Question 19
A research organisation plans to use a drone to drop a scientific instrument vertically from a stationary position above the ocean surface.
The acceleration $\left(\text{m s}^{-2}\right)$ of the falling instrument can be modelled by $9.8-0.1v$, where $v$ is its velocity $\left(\text{m s}^{-1}\right)$.
In order for the instrument sensors to activate, its speed as it hits the ocean surface must reach at least $20\ \text{m s}^{-1}$. However, if it hits with a speed above $50\ \text{m s}^{-1}$, the sensors will be damaged.
Determine the range of the drone’s flying height above the ocean surface to ensure that the sensors are activated but not damaged.
The height sets the speed it hits the water
Watch two drops. The graph on the right shows speed against distance fallen, which is exactly what the question needs.
Both drops start from rest and follow the same curve. The height only decides how far along the curve you get.
Choose a height and see how fast it hits
Move the slider to change the drone’s height. The model $v+98\ln(98-v)=-\frac{x}{10}+98\ln(98)$ gives you the speed when $x$ equals the height.
Work through the solution one mark at a time
Try each step yourself before you reveal it. This follows the QCAA’s first method, with downwards as positive and the origin at the point of release.
You are asked about height, which is a distance, and you are given speeds. There is no time in the question, so use the form of acceleration that links $v$ and $x$.
$$a=v\frac{dv}{dx}=9.8-0.1v$$
Writing $\frac{d}{dx}\left(\frac{1}{2}v^{2}\right)=9.8-0.1v$ is also accepted. If you take upwards as positive instead, your signs change but you get the same answer.
Marker: correctly establishes a differential equation in terms of $v$ and $x$
Separate the variables. Then rewrite the fraction so you can integrate it: $\frac{v}{9.8-0.1v}=\frac{-10v}{v-98}$, and $\frac{v}{v-98}=1+\frac{98}{v-98}$.
$$\begin{aligned}\int\frac{v}{9.8-0.1v}\,dv&=\int dx\\\int\left(1+\frac{98}{v-98}\right)dv&=\int-\frac{1}{10}\,dx\\v+98\ln\left|v-98\right|&=-\frac{x}{10}+c\end{aligned}$$
Because $v<98$, you can write the log as $\ln(98-v)$ if you would rather not use absolute value bars. Writing $\ln(v-98)$ without bars is not correct, because $v-98$ is negative. You keep follow-through marks from here if you slipped earlier.
Marker: determines a general solution of the differential equation
Put the origin at the point of release. The instrument starts from rest, so $v=0$ when $x=0$.
$$0+98\ln\left|0-98\right|=0+c\quad\Rightarrow\quad c=98\ln(98)$$
If you wrote the general solution as $v+98\ln(9.8-0.1v)$, you get $c=98\ln(9.8)$ instead, which is also fine.
Marker: determines a value for the constant
Put the constant back in. This model links the speed to the distance fallen.
$$v+98\ln\left|v-98\right|=-\frac{x}{10}+98\ln(98)$$
Let $h$ be the drone’s height above the ocean. The instrument hits the water when $x=h$.
Marker: determines a model for the velocity in terms of displacement
The slowest safe speed is $20\ \text{m s}^{-1}$. Substitute $v=20$ and $x=h$, then solve with your calculator.
$$20+98\ln\left|-78\right|=-\frac{h}{10}+98\ln(98)\quad\Rightarrow\quad h\approx23.7\text{ m}$$
23.6 m is also accepted if you rounded earlier.
Marker: determines displacement of the drop for the minimum acceptable speed
The fastest safe speed is $50\ \text{m s}^{-1}$. Do the same with $v=50$.
$$50+98\ln\left|-48\right|=-\frac{h}{10}+98\ln(98)\quad\Rightarrow\quad h\approx199.5\text{ m}$$
199.6 m is also accepted if you rounded earlier.
Marker: determines displacement of the drop for the maximum acceptable speed
Finish with a sentence that answers the question, with units.
The drone should fly between 23.7 m and 199.5 m above the ocean surface.
You only get this mark if you include the units. The QCAA’s second method puts the origin at the ocean surface, and the third works with $v$ and $t$ and then integrates for distance. All three give the same range.
Marker: communicates range of the drone’s flying height including units
Each phrase of the question gave you something
You are given acceleration as a function of velocity and asked for a distance. You have to choose the form of acceleration yourself, and there is no diagram or equation of motion to start from.
The choice I would check first is $v\frac{dv}{dx}$ against $\frac{dv}{dt}$. Both work, but $\frac{dv}{dt}$ makes you find the time and then integrate again for distance. The QCAA often set resisted motion questions like this, so it is worth asking what the question wants before you choose.
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.