479 Horizontal Motion: The horizontal component of velocity ... - Wuala
479 Horizontal Motion: The horizontal component of velocity ... - Wuala
479 Horizontal Motion: The horizontal component of velocity ... - Wuala
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91962_05_R1_p0<strong>479</strong>-0512 6/5/09 3:53 PM Page 492<br />
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently<br />
exist. No portion <strong>of</strong> this material may be reproduced, in any form or by any means, without permission in writing from the publisher.<br />
R1–18. At the instant shown, cars A and B travel at speeds<br />
<strong>of</strong> 55 mi>h and 40 mi>h, respectively. If B is increasing its<br />
speed by 1200 mi>h 2 , while A maintains its constant speed,<br />
determine the <strong>velocity</strong> and acceleration <strong>of</strong> B with respect to<br />
A. Car B moves along a curve having a radius <strong>of</strong> curvature<br />
<strong>of</strong> 0.5 mi.<br />
v B 40 mi/h<br />
B<br />
A 30<br />
v A 55 mi/h<br />
v B = -40 cos 30°i + 40 sin 30°j = {-34.64i + 20j} mi>h<br />
v A = {-55i} mi>h<br />
v B>A = v B - v A<br />
= (-34.64i + 20j) - (-55i) = {20.36i + 20j} mi>h<br />
y B>A = 220.36 2 + 20 2 = 28.5 mi>h<br />
u = tan -1 a 20<br />
20.36 b = 44.5° a<br />
Ans.<br />
Ans.<br />
(a B ) n = y2 B<br />
r = 402<br />
0.5 = 3200 mi>h2 (a B ) t = 1200 mi>h 2<br />
a B = (3200 sin 30° - 1200 cos 30°)i + (3200 cos 30° + 1200 sin 30°)j<br />
= {560.77i + 3371.28j} mi>h 2<br />
a A = 0<br />
a B = a A + a B>A<br />
560.77i + 3371.28j = 0 + a B>A<br />
a B>A = {560.77i + 3371.28j} mi>h 2<br />
a B>A = 2(560.77) 2 + (3371.28) 2 = 3418 mi>h 2 = 3.42 A10 3 B mi>h 2<br />
u = tan -1 a 3371.28<br />
560.77 b = 80.6° a<br />
Ans.<br />
Ans.<br />
492