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Revision as of 17:15, 12 April 2010 (edit) Ian (Talk | contribs) ← Previous diff		Current revision (12:16, 27 March 2011) (edit) (undo) Ian (Talk | contribs)
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	In this case we have		In this case we have
	<math>\mathbf{u}=4\mathbf{i}</math>		<math>\mathbf{u}=4\mathbf{i}</math>
-	,	+	and
-	<math>\mathbf{a}=0.9\mathbf{i}+0.7\mathbf{j}</math>	+	<math>\mathbf{a}=0\textrm{.}9\mathbf{i}+0\textrm{.}7\mathbf{j}</math>
-	and	+
-	<math>{{\mathbf{r}}_{0}}=400\mathbf{i}+350\mathbf{j}</math>.	+
-	Substituting these into the equation	+	Substituting these into the equation
-	<math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+400\mathbf{i}+350\mathbf{j}</math>	+	<math>\mathbf{v}=\mathbf{u}+\mathbf{a}t \ \</math>
-	gives the position vector of the ball at time <math>10</math> as:	+	gives the velocity of the jet-ski at time <math>t=10 \text{ s }</math> as:
-		+
-	<math>\begin{align}	+	<math>\mathbf{v}=4\mathbf{i}+(0\textrm{.}9\mathbf{i}+0\textrm{.}7\mathbf{j})\times 10</math>
-	& \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} \\	+	which gives
-	& =(8t)\mathbf{i}+(1\textrm{.}5+2t-5{{t}^{\ 2}})\mathbf{j} \ \text{m}	+
-	\end{align}</math>	+	<math>\mathbf{v}=13\mathbf{i}+7\mathbf{j} \ \text{ m}{{\text{s}}^{\text{-1}}}</math>

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Current revision

In this case we have \displaystyle \mathbf{u}=4\mathbf{i} and \displaystyle \mathbf{a}=0\textrm{.}9\mathbf{i}+0\textrm{.}7\mathbf{j}

Substituting these into the equation \displaystyle \mathbf{v}=\mathbf{u}+\mathbf{a}t \ \ gives the velocity of the jet-ski at time \displaystyle t=10 \text{ s } as:

\displaystyle \mathbf{v}=4\mathbf{i}+(0\textrm{.}9\mathbf{i}+0\textrm{.}7\mathbf{j})\times 10

which gives

\displaystyle \mathbf{v}=13\mathbf{i}+7\mathbf{j} \ \text{ m}{{\text{s}}^{\text{-1}}}