## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 3 - Vectors and Coordinate Systems - Exercises and Problems - Page 83: 30

#### Answer

(a) The displacement vector is $(1.73~\hat{i}-3.0~\hat{j})~cm$ (b) The displacement is zero.

#### Work Step by Step

(a) At 8:00, the minute hand points straight up. At 8:20, the minute hand makes a $30^{\circ}$ angle below the positive x-axis. We can find the x-component and the y-component of the displacement as: $\Delta x = (2.0~cm)~cos(30^{\circ})$ $\Delta x = 1.73~cm$ $\Delta y = -(2.0~cm)~sin(30^{\circ})-(2.0~cm)$ $\Delta y = -3.0~cm$ The displacement vector is $(1.73~\hat{i}-3.0~\hat{j})~cm$ (b) At 8:00 and 9:00, the minute hand points straight up, so the displacement is zero.

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