Question 14
An object is moving in a straight line with an acceleration represented by the differential equation $\dfrac{dv}{dt}=-\left(4+v^2\right)$, where $v$ is the object’s velocity $\left(\text{m s}^{-1}\right)$ over time, $t\ (\text{s})$, where $t\ge0$, until it comes to rest.
Determine the general solution of the differential equation.
[3 marks]The initial velocity of the object is $1.5\ \text{m s}^{-1}$.
Determine the time when the particle comes to rest.
[2 marks]The velocity falls until the object stops
The top line shows where the object is. The graph below it plots the velocity against time as the object moves.
The object stops about $0.22$ m along the track. The question only asks when it stops.
Change the starting velocity and see when the object stops
The grey curves are particular solutions for different starting velocities, and the black one is the start you chose. The orange point is where it comes to rest. Move along the curve with the second slider: the blue tangent has gradient $\frac{dv}{dt}$, which is the acceleration.
This is the question’s object. It comes to rest after about $0.32$ s. The tangent is steepest at the start, where $v$ is largest, so the object loses speed fastest at the beginning.
A faster start takes longer to stop, but every curve reaches zero before the dashed line. The time to stop is $\frac12\tan^{-1}\left(\frac{v(0)}{2}\right)$, and $\tan^{-1}$ never reaches $\frac{\pi}{2}$, so the time is always less than $\frac{\pi}{4}\approx0.785$ s.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The steps follow the QCAA marking guide.
The acceleration is given in terms of $v$ and the question is about time, so keep it as $\frac{dv}{dt}$. Divide both sides by $4+v^2$ to get every $v$ on the left, then integrate both sides with respect to $t$.
$$\frac{1}{4+v^2}\,\frac{dv}{dt}=-1$$
$$\int\frac{1}{4+v^2}\,dv=\int-1\,dt$$
The QCAA also accept the minus sign on the other side, as $\displaystyle\int\frac{-1}{4+v^2}\,dv=\int dt$.
Marker: correctly establishes a suitable integration result based on the separation of variables technique
The left side matches this rule from the formula book, with $a=2$.
$$\begin{aligned}&\int\frac{a}{a^2+x^2}\,dx\\&\quad=\tan^{-1}\left(\frac{x}{a}\right)+c\end{aligned}$$
The numerator needs to be $2$, so multiply inside the integral by $2$ and outside by $\frac12$.
$$\begin{aligned}&\int\frac{1}{4+v^2}\,dv\\&\quad=\frac12\int\frac{2}{2^2+v^2}\,dv\\&\quad=\frac12\tan^{-1}\left(\frac{v}{2}\right)\end{aligned}$$
Marker: determines one side of the general solution in terms of $v$
Integrate the right side, and put the one constant of integration there.
$$\int-1\,dt=-t+c$$
This mark doesn’t need the $+c$, and the marking guide accepts equivalent forms such as $\tan^{-1}\left(\frac{v}{2}\right)=-2t$ and $v=2\tan(-2t)$. I’d still write the constant, because part b) is all about finding it. Rearranged with the constant, the general solution is $v=2\tan(2c-2t)$.
Marker: determines the other side of the general solution in terms of $t$
The initial velocity means $v=1.5$ when $t=0$. Substitute both into the general solution. Check that your calculator is in radians first.
$$\frac12\tan^{-1}\left(\frac{1.5}{2}\right)=-0+c$$
Keep the unrounded value, $0.3218$, in your calculator. The QCAA also accept an equivalent equation for the constant, such as $2\tan(2c)=1.5$.
Marker: determines an expression that represents the integration constant
At rest means $v=0$. Since $\tan^{-1}(0)=0$, the left side is zero.
$$\frac12\tan^{-1}(0)=-t+0.3218$$
$$0=-t+0.3218$$
$$t\approx0.3218$$
The time to rest equals the constant from Step 4. If your general solution was wrong, you can still earn both part b) marks by follow-through.
Marker: determines the time when the particle is at rest
Each phrase of the question gave you something
The acceleration depends on $v$ alone, so the equation separates in one line, and the integral is the inverse tangent rule from the formula book. Part b) is two substitutions into the general solution.
If I had to guess where you might lose marks, it would be the $\frac12$ in front of the inverse tangent, or a calculator left in degrees. In degrees, $\tan^{-1}(0.75)$ comes out as about $36.87$, and the object would take more than 18 seconds to stop. A time that long should tell you something has gone wrong.
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 2022 QCAA Specialist Mathematics 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.
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