7. Position Vectors
From Mechanics
| Theory | Exercises |
Key Points
Velocities of particles can be expressed as vectors,
\displaystyle \mathbf{v}=V\cos \alpha \mathbf{i}+V\sin \alpha \mathbf{j}
A particle when moving with a constant velocity \displaystyle \mathbf{u} has a position vector,
\displaystyle \mathbf{r}=\mathbf{u}t+{{\mathbf{r}}_{0}}
where \displaystyle t is the time and \displaystyle {{\mathbf{r}}_{0}} is the position vector of the particle at the start.
If \displaystyle \mathbf{u}=b\mathbf{i}+c\mathbf{j}
and
\displaystyle {{\mathbf{r}}_{0}}=p\mathbf{i}+q\mathbf{j}
where \displaystyle b, c, p, q are numbers,
then this equation may be rewritten as
\displaystyle \mathbf{r}=\left( bt+p \right)\mathbf{i}+\left( ct+q \right)\mathbf{j}
This is another way of writing the equation for a particle with constant velocity.
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}
