2024 Paper 2, Q18
An object experiencing straight-line motion along a path has an acceleration $(\mathrm{m\,s^{-2}})$ defined by the function $a(t)=3\sin(2t)$, where $t$ is time (s) since the object begins moving.
When $t=0$, both displacement and velocity are zero.
On the path is a motion sensor that is able to detect motion up to 2 metres away. The object passes directly by the motion sensor when $t=3$.
Determine the average velocity of the object while it moves through the range of the sensor.
An object creeping past a sensor
The object never reverses, but it surges and stalls. The sensor sits where the object is at $t=3$, and its range is a 4 m window the object crawls through.
$a(t)=3\sin(2t)$, with $v(0)=0$ and $d(0)=0$
You need to integrate twice to get to displacement. Both constants of integration are zero.
Range 2 m either side of $d(3)$
The question never tells you where the sensor is. You have to work it out.
the orange band is the sensor’s 4 m window · loop runs t = 0 to 6 s at half speed
Average velocity is displacement divided by time
Use the displacement graph to find the two times the object enters and leaves the sensor’s range. Once you have those, average velocity is just the 4 m divided by the time taken. Be careful here. It is not the average of the velocity function, and it is not the velocity at $t=3$.
in range from t = 1.689 to t = 4.592
blue line: the average velocity, 1.38 m/s
QCAA marking guide · 5 marks
Integrate the acceleration and use $v(0)=0$:
$$v(t)=\int 3\sin(2t)\,dt=-\tfrac32\cos(2t)+c,\qquad v(0)=0\Rightarrow c=\tfrac32$$
$$v(t)=-\tfrac32\cos(2t)+\tfrac32$$
Marker: correctly determines the velocity formula.
Integrate again, with $d(0)=0$:
$$d(t)=-\tfrac34\sin(2t)+\tfrac32t$$
Marker: determines the displacement formula.
The sensor’s position is not given. It is wherever the object is at $t=3$, so:
$$d(3)=4.70956\ \mathrm{m}$$
So the sensor detects the object between $d=2.70956$ and $d=6.70956$.
Marker: determines the object displacement when $t=3$. (Appropriate rounding accepted, e.g. 4.7.)
Solve $d(t)$ at each edge of the band, then subtract:
$$t=1.68914\ \text{and}\ t=4.59214\ \Rightarrow\ \Delta t=2.90296\ \mathrm{s}$$
Marker: determines the time when the object is within sensor range. (Appropriate rounding accepted, e.g. 2.9.)
Average velocity is displacement divided by time, and the displacement across the sensor’s range is exactly 4 m, so:
$$\bar v=\frac{4}{2.90296}$$
1.38 m s$^{-1}$
Marker: determines the average velocity. (Appropriate rounding accepted, e.g. 1.4.)
Putting the answer together
The velocity at $t=3$ is only 0.06 m/s, because the object is almost stopped as it passes the sensor. The fastest it moves while in range is 3 m/s. Neither of these is the answer. Average velocity is the displacement divided by the time, which is why you never need to integrate to find the 4 m.
The question never gives you the sensor’s position. The only clue is that the object “passes directly by the motion sensor when $t=3$”, so the sensor must be at $d(3)$. This means you work forwards first, integrating twice to get the displacement, and then backwards, solving $d(t)$ at each edge of the range to find the two times. The 4 m comes for free, because it is just the width of the sensor’s range. I see a lot of students average the velocity function, or give $v(3)$ as their answer. That is more work, and it earns fewer marks!
Question wording and marking-guide steps are from the 2024 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.