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

Published by Pearson

# Chapter 2 - Kinematics in One Dimension - Exercises and Problems: 15

#### Answer

(a) $t = 354 ~days$ (b) $x = 4.59\times 10^{15}~m$ (c) $x = 0.485~ly$

#### Work Step by Step

(a) $t = \frac{v-v_0}{a}$ $t = \frac{3.0\times 10^8~m/s-0}{9.80~m/s^2}$ $t = 3.0612\times 10^7~s$ We can convert this time to days: $t = (3.0612\times 10^7~s)(\frac{1~day}{24\times 3600~s})$ $t = 354 ~days$ (b) $x = (\frac{v+v_0}{2})(t)$ $x = (\frac{3.0\times 10^8~m/s}{2})(3.0612\times 10^7~s)$ $x = 4.59\times 10^{15}~m$ (c) We can find the number of meters in one light year: $1~ly = (3.0\times 10^8~m/s)(365\times 24\times 3600~s)$ $1~ly = 9.46\times 10^{15}~m$ We can express the answer in part (b) as a fraction of a light year; $x = \frac{4.59\times 10^{15}~m}{9.46\times 10^{15}~m}$ $x = 0.485~ly$

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