College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 2 - Kinematics: Description of Motion - Learning Path Questions and Exercises - Exercises - Page 62: 35

Answer

a). Graph is shown, b). 10 m/s c). 237.5m d). 16.96 m/s

Work Step by Step

a). graph is as shown. b). $v=u+at$ $v=25-5\times 3=25-15=10 m/s$ c). in first 5s, s=$25 \times 5= 125m$ in second 3s, s=$ut+\frac{1}{2}at^{2}=25\times3-0.5\times 5\times 3^{2}=52.5m$ for the next 6s, s=$6\times 10=60m$ So, total displacement = $125+52.5+60=237.5m$ d). Average speed = $\frac{total\, distance}{total\, time}=\frac{237.5}{14}=16.96 m/s$
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