Solution 8.4a
From Mechanics
(Difference between revisions)
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| - | Here we have <math>\mathbf{a}=0.2\mathbf{i}+0.3\mathbf{j},\ \</math> <math>\mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \</math> and <math>\ \ \mathbf{ r}_{0}=6\mathbf{i}+2\mathbf{j}</math>. | + | Here we have <math>\mathbf{a}=0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j},\ \</math> <math>\mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \</math> and <math>\ \ \mathbf{ r}_{0}=6\mathbf{i}+2\mathbf{j}</math>. |
Using <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t\ \</math> at <math>\ \ t=10</math> gives | Using <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t\ \</math> at <math>\ \ t=10</math> gives | ||
| - | <math>\mathbf{v}=4\mathbf{i}– 8\mathbf{j}+(0.2\mathbf{i}+0.3\mathbf{j})\times 10\ \</math> or | + | <math>\mathbf{v}=4\mathbf{i}– 8\mathbf{j}+(0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j})\times 10\ \</math> or |
| - | <math>\mathbf{v}=6\mathbf{i}– 5\mathbf{j}</math> | + | <math>\mathbf{v}=6\mathbf{i}– 5\mathbf{j}\ \ \text{m}{{\text{s}}^{-1}}</math> |
| + | The speed is the magnitude of this velocity vector. | ||
| - | + | <math>v=\sqrt{{{6}^{2}}+{{\left( -5 \right)}^{2}}}=\sqrt{61}=7\textrm{.}8\ \text{m}{{\text{s}}^{-1}}</math> | |
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Revision as of 16:05, 13 April 2010
Here we have \displaystyle \mathbf{a}=0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j},\ \ \displaystyle \mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \ and \displaystyle \ \ \mathbf{ r}_{0}=6\mathbf{i}+2\mathbf{j}.
Using \displaystyle \mathbf{v}=\mathbf{u}+\mathbf{a}t\ \ at \displaystyle \ \ t=10 gives
\displaystyle \mathbf{v}=4\mathbf{i}– 8\mathbf{j}+(0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j})\times 10\ \ or
\displaystyle \mathbf{v}=6\mathbf{i}– 5\mathbf{j}\ \ \text{m}{{\text{s}}^{-1}}
The speed is the magnitude of this velocity vector.
\displaystyle v=\sqrt{{{6}^{2}}+{{\left( -5 \right)}^{2}}}=\sqrt{61}=7\textrm{.}8\ \text{m}{{\text{s}}^{-1}}
