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

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 481
© 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.
*R1–4. The flight path of a jet aircraft as it takes off is
defined by the parametric equations x = 1.25t 2 and
y = 0.03t 3 , where t is the time after take-off, measured in
seconds, and x and y are given in meters. If the plane starts
to level off at t = 40 s, determine at this instant (a) the
horizontal distance it is from the airport, (b) its altitude,
(c) its speed, and (d) the magnitude of its acceleration.
y
x
Position: When t = 40 s, its horizontal position is given by
x = 1.25A40 2 B = 2000 m = 2.00 km
Ans.
and its altitude is given by
y = 0.03A40 3 B = 1920 m = 1.92 km
Ans.
Velocity: When t = 40 s, the horizontal component of velocity is given by
y x = x # = 2.50t t = 40 s = 100 m>s
The vertical component of velocity is
y y = y # = 0.09t 2 t = 40 s = 144 m>s
Thus, the plane’s speed at t = 40 s is
y y = 2y 2 x + y 2 y = 2100 2 + 144 2 = 175 m>s
Ans.
Acceleration: The horizontal component of acceleration is
a x = x $ = 2.50 m>s 2
and the vertical component of acceleration is
a y = y $ = 0.18t t = 40 s = 7.20 m>s 2
Thus, the magnitude of the plane’s acceleration at t = 40 s is
a = 2a 2 x + a 2 y = 22.50 2 + 7.20 2 = 7.62 m>s 2
Ans.
481