Solution 19.8d
From Mechanics
(Difference between revisions)
(New page: The expression for the position vector <math>\mathbf{r}</math> has been obtained in part b). <math>\mathbf{r}=\left( 40t \right)\mathbf{i}+\left( -2t+400 \right)\mathbf{j}</math>) |
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<math>\mathbf{r}=\left( 40t \right)\mathbf{i}+\left( -2t+400 \right)\mathbf{j}</math> | <math>\mathbf{r}=\left( 40t \right)\mathbf{i}+\left( -2t+400 \right)\mathbf{j}</math> | ||
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| + | If the aeroplane is due east of the origin its position in the north-south direction must be zero, which means the <math>\mathbf{j}</math> component of the position vector must be zero. | ||
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| + | <math>\begin{align} | ||
| + | & -2t+400=0 \\ | ||
| + | & \\ | ||
| + | & t=200\text{ s} \\ | ||
| + | \end{align}</math> | ||
Revision as of 17:13, 11 October 2010
The expression for the position vector \displaystyle \mathbf{r} has been obtained in part b).
\displaystyle \mathbf{r}=\left( 40t \right)\mathbf{i}+\left( -2t+400 \right)\mathbf{j}
If the aeroplane is due east of the origin its position in the north-south direction must be zero, which means the \displaystyle \mathbf{j} component of the position vector must be zero.
\displaystyle \begin{align} & -2t+400=0 \\ & \\ & t=200\text{ s} \\ \end{align}
