Solution 8.10a
From Mechanics
(New page: We use, <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> to find an expression for the position of the aeroplane at a time t. Here, <math>{{\mathbf{...) |
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<math>\sqrt{{{B}^{2}}+{{B}^{2}}}=\sqrt{2}B</math> | <math>\sqrt{{{B}^{2}}+{{B}^{2}}}=\sqrt{2}B</math> | ||
| - | However the magnitude of the initial velocity is the initial | + | However the magnitude of the initial velocity is the initial speed which is |
| - | <math>8\sqrt{2}</math>. | + | <math>8\sqrt{2}\text{ m}{{\text{s}}^{\text{-1}}}</math>. Comparing these results shows <math>B=8</math> and thus |
| + | |||
| + | <math>\mathbf{u}=8\ \mathbf{i}+8 \ \mathbf{ j}\text{ m}{{\text{s}}^{\text{-1}}}</math> | ||
Revision as of 16:28, 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}=(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k})\text{ m}{{\text{s}}^{\text{-2}}}
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
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}}}
