## Physics: Principles with Applications (7th Edition)

Published by Pearson

# Chapter 3 - Kinematics in Two Dimensions; Vectors - Problems: 51

#### Answer

65km/h, 58$^o$ west of north. 65km/h, 32$^o$ south of east.

#### Work Step by Step

Call east the positive x direction and north the positive y direction. Let 1 denote Car 1, 2 denote Car 2, and s the street. The pair “12”, for example, represents Car 1’s motion relative to Car 2, i.e., as seen by Car 2. $$\vec{v_{1s}} = \vec{v_{12}} + \vec{v_{2s}}$$ $$(0, 35 km/h) = \vec{v_{12}} + (55 km/h, 0)$$ Evidently, $\vec{v_{12}} = (-55 km/h, 35 km/h)$. As expected, this is north and west (58$^o$). $\vec{v_{12}} = \sqrt {(-55 km/h)^2 + (35 km/h)^2} = 65.2km/h \approx65km/h.$ The velocity of car 2 relative to car 1 is exactly the opposite of this. $\vec{v_{21}} = (55 km/h, -35 km/h)$ $\vec{v_{21}} = \sqrt {(55 km/h)^2 + (-35 km/h)^2} = 65.2km/h \approx65km/h$ $and$ $32^o south of east$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.