2025 Paper 1, Q18
Two objects are launched simultaneously from different positions and travel along the same straight-line path. The objects are launched towards each other with the same initial speed.
The first object’s displacement (m) from the origin is given by $d=\frac13t^3-\frac12t^2+kt$, where $t$ is the time (s) since the objects were launched and $k$ is a constant, $k\ne 0$. The second object is moving with a constant acceleration of $4\ \mathrm{m\,s^{-2}}$.
The second object changes its direction, and at time $t=1$ s the objects have equal velocities and continue to travel in the same direction.
Compared to the first object, how much further does the second object travel between $t=1$ s and the next time the objects have equal velocities?
Two objects launched towards each other
No algebra yet. Object 2 is thrown back towards object 1, turns around half a second in, and by the time their speeds match again it has covered more ground.
$d=\tfrac13t^3-\tfrac12t^2+kt$
Differentiate to get its velocity. $k$ is unknown.
$a=4\ \mathrm{m\,s^{-2}}$, constant
Integrate to get its velocity. The constant $c$ is unknown.
Launched at the same instant, towards each other, with the same initial speed. That sentence is what ties $c$ to $k$.
white: object 1 · orange: object 2 · the loop runs t = 0 to t = 4 s, played at half speed
The answer is the area between two velocities
Once $k=2$ is recovered, the velocities cross at $t=1$ and $t=4$. Between those times the second object is always faster, and the gap in distance is the area between the curves.
solid: object 1 · dashed: object 2 · equal at t = 1 and t = 4
the shaded area, accumulating
QCAA marking guide · 6 marks
Differentiate the given displacement:
$$v_1(t)=\frac{d}{dt}\left(\tfrac13t^3-\tfrac12t^2+kt\right)=t^2-t+k$$
Marker: correctly determines the equation for the first object’s velocity.
The second object is given by its acceleration, so integrate, and keep the constant:
$$v_2(t)=\int 4\,dt=4t+c$$
Marker: correctly determines the equation for the second object’s velocity, including the constant $c$.
There are two constants, and the only way to find them is from the words of the question. “Launched towards each other with the same initial speed” gives $v_1(0)=-v_2(0)$, so $c=-k$. Then the equal velocities at $t=1$ give:
$$v_1(1)=v_2(1):\quad k=4+c=4-k\ \Rightarrow\ k=2,\ c=-2$$
Marker: determines the relationship between the constants $k$ and $c$. (Equivalent relationships accepted, e.g. $c=-k$ from $v_1(0)=-v_2(0)$.)
Find when the velocities agree again:
$$t^2-t+2=4t-2\ \Rightarrow\ t^2-5t+4=0\ \Rightarrow\ (t-1)(t-4)=0$$
$t=1$ is the time already given, so the next one is $t=4$.
Marker: determines the second time when the two velocities are the same.
Both objects travel in one direction across $1\le t\le 4$, so distance equals displacement and the difference can be integrated in one go:
$$d_2-d_1=\int_1^4\big(v_2-v_1\big)dt$$
Marker: uses a suitable method for determining the difference between the distances the objects have travelled in the given interval. (Subtracting each object’s distance separately is accepted.)
$$\int_1^4\left(-t^2+5t-4\right)dt=\left[-\tfrac{t^3}{3}+\tfrac{5t^2}{2}-4t\right]_1^4$$
4.5 m further
Marker: determines the difference between the distances of the two objects. Equivalent forms accepted, e.g. $\frac92$ or $4\frac12$.
Two sentences give you two constants
$v_1(0)=k$ and $v_2(0)=c$ point opposite ways with equal size, so $c=-k$. One sentence gives you one equation.
$k=4-k$, so $k=2$. Now every later number in the question exists.
With $k=2$ the velocities are $t^2-t+2$ and $4t-2$, crossing at $t=1$ and $t=4$. Between them the straight line sits above the parabola, and the area it gains is exactly 4.5 m.
This is inversion at its purest. There are two unknown constants, and the only equations for them are sentences. “Towards each other with the same initial speed” has to become $c=-k$, and you have to use $t=1$ as information rather than as the answer. If you never find $k$, you can still differentiate, integrate and set the velocities equal, but you end up with an expression in $k$ that is only worth three of the six marks. The 2025 subject report called this question out, so I would expect the QCAA to keep testing two-object questions like it.
Question wording and marking-guide steps are from the 2025 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.