Solution 8.9b
From Mechanics
We use \displaystyle \mathbf{v}=\mathbf{u}+\mathbf{a}t \ with
\displaystyle \mathbf{u}=15\mathbf{i}+18\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}
\displaystyle \mathbf{a}=-10\mathbf{j}
\displaystyle t=3\text{ s}
giving \displaystyle \mathbf{v}=15\mathbf{i}+18\mathbf{j}+(-10\mathbf{j} )\times 3
Simplifying \displaystyle \mathbf{v}=15\mathbf{i}+18\mathbf{j}-30\mathbf{j}=15\mathbf{i}-12\mathbf{j}
The speed \displaystyle S is the magnitude of this velocity.
\displaystyle S=\sqrt{{{15}^{2}}+{{\left( -12 \right)}^{2}}}=\sqrt{225+144}=19\textrm{.}2\ \text{m}{{\text{s}}^{-1}}