 x 103 sec

According to the definition of average velocity of a body we know vavg=net displacement total time…

Answered: A bike travels 6.53 x 103 m at 67.9…

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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A bike travels 6.53 x 10³m at 67.9 degrees North of East in a time of 2.56 x 10³seconds. The biker then has a drink for 1.23 x 10³seconds. She then rode a distance 6.05 x 10³m South in a time of 2.93 x 10³m seconds. She paused for 1.88 x 10³ seconds, and it takes her back to her starting point. What was the magnitude of the average velocity of their entire ride?

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Step 1

According to the definition of average velocity of a body we know

v_avg= $\frac{n e t d i s p l a c e m e n t}{t o t a l t i m e}$

now average velocity of the bike along x-direction is given by

(vx)_avg= $\frac{6.53 \times 10^{3} \cos (67.9) - 0}{(2.56 \times 10^{3} + 1.23 \times 10^{3})} = 0.648 m / s$

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