Question 14
The motion of an object that moves in a straight line is given by $v(x)=\cos^{-1}(2x)$ where $v$ is the velocity $\left(\text{m s}^{-1}\right)$ and $x$ is the displacement (m) from the origin.
Determine $a(x)$ where $a$ is the acceleration $\left(\text{m s}^{-2}\right)$ of the object.
[2 marks] b)Use the result from 14a) to determine $a(0)$, given $-2\pi\le a(0)\le0$. Express your answer in simplest form.
[2 marks]The object slows down as it moves to the right
The animation starts the object at $x=-0.5$, where $v=\pi$. Underneath, the same velocity is plotted against displacement.
The velocity is a function of displacement, so its graph is drawn against $x$. The gradient is a rate per metre, which is why you multiply by $v$.
Slide the object along and watch the tangent
Move the slider to change the displacement $x$. The orange tangent has gradient $\frac{dv}{dx}$, and the height of the point is $v$. Their product is the acceleration.
This is part b). At the origin the values are exact, because $\cos^{-1}(0)=\frac{\pi}{2}$ and $\sqrt{0.25}=0.5$.
Here $2x=\pm\frac{1}{2}$, so $\cos^{-1}(2x)$ has an exact value. The gradient is negative and $v$ is positive, so the acceleration is negative.
Near $x=-0.5$, $v$ approaches $\pi$ and the tangent gets very steep, so the acceleration is large and negative.
Near $x=0.5$ the tangent is very steep, but $v$ is very small. The product settles towards $a=-2$.
The gradient is negative everywhere and $v$ is positive, so the acceleration is always negative. The object slows down the whole way.
Why the range is given
Many angles have a cosine of $0$. $\cos^{-1}(0)$ means the one between $0$ and $\pi$. Try others and see where each one puts $a(0)=\frac{-\theta}{0.5}$. The shaded part of the number line is the given range.
This is the principal value, $\cos^{-1}(0)=\frac{\pi}{2}$, and $a(0)=-\pi$ lies inside the given range.
This angle also has a cosine of $0$, but it puts $a(0)$ outside $-2\pi\le a(0)\le0$. The range rules it out.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The steps follow the QCAA marking guide, which uses $a=v\frac{dv}{dx}$.
Velocity is given as a function of displacement, so the acceleration comes from $a=v\frac{dv}{dx}$. Start with $\frac{dv}{dx}$. The formula book gives the derivative below. I have called its constant $c$ so it doesn’t clash with the acceleration.
$$\frac{d}{dx}\cos^{-1}\left(\frac{x}{c}\right)=\frac{-1}{\sqrt{c^2-x^2}}$$
Here $2x=\frac{x}{0.5}$, so $c=0.5$ and $c^2=0.25$.
$$\frac{dv}{dx}=\frac{-1}{\sqrt{0.25-x^2}}$$
The chain rule gives the same derivative in another form, $\frac{dv}{dx}=\frac{-2}{\sqrt{1-4x^2}}$. Either form earns the mark, and it can be implied by your next line.
Marker: correctly determines $\frac{dv}{dx}$
Multiply by $v$.
$$\begin{aligned}a(x)&=v\frac{dv}{dx}\\&=\cos^{-1}(2x)\times\frac{-1}{\sqrt{0.25-x^2}}\end{aligned}$$
The QCAA accept equivalent expressions, such as $v\frac{dv}{dx}=\frac{-2\cos^{-1}(2x)}{\sqrt{1-4x^2}}$. If your $\frac{dv}{dx}$ was wrong, you can still earn this mark by follow-through.
Marker: determines an expression for acceleration as a function of displacement
Substitute $x=0$ into your answer to part a).
$$\begin{aligned}a(0)&=\frac{-\cos^{-1}(0)}{\sqrt{0.25}}\\[4pt]&=\frac{-\cos^{-1}(0)}{0.5}\end{aligned}$$
The numerator needs one exact value. $\cos^{-1}$ gives the angle between $0$ and $\pi$ whose cosine is the input.
$$\cos^{-1}(0)=\frac{\pi}{2}$$
This mark can be implied by your next line.
The marking guide also gives this mark for a correct value of an alternative inverse trigonometric expression, and its example is $\cos^{-1}(0)=\frac{-\pi}{2}$. That isn’t true as written, because $\cos^{-1}(0)=\frac{\pi}{2}$. The guide may mean the value of the whole numerator, $-\cos^{-1}(0)=\frac{-\pi}{2}$, or it may be accepting $\frac{-\pi}{2}$ here and leaving the range in part b) to rule it out. Whichever it means, write $\cos^{-1}(0)=\frac{\pi}{2}$.
Marker: determines a correct exact value for the inverse trigonometric expression on the numerator
Finish the division.
$$\begin{aligned}a(0)&=\frac{-\frac{\pi}{2}}{0.5}\\[4pt]&=-\pi\end{aligned}$$
Check it against the range: $-2\pi\le-\pi\le0$. The range is there to steer you to the principal value. Any other angle with a cosine of $0$, such as $-\frac{\pi}{2}$ or $\frac{3\pi}{2}$, puts $a(0)$ outside it. If you made an error in part a), you can still earn the marks in part b) by follow-through.
Marker: determines a reasonable solution for $a(0)$ (based on the given range $-2\pi\le a(0)\le0$)
Each phrase of the question gave you something
Velocity given as a function of displacement is a familiar setup, and the method is the same every time. Differentiate, then multiply by $v$. The derivative of $\cos^{-1}$ is in the formula book, and the exact value $\cos^{-1}(0)=\frac{\pi}{2}$ is one you already know.
If I had to guess where you might lose marks, it would be stopping at $\frac{dv}{dx}$. At the origin that is $-2$, a rate per metre, and you still have to multiply by $v=\frac{\pi}{2}$. The other risk is the chain rule. Leave out the factor of $2$ from $2x$ and you get $\frac{dv}{dx}=\frac{-1}{\sqrt{1-4x^2}}$, which gives $a(0)=-\frac{\pi}{2}$. That value sits inside the range too, so the range won’t catch it for you.
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 2020 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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