Solution 8.4a
From Mechanics
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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}=0.2\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>. |
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| + | Using <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t\ \</math> at <math>\ \ t=10</math> gives | ||
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| + | Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> | ||
Revision as of 15:48, 13 April 2010
Here we have \displaystyle \mathbf{a}=0.2\mathbf{i}+0.3\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
Using \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}
