Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 3 - Vectors and Coordinate Systems - Exercises and Problems - Page 84: 40

Answer

$5.18\;\rm \mu m/s, 333.4^\circ$

Work Step by Step

We know that the average velocity is given by $$v_{avg}=\dfrac{\vec R}{t_{tot}}\tag 1$$ whereas $\vec R$ is the resultant displacement and $t_{tot}$ is the time it takes for the entire trip. First, we need to find $\vec R$. From the given graph: $$\vec R=-(\vec{AB}+\vec{BC}+\vec{CD}+\vec{DE})=-\vec{EA}=\vec{AE}$$ $$\vec R=\vec{AE}=-(\vec{AE})_x\hat i-(\vec{AE})_y\hat j$$ $$\vec R= (40\;{\rm \mu m})\hat i+(-20\;{\rm \mu m})\hat j$$ $$\boxed{\vec R= (40\;{\rm \mu m})\hat i-(20\;{\rm \mu m})\hat j}$$ The magnitude of $\vec D$ is given by applying the Pythagorean theorem. $$|\vec R|=\sqrt{R_x^2+R_y^2}=\sqrt{40^2+(-20)^2}=\color{blue}{\bf 44.7}\;\mu m\tag 2$$ The direction of the average velocity is the direction of the resultant displacement. And the displacement direction is given by $$\tan\alpha_R=\dfrac{R_y}{R_x}$$ Thus, $$\alpha_R=\tan^{-1}\left[\dfrac{R_y}{R_x}\right]=\tan^{-1}\left[\dfrac{-20}{40}\right]=\bf -26.6^\circ$$ And since it is in the fourth quadrant; $$\alpha_R=\color{red}{\bf 333.4^\circ}$$ Now we need to find the total time it takes for the whole trip. $$t_{tot}=t_{\vec{AB}}+t_{\vec{BC}}+t_{\vec{CD}}+t_{\vec{DE}} $$ And since the velocity of E.coli bacterium is constant, the time it takes for each move is given by $$t=\dfrac{d}{v}$$ So, we need to find the length of each vector and divide it by its velocity. $$t_{tot}=\dfrac{\sqrt{50^2+10^2}}{20}+\dfrac{10} {20}+\dfrac{\sqrt{(90-50)^2+(30-20)^2}}{20}+\dfrac{\sqrt{(40-90)^2+(-20-30)^2}}{20}$$ $$t_{tot}=\color{blue}{\bf 8.65}\;\rm \mu s\tag 3$$ Plugging (2) and (3) into (1); $$v_{avg}=\dfrac{ 44.7}{8.65} =\color{red}{\bf 5.18}\;\rm \mu m/s$$
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