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Revision as of 17:10, 12 April 2010 (edit) Ian (Talk | contribs) (New page: In this case we have <math>\mathbf{u}=4\mathbf{i}</math> , <math>\mathbf{a}=0.9\mathbf{i}+0.7\mathbf{j}</math> and <math>{{\mathbf{r}}_{0}}=400\mathbf{i}+350\mathbf{j}</math>. Substit...) ← Previous diff		Revision as of 17:15, 12 April 2010 (edit) (undo) Ian (Talk | contribs) Next diff →
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	Substituting these into the equation		Substituting these into the equation
-	<math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math>	+	<math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+400\mathbf{i}+350\mathbf{j}</math>
	gives the position vector of the ball at time <math>10</math> as:		gives the position vector of the ball at time <math>10</math> as:
	<math>\begin{align}		<math>\begin{align}
-	& \mathbf{r}=(4\mathbf{i})t+\frac{1}{2}(0.9\mathbf{i}+0.7\mathbf{j}){{t}^{\ 2}}+1\textrm{.}5\mathbf{j} \\	+	& \mathbf{r}=(4\mathbf{i})\times 10+\frac{1}{2}(0.9\mathbf{i}+0.7\mathbf{j})\times {{10}^{\ 2}}+400\mathbf{i}+350\mathbf{j} \\
	& =8t\mathbf{i}+2t\mathbf{j}-5{{t}^{\ 2}}\mathbf{j}+1\textrm{.}5\mathbf{j} \\		& =8t\mathbf{i}+2t\mathbf{j}-5{{t}^{\ 2}}\mathbf{j}+1\textrm{.}5\mathbf{j} \\
	& =(8t)\mathbf{i}+(1\textrm{.}5+2t-5{{t}^{\ 2}})\mathbf{j} \ \text{m}		& =(8t)\mathbf{i}+(1\textrm{.}5+2t-5{{t}^{\ 2}})\mathbf{j} \ \text{m}
	\end{align}</math>		\end{align}</math>

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Revision as of 17:15, 12 April 2010

In this case we have \displaystyle \mathbf{u}=4\mathbf{i} , \displaystyle \mathbf{a}=0.9\mathbf{i}+0.7\mathbf{j} and \displaystyle {{\mathbf{r}}_{0}}=400\mathbf{i}+350\mathbf{j}.

Substituting these into the equation \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+400\mathbf{i}+350\mathbf{j} gives the position vector of the ball at time \displaystyle 10 as:

\displaystyle \begin{align} & \mathbf{r}=(4\mathbf{i})\times 10+\frac{1}{2}(0.9\mathbf{i}+0.7\mathbf{j})\times {{10}^{\ 2}}+400\mathbf{i}+350\mathbf{j} \\ & =8t\mathbf{i}+2t\mathbf{j}-5{{t}^{\ 2}}\mathbf{j}+1\textrm{.}5\mathbf{j} \\ & =(8t)\mathbf{i}+(1\textrm{.}5+2t-5{{t}^{\ 2}})\mathbf{j} \ \text{m} \end{align}