Answer
$$s = 5\sqrt {13} $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = 2t{\bf{i}} - 3t{\bf{j}},{\text{ }}\left[ {0,5} \right] \cr
& {\text{Differentiate }}{\bf{r}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {2t{\bf{i}} - 3t{\bf{j}}} \right] \cr
& {\bf{r}}'\left( t \right) = 2{\bf{i}} - 3{\bf{j}} \cr
& {\text{Find }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 3} \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {13} \cr
& {\text{Find the arc lenght }}s{\text{, using the formula }}s = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& s = \int_0^5 {\sqrt {13} } dt \cr
& {\text{Integrate}} \cr
& s = \left[ {\sqrt {13} t} \right]_0^5 \cr
& s = 5\sqrt {13} \cr
& \cr
& {\text{Graph}} \cr} $$
