Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 5 - Force and Motion-I - Problems - Page 119: 35b

Answer

${\theta_v}=28.0^{\circ}$

Work Step by Step

Given that, $v=(8.00ti^{\wedge}+3.00t^2j^{\wedge})$ We plug in the value of $t=1.415$ to obtain: $v=(8.00(1.415)i^{\wedge}+3.00(1.415)^2j^{\wedge})$ $v=(11.3i^{\wedge}+6.01j^{\wedge})$ Thus we obtain: ${\theta_v}=tan^{-1}\frac{v_y}{v_x}$ We plug in the known values to obtain: ${\theta_v}=tan^{-1}\frac{6.01}{11.3}$ ${\theta_v}=28.0^{\circ}$
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