Solution 8.9a

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Current revision (15:18, 19 April 2010) (edit) (undo)
 
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Here we use <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> with
Here we use <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> with
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<math>{{\mathbf{r}}_{0}}=1\textrm{.}5\mathbf{i} \text{ m}</math>
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<math>{{\mathbf{r}}_{0}}=1\textrm{.}5\mathbf{j} \text{ m}</math>
<math>\mathbf{u}=15\mathbf{i}+18\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math>
<math>\mathbf{u}=15\mathbf{i}+18\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math>
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<math>t=3\text{ s}</math>
<math>t=3\text{ s}</math>
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Giving <math>\mathbf{r}=(15\mathbf{i}+18\mathbf{j} ) \times 3+\frac{1}{2}(-10\mathbf{j}) \times {{3}^{\ 2}}+1\textrm{.}5\mathbf{j}</math>
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Simplifying <math>\mathbf{r}=45\mathbf{i}+54\mathbf{j}-45\mathbf{j} +1\textrm{.}5\mathbf{j}=45\mathbf{i}+10\textrm{.}5\mathbf{j} \text{ m} </math>

Current revision

Here we use \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} with

\displaystyle {{\mathbf{r}}_{0}}=1\textrm{.}5\mathbf{j} \text{ m}

\displaystyle \mathbf{u}=15\mathbf{i}+18\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}

\displaystyle \mathbf{a}=-10\mathbf{j}

\displaystyle t=3\text{ s}

Giving \displaystyle \mathbf{r}=(15\mathbf{i}+18\mathbf{j} ) \times 3+\frac{1}{2}(-10\mathbf{j}) \times {{3}^{\ 2}}+1\textrm{.}5\mathbf{j}

Simplifying \displaystyle \mathbf{r}=45\mathbf{i}+54\mathbf{j}-45\mathbf{j} +1\textrm{.}5\mathbf{j}=45\mathbf{i}+10\textrm{.}5\mathbf{j} \text{ m}

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