## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 3 - Radian Measure and the Unit Circle - Section 3.4 Linear and Angular Speed - 3.4 Exercises - Page 132: 50

#### Answer

It takes a point on the front of the train 28.6 seconds to go around this portion of the track.

#### Work Step by Step

We can convert the train's speed to feet per second: $v = 30.0~mph\times \frac{1~hr}{3600~s}\times \frac{5280~ft}{1~mi} = 44~ft/s$ We can convert the central angle to radians: $\theta = (40^{\circ})(\frac{\pi~rad}{180^{\circ}}) = 0.698~rad$ We can find the distance $d$ of the arc length: $d = \theta ~r$ $d = (0.698~rad)(1800~ft)$ $d = 1256.4~ft$ We can find the time it takes the train to complete this distance: $t = \frac{d}{v}$ $t = \frac{1256.4~ft}{44~ft/s}$ $t = 28.6~s$ It takes a point on the front of the train 28.6 seconds to go around this portion of the track.

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