2023 Paper 1, Q18
A person enters the lowest carriage of a miniature Ferris wheel with a six-metre diameter. The bottom carriage is one metre off the ground. When top speed is reached, it takes three seconds for a carriage to travel from the lowest to the highest point of the ride. It is claimed that:
The vertical motion of the Ferris wheel produces a maximum vertical acceleration on each rider that is more than half the acceleration of free fall.
Free fall occurs when gravity is the only force acting, resulting in an acceleration of $9.8\ \mathrm{ms^{-2}}$.
Evaluate the reasonableness of the claim.
One full turn of the wheel takes six seconds
Three seconds from the bottom to the top is half a turn, so the period is six seconds. Watch the height marker slide up the scale. It moves fast through the middle and is almost still at the ends.
one loop = one turn at top speed, then a short hold · the orange carriage is the rider
A point on a turning wheel goes up and down like a cosine curve
The wheel traces the graph as it turns. Build the model one parameter at a time, because each one comes from a single sentence in the question.
The diameter is 6, so the amplitude is 3. The bottom carriage is 1 m off the ground, so the midline is 4. Three seconds from the bottom to the top is half a turn, so the period is 6. Be careful with that last one! If you take the period as 3, your acceleration comes out four times too big and your verdict flips.
The claim is a height on the second derivative
Height is a cosine. Differentiate twice and the acceleration is a cosine again, with amplitude $\frac{\pi^2}{3}$. The 4.9 line sits well above anything the wheel can reach.
$h(t)=-3\cos\frac{\pi t}{3}+4$
$h''(t)=\frac{\pi^2}{3}\cos\frac{\pi t}{3}$
QCAA marking guide · 4 marks
Because the motion repeats, use a cosine graph. We use cosine because at $t=0$ the height is a minimum.
Marker: correctly recognises a periodic model is to be used for the vertical position of the carriage. The recognition could be in words or as an equation, and may be implied by subsequent working. Either cosine or sine is acceptable as the starting form.
Start from $h(t)=a\cos\!\left(b(t+c)\right)+d$. The amplitude is $a=3$, half the six metre diameter. Three seconds from the bottom to the top is half a period, so $T=6$ and $b=\frac{2\pi}{6}=\frac{\pi}{3}$. There is no phase shift, so $c=0$. The axle sits $1+3=4$ m above the ground, so $d=4$.
$$h(t)=-3\cos\!\left(\frac{\pi t}{3}\right)+4$$
Marker: correctly determines the model for the vertical position of a carriage on the Ferris wheel. The minus is not needed for this mark. A sine model with a phase shift is accepted, e.g. $h(t)=3\sin\!\left(\frac{\pi}{3}\left(t-\frac32\right)\right)+4$.
$$v(t)=h'(t)=\pi\sin\!\left(\frac{\pi t}{3}\right),\qquad a(t)=h''(t)=\frac{\pi^2}{3}\cos\!\left(\frac{\pi t}{3}\right)$$
The amplitude of the acceleration model is $\frac{\pi^2}{3}$, and that amplitude is the maximum acceleration:
$$\frac{\pi^2}{3}\approx\frac{9}{3}=3$$
Marker: determines an approximation of the maximum acceleration produced, based on the second derivative equation obtained. Follow-through marks are allowed for errors in prior working.
Half of free fall is $4.9\ \mathrm{ms^{-2}}$, and the wheel’s maximum is about $3.3\ \mathrm{ms^{-2}}$.
3.3 < 4.9, so the claim is not reasonable
Marker: provides appropriate statement of reasonableness. Equivalent statements accepted; the mark can only be awarded if previous evidence supports the conclusion.
A technology-free question with $\pi^2$ in it
There is no calculator, so use $\pi^2\approx 9$ and $\frac{\pi^2}{3}\approx 3$. That is all the accuracy you need, because 3 is clearly less than 4.9.
There is no need to solve $h'''=0$ or test endpoints. For $A\cos(Bt)$ the greatest value is $A$, so reading the amplitude off the second derivative finishes the question.
It is also worth knowing what this means physically. The biggest vertical acceleration happens at the very top and the very bottom, where the carriage is not rising at all for a moment. Riders feel it as the light-headed moment going over the top, and it is about a third of gravity, not half.
“Three seconds from the lowest to the highest point” is half a period, not a full period. The QCAA love to hand you half a period and see who notices. If you take $T=3$, $b$ doubles and the acceleration quadruples to about 13, which flips your decision. The word “vertical” also matters. The wheel’s total acceleration towards the centre stays the same size, but the question only asks about the vertical part, which is exactly $h''$. Nothing in the stem tells you to differentiate twice, or that the amplitude of the second derivative is the answer. That is what I would expect most students to miss.
Question wording and marking-guide steps are from the 2023 QCAA Mathematical Methods 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.