College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 10 - Problems - Page 399: 44


$v_m^2 = a_m~A$

Work Step by Step

We can write an expression for $v_m^2$: $v_m^2 = (A~\omega)^2 = A^2~\omega^2$ We can write an expression for $a_m$: $a_m = A~\omega^2$ Then: $v_m^2 = A^2~\omega^2 = (A~\omega^2)~A = a_m~A$
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