7. Lägesvektorer
Förberedande Mekanik
m |
|||
| Rad 8: | Rad 8: | ||
| - | + | == '''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 24: | 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 39: | Rad 35: | ||
'''Solution''' | '''Solution''' | ||
| + | |||
The tables below show the positions of the children at 1 second intervals for | The tables below show the positions of the children at 1 second intervals for | ||
<math>0\le t\le 4</math> | <math>0\le t\le 4</math> | ||
| Rad 77: | 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 100: | 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 113: | 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 \\ | ||
Versionen från 9 februari 2010 kl. 12.14
| 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 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}
