Question 19
The velocity vectors of two objects A and B $\left(\text{in m s}^{-1}\right)$ at time $t$ (in s) are given respectively by
$$\mathbf{v}_{A}=6\sin(3t)\,\hat{\mathbf{i}}+6\cos(3t)\,\hat{\mathbf{j}}\qquad\mathbf{v}_{B}=\cos(t)\,\hat{\mathbf{i}}-\sin(t)\,\hat{\mathbf{j}}$$
Objects A and B are initially at $(-2,\ 0,\ 2)$ and $(0,\ 1,\ -1)$ respectively.
Determine the position of Object A when it is 4 metres away from Object B for the first time.
Two objects circling at different heights
Watch both objects move and the gap between them change. The graph on the right shows that gap against time.
The height difference is always 3, so only the horizontal positions change the distance.
Find the first time they are 4 m apart
Move time forwards and watch the distance. It is $\sqrt{14-4\sin(2t)}$, which you get once you use a compound angle identity.
Work through the solution one mark at a time
Try each step yourself before you reveal it. There is no calculator, so the trig identity in step 5 is what makes the equation solvable.
Integrate A’s velocity to get its position, then use the starting point to find the constant vector.
$$\mathbf{r}_{A}=\int\mathbf{v}_{A}\,dt=-2\cos(3t)\,\hat{\mathbf{i}}+2\sin(3t)\,\hat{\mathbf{j}}+\mathbf{c}_{A}$$
$$t=0:\ -2\hat{\mathbf{i}}+2\hat{\mathbf{k}}=-2\hat{\mathbf{i}}+\mathbf{c}_{A}\ \Rightarrow\ \mathbf{c}_{A}=2\hat{\mathbf{k}}$$
$$\mathbf{r}_{A}=-2\cos(3t)\,\hat{\mathbf{i}}+2\sin(3t)\,\hat{\mathbf{j}}+2\hat{\mathbf{k}}$$
Marker: correctly determines the expression for the position of Object A
Do the same for B. Be careful with the signs when you integrate the sine and cosine.
$$\mathbf{r}_{B}=\sin(t)\,\hat{\mathbf{i}}+\cos(t)\,\hat{\mathbf{j}}+\mathbf{c}_{B}$$
$$t=0:\ \hat{\mathbf{j}}-\hat{\mathbf{k}}=\hat{\mathbf{j}}+\mathbf{c}_{B}\ \Rightarrow\ \mathbf{c}_{B}=-\hat{\mathbf{k}}$$
$$\mathbf{r}_{B}=\sin(t)\,\hat{\mathbf{i}}+\cos(t)\,\hat{\mathbf{j}}-\hat{\mathbf{k}}$$
Marker: correctly determines the expression for the position of Object B
The vector from A to B is $\mathbf{r}_{B}-\mathbf{r}_{A}$. Subtract each component.
$$\mathbf{r}_{B}-\mathbf{r}_{A}=\left(\sin(t)+2\cos(3t)\right)\hat{\mathbf{i}}+\left(\cos(t)-2\sin(3t)\right)\hat{\mathbf{j}}-3\hat{\mathbf{k}}$$
Using $\mathbf{r}_{A}-\mathbf{r}_{B}$ is just as good. You keep follow-through marks from here if you slipped earlier.
Marker: determines an expression to represent the relative position of Objects A and B
Square and add the components to get the square of the distance. Expand each bracket carefully.
$$\begin{aligned}\left|\mathbf{r}_{B}-\mathbf{r}_{A}\right|^{2}&=\sin^{2}t+4\sin t\cos3t+4\cos^{2}3t\\&\quad+\cos^{2}t-4\cos t\sin3t+4\sin^{2}3t+9\end{aligned}$$
Group the squares: $\sin^{2}t+\cos^{2}t=1$ and $4\left(\cos^{2}3t+\sin^{2}3t\right)=4$, so the squares add up to $1+4+9=14$.
$$=14-4\left(\sin3t\cos t-\cos3t\sin t\right)$$
Marker: determines an expression to represent the distance (or square of the distance) between the objects
The bracket is the expansion of $\sin(A-B)$ with $A=3t$ and $B=t$. That turns it into one trig function.
$$\sin3t\cos t-\cos3t\sin t=\sin(3t-t)=\sin(2t)$$
$$\left|\mathbf{r}_{B}-\mathbf{r}_{A}\right|=\sqrt{14-4\sin(2t)}$$
Marker: uses a trigonometric identity to determine an expression in terms of a single trigonometric function that represents the distance (or square of the distance) between the objects
Set the distance equal to 4 and solve. You want the first positive solution.
$$\sqrt{14-4\sin(2t)}=4\ \Rightarrow\ 14-4\sin(2t)=16\ \Rightarrow\ \sin(2t)=-\frac{1}{2}$$
$$2t=\frac{7\pi}{6}\ \Rightarrow\ t=\frac{7\pi}{12}\text{ s}$$
Sine is first negative in the third quadrant, so $\frac{7\pi}{6}$ is the smallest positive angle. The next one, $\frac{11\pi}{6}$, gives the second time they are 4 m apart.
Marker: determines the first time that Object A is 4 metres away from Object B
Substitute $t=\frac{7\pi}{12}$ into $\mathbf{r}_{A}$, so $3t=\frac{7\pi}{4}$.
$$\mathbf{r}_{A}=-2\cos\left(\tfrac{7\pi}{4}\right)\hat{\mathbf{i}}+2\sin\left(\tfrac{7\pi}{4}\right)\hat{\mathbf{j}}+2\hat{\mathbf{k}}$$
$\mathbf{r}_{A}=-\sqrt{2}\,\hat{\mathbf{i}}-\sqrt{2}\,\hat{\mathbf{j}}+2\hat{\mathbf{k}}$ m
You can also give it as the point $\left(-\sqrt{2},\ -\sqrt{2},\ 2\right)$. Remember that $\cos\left(\frac{7\pi}{4}\right)=\frac{\sqrt{2}}{2}$ and $\sin\left(\frac{7\pi}{4}\right)=-\frac{\sqrt{2}}{2}$.
Marker: determines position of Object A
Each phrase of the question gave you something
You are given velocities, not positions, and the $\hat{\mathbf{k}}$ components only appear in the starting points. After you expand the distance, you have to spot a compound angle identity to get an equation you can solve without a calculator.
The step I would check first is the constants of integration. If you forget the $2\hat{\mathbf{k}}$ and $-\hat{\mathbf{k}}$, you lose the 9 inside the square root and every answer after that is wrong. The QCAA often start objects away from the origin for exactly this reason.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2021 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.