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: 18c

Answer

$ -12 m/s^2 $

Work Step by Step

The position is given by $ x = 12t^2 - 2t^3 $ Since the velocity equation is the derivative of the position equation we need to find $\frac{d}{dt} (12t^2 - 2t^3) $ Since $\frac{d}{dt} t^n = n*t^(n-1)$ then $\frac{d}{dt} (12t^2 - 2t^3) = 12(2)t - 2(3)t^2 = 24t - 6t^2 $ Therefore $v = 24t - 6t^2$ The acceleration is given by the derivative of the velocity equation This is $ \frac{d}{dt} (24t - 6t^2) = 24 - 6(2)t = 24 - 12t $ Therefore, to find acceleration at t =3, we plug in t = 3 into the equation This gives us $ 24 -12(3) = 24-36 = -12 $ Therefore the acceleration is $-12m/s^2$
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