College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 4 - Problems - Page 152: 78

Answer

The x-component of the position vector is $12.0~m$ east The y-component of the position vector is $40.0~m$ north

Work Step by Step

We can find the x-component of the position vector at $t = 2.0~s$: $x = x_0+v_{0x}~t+\frac{1}{2}a_xt^2$ $x = 2.0~m+0+\frac{1}{2}(5.0~m/s^2)(2.0~s)^2$ $x = 12.0~m$ (east) The x-component of the position vector is $12.0~m$ east We can find the y-component of the position vector at $t = 2.0~s$: $y = y_0+v_{0y}~t+\frac{1}{2}a_yt^2$ $y = 0+(20~m/s)(2.0~s)+0$ $y = 40.0~m$ (north) The y-component of the position vector is $40.0~m$ north
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