Solution 8.10a

From Mechanics

Revision as of 17:42, 19 April 2010

We use, \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} to find an expression for the position of the aeroplane at a time t. Here,

\displaystyle {{\mathbf{r}}_{0}}=10\mathbf{k}\text{ m}

\displaystyle \mathbf{a}=c\text{ m}{{\text{s}}^{\text{-2}}}

We need to calculate \displaystyle \mathbf{u}.

If a vector, like the starting velocity \displaystyle \mathbf{u} in this problem, points north east this means it has the same component in the east and north directions. In other words its \displaystyle \mathbf{i} component and its \displaystyle \mathbf{j} component are the same. Thus \displaystyle \mathbf{u} is of the form

\displaystyle \mathbf{u}=B\ \mathbf{i}+B \ \mathbf{ j}

This has magnitude \displaystyle \sqrt{{{B}^{2}}+{{B}^{2}}}=\sqrt{2}B \text{ m}{{\text{s}}^{\text{-1}}}

However the magnitude of the initial velocity is the initial speed which is

\displaystyle 8\sqrt{2}\text{ m}{{\text{s}}^{\text{-1}}}. Comparing these results shows \displaystyle B=8 and thus

\displaystyle \mathbf{u}=8\ \mathbf{i}+8 \ \mathbf{ j}\text{ m}{{\text{s}}^{\text{-1}}}

We can now substitute in the equation \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} giving,

\displaystyle \mathbf{r}=(8\ \mathbf{i}+8 \ \mathbf{ j})t+\frac{1}{2}(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}){{t}^{\ 2}}+10\mathbf{k}