7. Lägesvektorer
Förberedande Mekanik
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m (flyttade 7. Position Vectors till 7. Lägesvektorer) |
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| (5 mellanliggande versioner visas inte.) | |||
| Rad 1: | Rad 1: | ||
| - | 7. Position Vectors | + | __NOTOC__ |
| + | {| border="0" cellspacing="0" cellpadding="0" height="30" width="100%" | ||
| + | | style="border-bottom:1px solid #797979" width="5px" | | ||
| + | {{Selected tab|[[7. Position Vectors|Theory]]}} | ||
| + | {{Not selected tab|[[7. Exercises|Exercises]]}} | ||
| + | | style="border-bottom:1px solid #797979" width="100%"| | ||
| + | |} | ||
| - | Key Points | + | |
| + | == '''Key Points''' == | ||
Velocities can be expressed as vectors, for example | Velocities can be expressed as vectors, for example | ||
<math>\mathbf{v}=V\cos \alpha \mathbf{i}+V\sin \alpha \mathbf{j}</math> | <math>\mathbf{v}=V\cos \alpha \mathbf{i}+V\sin \alpha \mathbf{j}</math> | ||
| - | |||
When moving with a constant velocity | When moving with a constant velocity | ||
| Rad 15: | Rad 21: | ||
'''[[Example 7.1]]''' | '''[[Example 7.1]]''' | ||
| - | Two children, A and B, start at the origin and run so that their position vectors in metres at time t seconds are given by: | + | |
| + | Two children, A and B, start at the origin and run so that their position vectors in metres at time <math>t</math> seconds are given by: | ||
| - | |||
<math>{{\mathbf{r}}_{A}}=2t\mathbf{i}+t\mathbf{j}</math> | <math>{{\mathbf{r}}_{A}}=2t\mathbf{i}+t\mathbf{j}</math> | ||
and | and | ||
<math>{{\mathbf{r}}_{B}}=(6t-{{t}^{2}})\mathbf{i}+t\mathbf{j}</math> | <math>{{\mathbf{r}}_{B}}=(6t-{{t}^{2}})\mathbf{i}+t\mathbf{j}</math> | ||
| - | |||
Plot the paths of the two children for | Plot the paths of the two children for | ||
| Rad 30: | Rad 35: | ||
'''Solution''' | '''Solution''' | ||
| - | The tables below show the positions of the children at 1 second intervals for | + | |
| + | The tables below show the positions in metres of the children at 1 second intervals for | ||
<math>0\le t\le 4</math> | <math>0\le t\le 4</math> | ||
. | . | ||
| - | + | {| width="100%" cellspacing="10px" align="center" | |
| - | Time Position of A ( | + | |align="left"| Time |
| - | 0 | + | | valign="top"|Position of A (<math>{{\mathbf{r}}_{A}}</math>) |
| - | <math>{{\mathbf{r}}_{A}}=0\mathbf{i}+0\mathbf{j}</math> | + | | valign="top"|Position of B (<math>{{\mathbf{r}}_{B}}</math>) |
| - | + | |- | |
| - | <math>{{\mathbf{r}}_{B}}=(6\times 0-{{0}^{2}})\mathbf{i}+0\mathbf{j}=0\mathbf{i}+0\mathbf{j}</math> | + | |0 |
| - | + | | valign="top"| <math>{{\mathbf{r}}_{A}}=0\mathbf{i}+0\mathbf{j}</math> | |
| - | 1 | + | | valign="top"|<math>{{\mathbf{r}}_{B}}=(6\times 0-{{0}^{2}})\mathbf{i}+0\mathbf{j}=0\mathbf{i}+0\mathbf{j}</math> |
| - | <math>{{\mathbf{r}}_{A}}=2\mathbf{i}+1\mathbf{j}</math> | + | |- |
| - | + | | 1 | |
| - | <math>{{\mathbf{r}}_{B}}=(6\times 1-{{1}^{2}})\mathbf{i}+1\mathbf{j}=5\mathbf{i}+1\mathbf{j}</math> | + | |valign="top"| <math>{{\mathbf{r}}_{A}}=2\mathbf{i}+1\mathbf{j}</math> |
| - | + | |valign="top"| <math>{{\mathbf{r}}_{B}}=(6\times 1-{{1}^{2}})\mathbf{i}+1\mathbf{j}=5\mathbf{i}+1\mathbf{j}</math> | |
| - | 2 | + | |- |
| - | <math>{{\mathbf{r}}_{A}}=4\mathbf{i}+2\mathbf{j}</math> | + | | 2 |
| - | + | |valign="top"|<math>{{\mathbf{r}}_{A}}=4\mathbf{i}+2\mathbf{j}</math> | |
| - | <math>{{\mathbf{r}}_{B}}=(6\times 2-{{2}^{2}})\mathbf{i}+2\mathbf{j}=8\mathbf{i}+2\mathbf{j}</math> | + | |valign="top"|<math>{{\mathbf{r}}_{B}}=(6\times 2-{{2}^{2}})\mathbf{i}+2\mathbf{j}=8\mathbf{i}+2\mathbf{j}</math> |
| - | + | |- | |
| - | 3 | + | |3 |
| - | <math>{{\mathbf{r}}_{A}}=6\mathbf{i}+3\mathbf{j}</math> | + | |valign="top"|<math>{{\mathbf{r}}_{A}}=6\mathbf{i}+3\mathbf{j}</math> |
| - | + | |valign="top"|<math>{{\mathbf{r}}_{B}}=(6\times 3-{{3}^{2}})\mathbf{i}+3\mathbf{j}=9\mathbf{i}+3\mathbf{j}</math> | |
| - | <math>{{\mathbf{r}}_{B}}=(6\times 3-{{3}^{2}})\mathbf{i}+3\mathbf{j}=9\mathbf{i}+3\mathbf{j}</math> | + | |- |
| - | + | |4 | |
| - | 4 | + | |valign="top"|<math>{{\mathbf{r}}_{A}}=8\mathbf{i}+4\mathbf{j}</math> |
| - | <math>{{\mathbf{r}}_{A}}=8\mathbf{i}+4\mathbf{j}</math> | + | |valign="top"|<math>{{\mathbf{r}}_{B}}=(6\times 4-{{4}^{2}})\mathbf{i}+4\mathbf{j}=8\mathbf{i}+4\mathbf{j}</math> |
| - | + | |} | |
| - | <math>{{\mathbf{r}}_{B}}=(6\times 4-{{4}^{2}})\mathbf{i}+4\mathbf{j}=8\mathbf{i}+4\mathbf{j}</math> | + | |
| Rad 69: | Rad 74: | ||
'''[[Example 7.2]]''' | '''[[Example 7.2]]''' | ||
| + | |||
A boat moves so that at time t its position vector is r, where | A boat moves so that at time t its position vector is r, where | ||
<math>\mathbf{r}=(9t-180)\mathbf{i}+(5t-450)\mathbf{j}</math> | <math>\mathbf{r}=(9t-180)\mathbf{i}+(5t-450)\mathbf{j}</math> | ||
| - | and i and j are unit vectors directed due east and north respectively. | + | and <math>\mathbf{i}</math> and <math>\mathbf{j}</math> are unit vectors directed due east and north respectively. |
| - | + | a) Find the time when the boat is due east of the origin. | |
| - | + | ||
| - | + | b) Find the time when it is due south of the origin. | |
| + | |||
| + | c) Find the position of the boat when it is south east of the origin. | ||
'''Solution''' | '''Solution''' | ||
| - | + | ||
| + | a) When the boat is due east of the origin the position vector will contain only | ||
<math>\mathbf{i}</math> | <math>\mathbf{i}</math> | ||
terms and no | terms and no | ||
<math>\mathbf{j}</math> | <math>\mathbf{j}</math> | ||
terms. | terms. | ||
| - | + | ||
| - | + | ||
<math>\begin{align} | <math>\begin{align} | ||
& 5t-450=0 \\ | & 5t-450=0 \\ | ||
| Rad 92: | Rad 100: | ||
\end{align}</math> | \end{align}</math> | ||
| - | + | b) When the boat is due south of the origin the position vector will contain only | |
| - | + | ||
<math>\mathbf{j}</math> | <math>\mathbf{j}</math> | ||
terms and no | terms and no | ||
<math>\mathbf{i}</math> | <math>\mathbf{i}</math> | ||
terms. | terms. | ||
| - | |||
<math>\begin{align} | <math>\begin{align} | ||
| Rad 105: | Rad 111: | ||
\end{align}</math> | \end{align}</math> | ||
| - | + | c) When the boat is south east of the origin, its position vector will be of the form | |
| - | + | ||
<math>k\mathbf{i}-k\mathbf{j}</math> | <math>k\mathbf{i}-k\mathbf{j}</math> | ||
. | . | ||
| - | + | ||
| - | + | ||
<math>\begin{align} | <math>\begin{align} | ||
& -(9t-180)=5t-450 \\ | & -(9t-180)=5t-450 \\ | ||
Nuvarande version
| Theory | Exercises |
Key Points
Velocities can be expressed as vectors, for example
\displaystyle \mathbf{v}=V\cos \alpha \mathbf{i}+V\sin \alpha \mathbf{j}
When moving with a constant velocity
\displaystyle \mathbf{r}=\mathbf{u}t+{{\mathbf{r}}_{0}}
Two children, A and B, start at the origin and run so that their position vectors in metres at time \displaystyle t seconds are given by:
\displaystyle {{\mathbf{r}}_{A}}=2t\mathbf{i}+t\mathbf{j} and \displaystyle {{\mathbf{r}}_{B}}=(6t-{{t}^{2}})\mathbf{i}+t\mathbf{j}
Plot the paths of the two children for \displaystyle 0\le t\le 4 . What happens when \displaystyle t=4 ?
Solution
The tables below show the positions in metres of the children at 1 second intervals for \displaystyle 0\le t\le 4 .
| Time | Position of A (\displaystyle {{\mathbf{r}}_{A}}) | Position of B (\displaystyle {{\mathbf{r}}_{B}}) |
| 0 | \displaystyle {{\mathbf{r}}_{A}}=0\mathbf{i}+0\mathbf{j} | \displaystyle {{\mathbf{r}}_{B}}=(6\times 0-{{0}^{2}})\mathbf{i}+0\mathbf{j}=0\mathbf{i}+0\mathbf{j} |
| 1 | \displaystyle {{\mathbf{r}}_{A}}=2\mathbf{i}+1\mathbf{j} | \displaystyle {{\mathbf{r}}_{B}}=(6\times 1-{{1}^{2}})\mathbf{i}+1\mathbf{j}=5\mathbf{i}+1\mathbf{j} |
| 2 | \displaystyle {{\mathbf{r}}_{A}}=4\mathbf{i}+2\mathbf{j} | \displaystyle {{\mathbf{r}}_{B}}=(6\times 2-{{2}^{2}})\mathbf{i}+2\mathbf{j}=8\mathbf{i}+2\mathbf{j} |
| 3 | \displaystyle {{\mathbf{r}}_{A}}=6\mathbf{i}+3\mathbf{j} | \displaystyle {{\mathbf{r}}_{B}}=(6\times 3-{{3}^{2}})\mathbf{i}+3\mathbf{j}=9\mathbf{i}+3\mathbf{j} |
| 4 | \displaystyle {{\mathbf{r}}_{A}}=8\mathbf{i}+4\mathbf{j} | \displaystyle {{\mathbf{r}}_{B}}=(6\times 4-{{4}^{2}})\mathbf{i}+4\mathbf{j}=8\mathbf{i}+4\mathbf{j} |
From the values that we have obtained it is clear that the two children will have the same position when
\displaystyle t=4
and unless they take evasive action will collide. The paths are shown in the diagram below.
A boat moves so that at time t its position vector is r, where
\displaystyle \mathbf{r}=(9t-180)\mathbf{i}+(5t-450)\mathbf{j}
and \displaystyle \mathbf{i} and \displaystyle \mathbf{j} are unit vectors directed due east and north respectively.
a) Find the time when the boat is due east of the origin.
b) Find the time when it is due south of the origin.
c) Find the position of the boat when it is south east of the origin.
Solution
a) When the boat is due east of the origin the position vector will contain only \displaystyle \mathbf{i} terms and no \displaystyle \mathbf{j} terms.
\displaystyle \begin{align} & 5t-450=0 \\ & t=\frac{450}{5}=90\text{ s} \\ \end{align}
b) When the boat is due south of the origin the position vector will contain only \displaystyle \mathbf{j} terms and no \displaystyle \mathbf{i} terms.
\displaystyle \begin{align} & 9t-180=0 \\ & t=\frac{180}{9}=20\text{ s} \\ \end{align}
c) When the boat is south east of the origin, its position vector will be of the form \displaystyle k\mathbf{i}-k\mathbf{j} .
\displaystyle \begin{align} & -(9t-180)=5t-450 \\ & -9t+180=5t-450 \\ & 630=14t \\ & t=\frac{630}{14}=45\text{ s} \\ \end{align}
