## Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson

# Chapter 3 - Exercises and Problems - Page 50: 31

#### Answer

Please see the work below.

#### Work Step by Step

We know that the displacement in your direction is: $x=vt+\frac{1}{2}(acos(35))t^2$ We plug in the known values to obtain: $x=(6.5)(6.3)+\frac{1}{2}(0.48)cos(35)(6.3)^2$ $x=48.75m$ The displacement in the perpendicular direction is: $y=(\frac{1}{2})(a \space sin(35))t^2$ We plug in the known values to obtain: $y=(\frac{1}{2})(0.48)sin(35)(6.3)^2=5.464m$ The magnitude of the displacement is $d=\sqrt{x^2+y^2}$ We plug in the known values to obtain: $d=\sqrt{(48.75)^2+(5.464)^2}=49m$ The direction of your displacement is: $\theta=tan^{-1}\frac{y}{x}$ We plug in the known values to obtain: $\theta=tan^{-1}(\frac{5.464}{48.75})=6.4^{\circ}$

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