Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 3 - Kinematics in Two Dimensions - Check Your Understanding - Page 70: 13

Answer

$\therefore\vec{V_{AB}}$ = $\text{30+40 = 70 m/s}$ $\therefore \vec{V_{AB}}$ = $\text{50+(-20) = 30 m/s}$ $\therefore\vec{V_{AB}}$ = $\text{(-20)+60 = 40 m/s}$ $\therefore\vec{V_{AB}}$ = $\text{(-50)-10 = -60 m/s}$

Work Step by Step

$\vec{V_{AB}}$ = $\vec{V_{A}} - \vec{V_{B}}$ Similarly $\vec{V_{AC}}$ = $\vec{V_{A}} - \vec{V_{C}}$ $\vec{V_{CB}}$ = $\vec{V_{C}} - \vec{V_{B}}$ To find:- $\text{Case a)}$ $\vec{V_{AB}}$=? But $\vec{V_{AB}}$ = $\vec{V_{A}} - \vec{V_{C}} + \vec{V_{C}} - \vec{V_{B}}$ $\vec{V_{AB}}$ = $\vec{V_{AC}}+ \vec{V_{CB}}$ $\vec{V_{AB}}$ = $\text{30+40 = 70 m/s}$ Similarly, $\text{Case b)}$ $\vec{V_{AB}}$=? $\vec{V_{AB}}$ = $\vec{V_{A}} - \vec{V_{C}} + \vec{V_{C}} - \vec{V_{B}}$ $\vec{V_{AB}}$ = $\vec{V_{AC}}+ \vec{V_{CB}}$ $\vec{V_{AB}}$ = $\text{50+(-20) = 30 m/s}$ $\text{Case c)}$ $\vec{V_{CB}}$=? Similarly $\vec{V_{CB}}$ = $\vec{V_{C}} - \vec{V_{A}} + \vec{V_{A}} - \vec{V_{B}}$ $\vec{V_{AB}}$ = $-\vec{V_{AC}}+ \vec{V_{CB}}$ $\vec{V_{AB}}$ = $\text{(-20)+60 = 40 m/s}$ $\text{Case d)}$ $\vec{V_{AC}}$=? Similarly $\vec{V_{AC}}$ = $\vec{V_{A}} - \vec{V_{B}} + \vec{V_{B}} - \vec{V_{C}}$ $\vec{V_{AB}}$ = $\vec{V_{AB}} - \vec{V_{CB}}$ $\vec{V_{AB}}$ = $\text{(-50)-10 = -60 m/s}$
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