Answer
$Q_{PB} = 0.4m/s^2$
Work Step by Step
$Q_{PB} = (0.4 \times cos (60^{\circ})i)m/s^2 + (0.4 \times sin (60^{\circ})j)m/s^2$
$Q_{PB} = (0.2i + 0.346j )m/s^2$
$|Q_{PB}| =\sqrt {0.2^2 +0.346^2}$
$Q_{PB} = 0.4m/s^2$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
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 - Page 89: 78c
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