Solution 8.8b

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(New page: We must find the position of the particle after 30 s. Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> with <math>\ \mathbf{a}=\mathbf{i}+2\...)
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Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math>
Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math>
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with <math>\ \mathbf{a}=\mathbf{i}+2\mathbf{j} \text{ m}{{\text{s}}^{\text{-2}}}</math>,
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with <math>\ \mathbf{a}=\mathbf{i}+2\mathbf{j} \text{ m}{{\text{s}}^{\text{-2}}}</math> which was calculated in part a),
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<math>\ \mathbf{u}=<math>4\mathbf{i}+6\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math> and
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<math>\ \mathbf{u}=4\mathbf{i}+6\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math> and
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<math>{{\mathbf{r}}_{0}}=0</math> gives,
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<math>{{\ \mathbf{r}}_{0}}=0</math> gives,
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<math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math>
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<math>\mathbf{r}=(4\mathbf{i}+6\mathbf{j}) \times 30+\frac{1}{2}(\mathbf{i}+2\mathbf{j}) \times {{30}^{\ 2}}</math>

Revision as of 15:23, 18 April 2010

We must find the position of the particle after 30 s.

Using \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} with \displaystyle \ \mathbf{a}=\mathbf{i}+2\mathbf{j} \text{ m}{{\text{s}}^{\text{-2}}} which was calculated in part a), \displaystyle \ \mathbf{u}=4\mathbf{i}+6\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}} and \displaystyle {{\ \mathbf{r}}_{0}}=0 gives,

\displaystyle \mathbf{r}=(4\mathbf{i}+6\mathbf{j}) \times 30+\frac{1}{2}(\mathbf{i}+2\mathbf{j}) \times {{30}^{\ 2}}

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