479 Horizontal Motion: The horizontal component of velocity ... - Wuala

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 488
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
R1–13. The position of a particle is defined by
r = 551cos 2t2i + 41sin 2t2j6 m, where t is in seconds and
the arguments for the sine and cosine are given in radians.
Determine the magnitudes of the velocity and acceleration
of the particle when t = 1 s. Also, prove that the path of the
particle is elliptical.
Velocity: The velocity expressed in Cartesian vector form can be obtained by
applying Eq. 12–7.
v = dr
dt
= {-10 sin 2ri + 8 cos 2rj} m>s
When t = 1 s, v = -10 sin 2(1)i + 8 cos 2(1) j = (-9.093i - 3.329j} m>s. Thus, the
magnitude of the velocity is
y = 2y 2 x + y 2 y = 2(-9.093) 2 + (-3.329) 2 = 9.68 m>s
Ans.
Acceleration: The acceleration express in Cartesian vector form can be obtained by
applying Eq. 12–9.
a = dv
dt
= {-20 cos 2ri - 16 sin 2rj} m>s2
When t = 1 s, a = -20 cos 2(1) i - 16 sin 2(1) j = {8.323i - 14.549j} m>s 2 . Thus,
the magnitude of the acceleration is
a = 2a 2 x + a 2 y = 28.323 2 + (-14.549) 2 = 16.8 m>s 2
Ans.
Travelling Path: Here, x = 5 cos 2t and y = 4 sin 2t. Then,
x 2
25 = cos2 2t
y 2
16 = sin2 2t
[1]
[2]
Adding Eqs [1] and [2] yields
x 2
25 + y2
16 = cos2 2r + sin 2 2t
However, cos 2 2 r + sin 2 2t = 1. Thus,
x 2
25 + y2
= 1 (Equation of an Ellipse)
16
(Q.E.D.)
488