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

Published by Pearson

# Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems - Page 199: 10

#### Answer

The correct banking angle of the road is $7.3^{\circ}$.

#### Work Step by Step

We convert the speed to units of m/s: $v = (90~km/h)(\frac{1000~m}{1~km})(\frac{1~hr}{3600~s})$ $v = 25~m/s$ We then use the equation for the speed around a banked curve to find the banking angle; $v = \sqrt{rg~tan(\theta)}$ $v^2 = rg~tan(\theta)$ $tan(\theta) = \frac{v^2}{rg}$ $\theta = arctan(\frac{v^2}{rg})$ $\theta = arctan(\frac{(25~m/s)^2}{(500~m)(9.80~m/s^2)})$ $\theta = 7.3^{\circ}$ The correct banking angle of the road is $7.3^{\circ}$.

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