Answer
$v_B=20.8$ ft/s
Work Step by Step
$s=s_0+v_0t$
$d_{AC}=0+50\cos 60 t$
$v=v_0+a_c t$
$-50 \sin 60 = 50 \sin 60 - 32.2 t$
$t=2.69$ s
$d_{AC}=67.24$ ft
$d_{BC}=\sqrt{30^2+(67.24-20)^2}=55.96$ ft
$v_B=\frac{55.96}{2.69}=20.8$ ft/s
Engineering Mechanics: Statics & Dynamics (14th Edition)
by Hibbeler, Russell C.
Published by
Pearson
ISBN 10:
0133915425
ISBN 13:
978-0-13391-542-6
Chapter 12 - Kinematics of a Particle - Section 12.10 - Relative-Motion of Two Particles using Translating Axes - Problems - Page 104: 232
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