Solution 8.6b

From Mechanics

Revision as of 19:42, 15 April 2010

We assume the starting point is the origin. This means \displaystyle {{\mathbf{r}}_{0}}=0

The boat first accelerates to a point \displaystyle A say. We first must calculate the position of this point \displaystyle {{\mathbf{r}}_{A}}.

Using \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} with,

\displaystyle t=10, \displaystyle \mathbf{u}=\mathbf{i}+2\mathbf{j} and from part a) \displaystyle \mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}\ \text{m}{{\text{s}}^{-2}}

\displaystyle {{\mathbf{r}}_{A}}=(\mathbf{i}+2\mathbf{j}) \times 10+\frac{1}{2}(0\textrm{.}5\mathbf{i}-\mathbf{j}) \times {{10}^{\ 2}}+0=35\mathbf{i}-50\mathbf{j}

The next stage has

\displaystyle t=40, \displaystyle \mathbf{a}=0, \displaystyle {{\mathbf{r}}_{0}}=35\mathbf{i}+70\mathbf{j}, and \displaystyle \mathbf{u}=6\mathbf{i}-8\mathbf{j} as the final position and velocity of the first stage is the initial position and velocity of the second stage.

Using once again \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} we get

\displaystyle \mathbf{r}=(6\mathbf{i}-8\mathbf{j}) \times 40+0+(35\mathbf{i}-50\mathbf{j})=275\mathbf{i}-355\mathbf{j}

This is the boat´s final position.

The distance from the starting point is the magnitude of this vector,

\displaystyle \sqrt{{{275}^{2}}+{{\left( -355 \right)}^{2}}}=450\ \text{m}