Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 3 - Vectors in Physics - Problems and Conceptual Exercises - Page 79: 41

Answer

$59^{\circ}$ north of east, $4.9\frac{m}{s}$

Work Step by Step

The magnitude and direction of the average velocity can be determined as: $\theta=tan^{-1}(\frac{r_y}{r_x})=tan^{\frac{2500ft}{1500ft}}=59^{\circ}north\space of\space east$ and the magnitude of the displacement is $r=\sqrt{(r_x)^2+(r_y)^2}$ $r=\sqrt{(1500ft)^2+(2500ft)^2}=2900ft\times 0.305\frac{m}{ft}=890m$ Now the magnitude of the average velocity is given as $v_{avg}=\frac{\Delta r}{\Delta t}$ We plug in the known values to obtain: $v_{avg}=\frac{890}{3.0min\times 60\frac{s}{min}}=4.9\frac{m}{s}$
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