Fundamentals of Physics Extended (10th Edition)

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

Chapter 2 - Motion Along a Straight Line - Problems: 20a

Answer

$t = \sqrt (4/3)$ and t = $-\sqrt(4/3)$

Work Step by Step

The position is given by $ x = 20t - 5t^3 $ To find the velocity, we need to take the derivative of the position equation or $ \frac{d}{dt} (20t - 5t^3) $ The position is given by $ x = 20t - 5t^3 $ Since the velocity equation is the derivative of the position equation we need to find $\frac{d}{dt} (20t - 5t^3) $ Since $\frac{d}{dt} t^n = n*t^(n-1)$ then $ v = 20 - 5(3)t^2 = 20 - 15t^2 $ To find when the velocity is 0, we plug in 0 for v and solve for t This gives us $ 0 = 20 - 15t^2 $ so then $ 20 = 15t^2 $ or $ (4/3) = t^2 $ Taking the square root of both sides we find that $ t = \sqrt (4/3)$ and $-\sqrt(4/3)$
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