Question 16
A particle is moving in a straight line. The velocity (ms−1) of the particle is given by
$v(t)=\dfrac{20\sin(2t)}{6-5\cos(2t)}$, $t\ge0$, where $t$ is time (s) after moving from its initial position.
The initial position of the particle is +6.0 m from the origin.
Use calculus methods to determine an equation for the position of the particle from the origin at any time $t$.
[3 marks] b)Determine the position of the particle relative to the origin when it first reaches maximum velocity.
[3 marks]Part a) needs the antiderivative of $\frac{f'(x)}{f(x)}$ as a log, which is not in the 2025 syllabus, but finding $c$ from the initial position and all of part b) still transfer.
The particle swings out from 6 m and back
Watch where the particle is when its velocity first peaks. That moment is part b).
The velocity peaks early, at about 0.29 s. Part b) asks where the particle is at that moment, and the position equation from part a) answers it.
Find the multiple of the log that differentiates back to the velocity
Part a) turns on one structure. Try multiples of $\ln\left(6-5\cos(2t)\right)$ and compare their derivative with $v(t)$. Then slide time along and find where the particle is when the velocity first peaks.
Choose a value of $k$. The dashed curve is the derivative of your $s(t)$, so the right choice lands exactly on $v(t)$.
The chain rule gives $\frac{d}{dt}\ln\left(6-5\cos(2t)\right)=\frac{10\sin(2t)}{6-5\cos(2t)}$. That is half of $v(t)$, so the dashed curve only reaches half as high.
The derivative of the denominator is $10\sin(2t)$, and $20\sin(2t)=2\times10\sin(2t)$. So $v(t)$ is 2 times $\frac{f'(t)}{f(t)}$, and the dashed curve sits exactly on it.
This is what you get if you forget the chain rule inside $\cos(2t)$. The denominator would seem to differentiate to $5\sin(2t)$, and the dashed curve comes out twice as high as $v(t)$.
Copying the 20 from the numerator ignores the $10\sin(2t)$ that differentiating the denominator already produces. The dashed curve is ten times too high and runs off the graph.
Work through the solution one mark at a time
Try each step yourself before you reveal it. Part a) is by hand. In part b) your calculator finds the time of the maximum and the definite integral.
Position is the integral of velocity. Look at the denominator and differentiate it. Its derivative is $10\sin(2t)$, and the numerator is 2 times that, so take the 2 outside.
$$\begin{aligned}s(t)&=\int\frac{20\sin(2t)}{6-5\cos(2t)}\,dt\\&=2\int\frac{10\sin(2t)}{6-5\cos(2t)}\,dt\end{aligned}$$
$$\begin{aligned}\text{Let }f(t)&=6-5\cos(2t)\\f'(t)&=10\sin(2t)\\s(t)&=2\int\frac{f'(t)}{f(t)}\,dt\end{aligned}$$
This uses $\int\frac{f'(x)}{f(x)}\,dx=\ln f(x)$. This mark may be implied by your later working.
Marker: correctly manipulates the integrand to obtain a numerator that is the derivative of the denominator
Integrate. The denominator is never smaller than $6-5=1$, so the log is defined for every $t$.
$$\begin{aligned}s(t)&=2\ln f(t)\\&=2\ln\left(6-5\cos(2t)\right)+c\end{aligned}$$
The $+\,c$ is not needed for this mark, but you need it for the next one.
Marker: correctly determines position formula
The particle starts 6.0 m from the origin, so $s(0)=6$. Use it to find $c$.
$$\begin{aligned}6&=2\ln\left(6-5\cos(0)\right)+c\\6&=2\ln(1)+c\\c&=6\end{aligned}$$
If you made a slip earlier, you can still earn this mark by follow-through.
Marker: determines the position formula relative to the origin, i.e. allows for starting at +6 m
Graph $v(t)$ for $t\ge0$ on your calculator. The velocity starts positive, so the particle heads away from the origin. Use the maximum feature on the first peak.
$$t\approx0.292843\ \text{s}$$
Equivalent values are accepted, such as 0.29. Other methods are accepted too. For example, setting $v'(t)=0$ gives $\cos(2t)=\frac{5}{6}$, so $t=\frac{1}{2}\cos^{-1}\left(\frac{5}{6}\right)$. The velocity is periodic, so take the first peak.
Marker: correctly determines first time when velocity is a maximum
The change in position from the start to that time is the definite integral of the velocity. Use your calculator’s definite integral function.
$$\int_0^{0.292843}v(t)\,dt\approx1.21227\ \text{m}$$
Substituting $t\approx0.292843$ into your formula from part a) is accepted instead. Follow-through applies if part a) or your time had a slip. Keep the unrounded time here. Using 0.29 gives about 7.195 m at the end.
Marker: identifies the required integral (or the formula established in 16a) and the max velocity time is substituted
The integral is the change from the starting point. The particle started at $s(0)=6$, so add it.
$$1.21227+6=7.21227\ \text{m}$$
The formula from part a) gives the same answer, $2\ln\left(\frac{11}{6}\right)+6\approx7.21227$, with the 6 already included.
Marker: determines the position of the particle
Each phrase of the question gave you something
Part a) only works once you spot that the numerator is a multiple of the derivative of the denominator. Nothing in the question tells you that, so you have to see the structure yourself. Part b) then joins two ideas. You find when the velocity first peaks, and carry that time into the position equation or a definite integral, remembering the 6.0 m start. It is still familiar because the wording names each step. “Calculus methods”, “position of the particle from the origin” and “first reaches maximum velocity” all point at a routine you have practised.
If I had to guess where you might lose marks, it would be the starting position. Integrating the velocity gives the change from where the particle began, so the 6.0 m has to go in, as $c=6$ in part a) or added after the definite integral in part b). The last mark in part a) is for exactly this, and part b) needs it again if you use the definite integral. The other trap is the word “first”. The velocity repeats every $\pi$ seconds, so it peaks again at about 3.43 s. Take the first peak and keep its time unrounded.
Only mark this done when you could do it without help. Reading the solution does not count.
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.
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