Answer
$11\ m$
Work Step by Step
$\vec{v}=\vec{b}-\vec{a}$
$=[6.0m-4.0m]\hat{i}+[8.0m-(-3.0m)]\hat{j}$
$=(2.0m)\hat{i}+(11m)\hat{j}$
Using (3-6)
$\vec{a}=a_{x}\hat{\mathrm{i}}+a_{y}\hat{\mathrm{j}} ,$
$a=\sqrt{a_{x}^{2}+a_{y}^{2}}$ and $\displaystyle \tan\theta=\frac{a_{y}}{a_{x}}$ (3-6)
$ v=|\vec{a}-\vec{b}|=\sqrt{(2.0m)^{2}+(11m)^{2}}=11\ m\quad$(2 sig.digits)
