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Revision as of 15:09, 18 April 2010 (edit) Ian (Talk | contribs) ← Previous diff		Current revision (15:18, 19 April 2010) (edit) (undo) Ian (Talk | contribs)
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-	Here we use the equation <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t</math> with	+	Here we use <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> with
-	<math>\mathbf{u}=<math>4\mathbf{i}+6\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math> and <math>\mathbf{v}=24\mathbf{i}+46\mathbf{j\text{ m}{{\text{s}}^{\text{-1}}}}</math>.	+	<math>{{\mathbf{r}}_{0}}=1\textrm{.}5\mathbf{j} \text{ m}</math>
-	Also <math>t=20\text{ s}</math> whichh gives,	+	<math>\mathbf{u}=15\mathbf{i}+18\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math>
-	<math>24\mathbf{i}+46\mathbf{j}=4\mathbf{i}+6\mathbf{j}+\mathbf{a} \times 20</math>	+	<math>\mathbf{a}=-10\mathbf{j}</math>
-	So,	+	<math>t=3\text{ s}</math>
-	<math>20\mathbf{a}=20\mathbf{i}+40\mathbf{j}</math>	+	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>
-	giving,	+
-	<math>\mathbf{a}=\mathbf{i}+2\mathbf{j} \text{ m}{{\text{s}}^{\text{-2}}}</math>	+	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>

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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}