Question 19
Object A is released from the origin with constant velocity, $\mathbf{v}_{A}$, such that its position after $t$ seconds is given by
$$\mathbf{r}_{A}=2\sqrt{3}\,t\,\hat{\mathbf{i}}+3t\,\hat{\mathbf{j}}+2t\,\hat{\mathbf{k}},\quad t\ge0$$
At a later time, object B is released from point $P\left(3\sqrt{3},\ 6,\ 0\right)$ and travels towards point $Q\left(5\sqrt{3},\ 8,\ 4\right)$ with constant velocity, $\mathbf{v}_{B}$, such that $\left|\mathbf{v}_{B}\right|=\sqrt{2}\left|\mathbf{v}_{A}\right|$.
Given that objects A and B collide, determine the time between the release of the two objects.
Assume all positions are given in metres and all velocities are given in metres per second.
B starts late but arrives at the same place at the same time
Watch both objects travel through space. You will see A leave first, then B set off from $P$ so that they reach the same point together.
The timeline on the right shows you that B only moves for the last 1.2 seconds of A's journey.
Change B's release delay until the two objects meet
The first slider sets how long after A you release B. The second one plays time forwards, so you can check whether they are in the same place at the same moment.
Work through the solution one mark at a time
Try each step yourself before you reveal it. This follows the QCAA's first method, which finds where the objects meet before it deals with time.
You differentiate $\mathbf{r}_{A}$ to get A's velocity, then find its magnitude. B is $\sqrt{2}$ times as fast.
$$\mathbf{v}_{A}=2\sqrt{3}\,\hat{\mathbf{i}}+3\,\hat{\mathbf{j}}+2\,\hat{\mathbf{k}}\qquad\left|\mathbf{v}_{A}\right|=\sqrt{12+9+4}=5$$
$$\left|\mathbf{v}_{B}\right|=5\sqrt{2}\ \text{m s}^{-1}$$
You can also write this as $\sqrt{50}$.
Marker: correctly determines the value of $\left|\mathbf{v}_{B}\right|$
B moves in a straight line from $P$ towards $Q$, so you can write its path as the equation of a line. Use $P$ as the point and $\overrightarrow{PQ}$ as the direction.
$$\overrightarrow{PQ}=\mathbf{q}-\mathbf{p}=2\sqrt{3}\,\hat{\mathbf{i}}+2\,\hat{\mathbf{j}}+4\,\hat{\mathbf{k}}$$
$$\mathbf{r}_{B}=\left(3\sqrt{3}+2\sqrt{3}\,l\right)\hat{\mathbf{i}}+\left(6+2l\right)\hat{\mathbf{j}}+4l\,\hat{\mathbf{k}},\quad l\in\mathbb{R}$$
Be careful here. $l$ is just a parameter along the line, and you cannot treat it as time, because $\overrightarrow{PQ}$ is not B's velocity. You keep follow-through marks here.
Marker: determines a vector in terms of a parameter representing the position of object B from the origin as it moves
Let the collision happen at $t=t_{1}$. At that moment both objects are at the same point, so you equate each component of $\mathbf{r}_{A}$ and $\mathbf{r}_{B}$.
$$\begin{aligned}3\sqrt{3}+2\sqrt{3}\,l&=2\sqrt{3}\,t_{1}\quad\ldots(1)\\6+2l&=3t_{1}\quad\ldots(2)\\4l&=2t_{1}\quad\ldots(3)\end{aligned}$$
You only need two of these for the mark. This mark can be implied by your later working.
Marker: determines at least two simultaneous equations based on the collision of the objects
Equation (3) gives you $l$ in terms of $t_{1}$. Substitute it into (1).
$$l=\frac{t_{1}}{2}\quad\Rightarrow\quad3\sqrt{3}+\sqrt{3}\,t_{1}=2\sqrt{3}\,t_{1}\quad\Rightarrow\quad t_{1}=3\text{ s}$$
So the collision happens 3 seconds after A is released, at $6\sqrt{3}\,\hat{\mathbf{i}}+9\,\hat{\mathbf{j}}+6\,\hat{\mathbf{k}}$. You can check this point in equation (2).
Marker: determines time from release to collision for object A
Because $l$ is not time, you find B's travel time from distance and speed instead. Work out how far B goes from $P$ to the collision point.
$$\begin{aligned}&\sqrt{\left(6\sqrt{3}-3\sqrt{3}\right)^{2}+(9-6)^{2}+(6-0)^{2}}\\&=\sqrt{27+9+36}=\sqrt{72}=6\sqrt{2}\text{ m}\end{aligned}$$
Marker: determines distance that object B travels to reach collision point
You divide the distance by B's speed to get its travel time, then take that away from A's time.
$$t_{B}=\frac{6\sqrt{2}}{5\sqrt{2}}=1.2\text{ s}\qquad3-1.2=1.8\text{ s}$$
B was released 1.8 s after A
You can also leave it as $3-\frac{\sqrt{72}}{\sqrt{50}}$.
Marker: determines time between the release of the two objects
Each phrase of the question gave you something
You are told when A starts but not when B starts, so you cannot write both positions with the same $t$. You have to find where they meet first, and only then work out how long B took to get there.
The trap I would point out is using the parameter in B's line as if it were time. The QCAA like to give you a speed for the second object instead of its velocity, so you have to find the distance B travels and divide by that speed.
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 2023 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.