Solution 8.10b

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<math>\sqrt{{{8\textrm{.}4}^{2}}+{{8\textrm{.}8}^{2}}+{{4}^{2}}}=\sqrt{70\textrm{.}56+77\textrm{.}44+16}=\sqrt{164}=12\textrm{.}8\ \text{m}{{\text{s}}^{-1}}</math>
<math>\sqrt{{{8\textrm{.}4}^{2}}+{{8\textrm{.}8}^{2}}+{{4}^{2}}}=\sqrt{70\textrm{.}56+77\textrm{.}44+16}=\sqrt{164}=12\textrm{.}8\ \text{m}{{\text{s}}^{-1}}</math>
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The horisontal speed is,
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<math>\sqrt{{{8\textrm{.}4}^{2}}+{{8\textrm{.}8}^{2}}}=\sqrt{70\textrm{.}56+77\textrm{.}44}=\sqrt{148}=12\textrm{.}2\ \text{m}{{\text{s}}^{-1}}</math>

Revision as of 10:30, 20 April 2010

We use the equation \displaystyle \mathbf{v}=\mathbf{u}+\mathbf{a}t \ where,

\displaystyle \mathbf{a}=(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k})\text{ m}{{\text{s}}^{\text{-2}}}

and from part a)

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

\displaystyle t=40 \ \text{s} .

giving,

\displaystyle \mathbf{v}=8\ \mathbf{i}+8 \ \mathbf{ j}+((0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}))\times 40 =8\textrm{.}4\mathbf{i}+8\textrm{.}8\mathbf{j}+4\mathbf{k}

This is the velocity of the aeroplane. The speed is the magnitude of the velocity,

\displaystyle \sqrt{{{8\textrm{.}4}^{2}}+{{8\textrm{.}8}^{2}}+{{4}^{2}}}=\sqrt{70\textrm{.}56+77\textrm{.}44+16}=\sqrt{164}=12\textrm{.}8\ \text{m}{{\text{s}}^{-1}}

The horisontal speed is, \displaystyle \sqrt{{{8\textrm{.}4}^{2}}+{{8\textrm{.}8}^{2}}}=\sqrt{70\textrm{.}56+77\textrm{.}44}=\sqrt{148}=12\textrm{.}2\ \text{m}{{\text{s}}^{-1}}

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