Question
HLPaper 1
The position of a helicopter relative to a communications tower at the top of a mountain at time $t$ (hours) can be described by the vector equation below.
$$ \mathbf{r} = \begin{pmatrix} 20 \\ -25 \\ 0 \end{pmatrix} + t \begin{pmatrix} 4.2 \\ 5.8 \\ -0.5 \end{pmatrix} $$
The entries in the column vector give the displacements east and north from the communications tower and above/below the top of the mountain respectively, all measured in kilometres.
1.
Find the speed of the helicopter.
[2]Verified
Solution
- Attempt to find the magnitude of the velocity vector: $|\mathbf{v}| = \sqrt{4.2^2 + 5.8^2 + (-0.5)^2}$ M1
- $7.18 \ (7.1784\dots) \ \text{km\,h}^{-1}$ A1
2 marks total
2.
Find the distance of the helicopter from the communications tower at $t=0$.
[2]Verified
Solution
- Identify position vector at $t=0$: $\mathbf{r} = \begin{pmatrix} 20 \\ -25 \\ 0 \end{pmatrix}$
- Attempt to find distance from origin: $|\mathbf{r}| = \sqrt{20^2 + (-25)^2}$ M1
- $\sqrt{1025} = 32.0 \ (32.0156\dots) \ \text{km}$ A1
2 marks total
3.
Find the bearing on which the helicopter is travelling.
[2]Verified
Solution
- Attempt to calculate the bearing: $\arctan\left(\frac{4.2}{5.8}\right)$ or $90^\circ - \arctan\left(\frac{5.8}{4.2}\right)$ M1
- $035.9^\circ \ (35.909\dots)$ A1
2 marks total
