Fundamentals of Physics Extended (10th Edition)

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

Chapter 4 - Motion in Two and Three Dimensions - Problems: 2b

Answer

magnitude = $\sqrt(89) = 9.43m $

Work Step by Step

The magnitude of a vector is found by the following: magnitude = $\sqrt (x^2 +y^2+z^2) $ where the x, y and z represent the x, y and z components (i, j and k) respectively. In part 1, we found that the position vector in this case is r = -5m i + 8m j Therefore we have x = -5 and y = 8 and z = 0 Plugging these in we have magnitude = $\sqrt ((-5)^2 +8^2+0^2) = \sqrt(25+64) = \sqrt(89) $
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