Answer
$| \vec{a}_{avg}|=0.471 m/s^{2}$
Work Step by Step
We are given (see part (a)):
$\vec{v}_{1}=(-10.0m/s)\hat{j}\qquad \vec{v_{2}}=(10.0m/s)\hat{i},\ \quad t=30.0s.$
Eq. 4-18 gives
$ \displaystyle \vec{a}_{avg}=\frac{\vec{v}_{2}-\vec{v}_{1}}{\Delta t}=\frac{\Delta\vec{v}}{\Delta t}$
$=\displaystyle \frac{(10m/s)\hat{ i}-(-10m/s)\hat{j}}{30.0s}$
$=(0.333m/s^{2})\hat{ i}+(0.333m/s^{2})\hat{j}$
$| \vec{a}_{avg}|=\sqrt{(0.333m/s^{2})^{2}+(0.333m/s^{2})^{2}}=0.471 m/s^{2}$
