Question 7
A marble moves in one direction in a straight line with velocity $v=2\ln(t+1)$ (in metres per second) where $t$ is time (in seconds) since the marble passed through the origin.
Determine the distance from the origin the marble has rolled after 4 seconds.
Not quite. 0.40 is the acceleration at 4 seconds: $a=\dfrac{2}{t+1}=\dfrac{2}{5}=0.4$. Differentiating velocity gives acceleration. To get distance you go the other way and integrate. In 2022, 14.7% of Queensland students chose (A).
Not quite. The velocity is at least $2\ln4\approx2.77$ m/s for the whole of the last second, so the marble covers more than 1.60 m in that second alone. Integrate the velocity from 0 to 4. In 2022, 9.2% of Queensland students chose (B).
Not quite. 3.22 is the velocity at 4 seconds, $2\ln5\approx3.22$ m/s. That is how fast the marble is going, not how far it has gone. Distance is the area under the velocity graph. In 2022, 23.8% of Queensland students chose (C).
That’s it. Step through the hints below to see why. In 2022, 51.1% of Queensland students chose (D).
After s the marble has rolled about m, the shaded area. Keep going to 4 seconds.
At the start the marble is at the origin and its velocity is $2\ln1=0$, so nothing is shaded yet. Drag right.
At 4 seconds the shaded area is about 8.09 m. That is the distance rolled. The height of the curve, 3.22, is the velocity at that moment, option (C). It tells you how fast, not how far.
Past 4 seconds the area keeps growing, because the marble keeps moving forward. Come back to 4 seconds.
The marble moves in one direction, and $v=2\ln(t+1)\ge0$ for $t\ge0$. It never turns back, so the distance from the origin is the total distance rolled.
Distance is the area under the velocity–time graph, so you integrate the velocity.
It starts at the origin when $t=0$, so integrate from 0 to 4.
$$d=\int_0^4 2\ln(t+1)\,dt$$
Paper 2 is technology-active, so let your calculator evaluate it.
$$\int_0^4 2\ln(t+1)\,dt\approx8.094$$
So the marble has rolled about 8.09 m. For an exact form, check by differentiating that $(t+1)\ln(t+1)-t$ has derivative $\ln(t+1)$. Then the distance is $2(5\ln5-4)$.
$$d=\int_0^4 2\ln(t+1)\,dt$$
$$d\approx8.094\text{ m}$$
The velocity is never negative, so the marble only moves away from the origin. The area under the velocity graph from 0 to 4 is the distance, about 8.09 m.
How Queensland students answered in 2022
From the QCAA Mathematical Methods subject report for 2022. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2022 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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