Ian at 15:20, 14 April 2010

2010-04-14T15:20:46Z

←Older revision		Revision as of 15:20, 14 April 2010
Line 21:		Line 21:

	<math>t=5\ \text{s}</math>		<math>t=5\ \text{s}</math>
		+
		+	At this time the velocity is
		+
		+	<math>\mathbf{v}=(80+5 )\mathbf{i}+4 \times 5\ \mathbf{j}=85\mathbf{i}+20\mathbf{j}</math>
		+
		+	The speed is the magnitude of this velocity
		+
		+
		+	<math>\sqrt{{{85}^{2}}+{{20}^{2}}}=87\textrm{.}3\ \text{m}{{\text{s}}^{-1}}</math>

Ian: New page: From part a) we have $\mathbf{v}=(80+t\ )\mathbf{i}+4t\ \mathbf{j}$ . and $\mathbf{r}=(80t+\frac{1}{2}{t}^{\ 2}\ )\mathbf{i}+2{t}^{\ 2}\ \mathbf{j}$ We see the ...

2010-04-14T15:13:41Z

New page: From part a) we have <math>\mathbf{v}=(80+t\ )\mathbf{i}+4t\ \mathbf{j}</math>. and <math>\mathbf{r}=(80t+\frac{1}{2}{t}^{\ 2}\ )\mathbf{i}+2{t}^{\ 2}\ \mathbf{j}</math> We see the ...

New page

From part a) we have

<math>\mathbf{v}=(80+t\ )\mathbf{i}+4t\ \mathbf{j}</math>.

and

<math>\mathbf{r}=(80t+\frac{1}{2}{t}^{\ 2}\ )\mathbf{i}+2{t}^{\ 2}\ \mathbf{j}</math>

We see the height is

Thus the time when the aeroplane is at a height <math>50 \text{ m} </math>.

We see the height is <math>2{t}^{\ 2}</math> for any time <math>t</math>.

Thus the time when the aeroplane is at a height <math>50 \text{ m} </math> is given by

<math>2{{t}^{\ 2}}=50</math>

or

<math>t=5\ \text{s}</math>

Solution 8.5b - Revision history

Ian at 15:20, 14 April 2010

Ian: New page: From part a) we have \mathbf{v}=(80+t\ )\mathbf{i}+4t\ \mathbf{j}. and \mathbf{r}=(80t+\frac{1}{2}{t}^{\ 2}\ )\mathbf{i}+2{t}^{\ 2}\ \mathbf{j} We see the ...

Ian: New page: From part a) we have $\mathbf{v}=(80+t\ )\mathbf{i}+4t\ \mathbf{j}$ . and $\mathbf{r}=(80t+\frac{1}{2}{t}^{\ 2}\ )\mathbf{i}+2{t}^{\ 2}\ \mathbf{j}$ We see the ...