Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems: 44

Answer

$r_f = (708\hat{i}-400\hat{j}+156\hat{k})\times 10^3~km$

Work Step by Step

$r = (600\hat{i}-400\hat{j}+200\hat{k})\times 10^3~km$ We can find the final position of each component of $r$ separately. $r_x = r_{x0}+v_{0x}t+\frac{1}{2}a_xt^2$ $r_x = (600\times 10^6~m)+(9500~m/s)(35~min)(60~s/min)+\frac{1}{2}(40~m/s^2)[(35~min)(60~s/min)]^2$ $r_x = 708\times 10^6~m$ There is no component of velocity or acceleration in the y-direction. $r_y = -400\times 10^6~m$ There is a component of acceleration in the z-direction. $r_z = z_0+\frac{1}{2}a_zt^2$ $r_z = (200\times 10^6~m)+\frac{1}{2}(-20~m/s^2)[(35~min)(60~s/min)]^2$ $r_z = 156\times 10^6~m$ We can add the components to find the final position of the spaceship. $r_f = (708\hat{i}-400\hat{j}+156\hat{k})\times 10^3~km$
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