2022 Paper 2, Q17
A snail is travelling along a straight path from point A. The snail’s velocity $(\mathrm{cm\,min^{-1}})$ is modelled by $v(t)=1.4\ln\!\left(1+t^2\right)$, where $t$ is time (in minutes) for $0\le t\le 15$.
An ant passes point A 12 minutes after the snail and follows the snail’s path. The ant moves with a constant acceleration of $2\ \mathrm{cm\,min^{-2}}$ and passes the snail at $t=15$ minutes.
Determine the ant’s velocity at point A.
A twelve minute head start, caught up in three
The snail plods for twelve minutes before the ant even reaches point A. The ant then has three minutes to cover everything the snail covered in fifteen.
$v(t)=1.4\ln\!\left(1+t^2\right)$
Integrate from 0 to 15 to find how far it goes, which is 76.04 cm.
$a=2\ \mathrm{cm\,min^{-2}}$, from $t=12$ to $t=15$
Its speed at A is the unknown constant of integration.
They are on the same path, but their clocks start 12 minutes apart, and they finish level. That is the whole equation.
white: the snail · orange: the ant · the loop runs t = 0 to 15 minutes, then holds
Two areas that have to be equal
“Passes the snail at $t=15$” means the two objects have covered the same distance from A. Under the velocity graph that is two areas, on different intervals, set equal.
black: snail · orange: ant, only from t = 12
they meet only at the very last moment
QCAA marking guide · 4 marks
$$\int_0^{15}1.4\ln\!\left(1+t^2\right)dt=76.0431\ \mathrm{cm}$$
Marker: correctly determines the total displacement of the snail.
The ant’s velocity is $v(t)=2t+c$. “Passes the snail at $t=15$” means its displacement from A over its three minutes equals the snail’s over fifteen:
$$\int_{12}^{15}\left(2t+c\right)dt=76.0431$$
Marker: establishes an equation linking the ant and the snail. (A time reference based on the ant is accepted: $\int_0^3(2t+c)\,dt=76.0431$.)
$$\left[t^2+ct\right]_{12}^{15}=81+3c=76.0431\ \Rightarrow\ c=-1.6523$$
Marker: determines the constant. (Analytic or numerical solution accepted.)
Point A is where the ant is at $t=12$, so the question wants $v(12)$, not $c$:
$$v(12)=2\times12-1.6523$$
22.35 cm min$^{-1}$
Marker: determines velocity of the ant. Equivalent decimals accepted, e.g. 22.3. On the ant’s own clock the constant is 22.3477.
Whose clock are you on?
The ant runs from $t=12$ to $t=15$ and $v=2t-1.6523$. At A, $t=12$, so $v=22.35$.
The ant runs from $t=0$ to $t=3$ and $v=2t+22.3477$. At A, $t=0$, so $v=22.35$ again.
Both clocks give the same answer, and both are accepted. But the constant $c$ is a different number in each. Quoting $c$ on the snail’s clock as the ant’s speed at A gives $-1.65\ \mathrm{cm\,min^{-1}}$, a negative velocity for an ant that is clearly chasing.
You have to turn “passes the snail” into an equation, and it is an equation between two areas measured over different intervals, fifteen minutes of snail against three minutes of ant. Nothing in the question mentions equal displacement. The second trap is the clock. The constant of integration depends on which starting time you use, and only one of the two possible numbers answers the question that was actually asked. The QCAA love two-object questions for exactly this reason, so always write down whose clock you are using before you integrate. I would do this even when it seems obvious.
Question wording and marking-guide steps are from the 2022 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.