Answer
displacement $10$ ft
distance $26$ ft
Work Step by Step
displacement
= $\int_0^5(12-4t)dt$
= $(12t-2t^{2})|_0^5$
= $10$ ft
distance
= $\int_0^3(12-4t)dt$ + $\int_3^5(12-4t)dt$
= $(12t-2t^{2})|_0^3$ + $(12t-2t^{2})|_3^5$
= $26$ ft
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 5 - The Integral - 5.6 Net Change as the Integral of a Rate of Change - Exercises - Page 268: 9
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