## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 3 - Problems - Page 112: 30

#### Answer

(a) The average speed is $102~km/h$ (b) The average velocity is $90.8~km/h$

#### Work Step by Step

(a) We can find the distance traveled directly west: $(108~km/h)(\frac{1}{3}~h) = 36.0~km$ We can find the distance traveled south of west: $(90.0~km/h)(\frac{1}{6}~h) = 15.0~km$ We can find the average speed for the trip: $average~speed = \frac{distance}{time}$ $average~speed = \frac{51.0~km}{\frac{1}{2}~h}$ $average~speed = 102~km/h$ The average speed is $102~km/h$ (b) We can find the west component of the displacement: $36.0~km+(15.0~km)~cos~60.0^{\circ} = 43.5~km$ We can find the south component of the displacement: $(15.0~km)~sin~60.0^{\circ} = 13.0~km$ We can find the magnitude of the displacement: $\sqrt{(43.5~km)^2+(13.0~km)^2} = 45.4~km$ We can find the average velocity $v_{av}$: $v_{av} = \frac{displacement}{time}$ $v_{av} = \frac{45.4~km}{\frac{1}{2}~h}$ $v_{av} = 90.8~km/h$ The average velocity is $90.8~km/h$

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