Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.5 Applications of Vectors - 7.5 Exercises - Page 338: 30a

Answer

The tangential velocity is 55.64 mi/s

Work Step by Step

We can express the angle $\theta$ in radians: $\theta = (10.34'')\times \frac{1^{\circ}}{3600''}\times \frac{\pi~rad}{180^{\circ}} = 0.0000501297~rad$ We can find the tangential velocity in units of mi/year: $v_t = \theta~r$ $v_t = (0.0000501297~rad)(35\times 10^{12}~mi/yr)$ $v_t = 1.7545407\times 10^9~mi/yr$ We can convert $v_t$ to units of mi/s: $v_t = (1.7545407\times 10^9~mi/yr)\times \frac{1~yr}{(365)~ (24)~ (3600)~s}$ $v_t = 55.64~mi/s$ The tangential velocity is 55.64 mi/s
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