How many QCAA Specialist Mathematics questions cover Modelling motion?

AusGrader has 87 QCAA Specialist Mathematics questions on Modelling motion, all with instant AI grading and detailed marking feedback.

QCAA Specialist Mathematics — Modelling motion Questions & Answers

Q12

2021

SCSA

Paper 2

6 marks

Q12

The horizontal displacement of a Ferris wheel cabin exhibits simple harmonic motion. The maximum horizontal speed is $\frac{\pi}{2}$ metres per second and its period of motion is exactly 60 seconds.

Let $x(t) = A\cos(nt)$ be the horizontal displacement after $t$ seconds.

Q12a

3 marks

Determine the values of $A$ and $n$ .

Reveal Answer

Period $T = 60 = \frac{2\pi}{n} \implies \therefore n = \frac{\pi}{30} \quad (0.1047....)$

$v(t) = -A\left(\frac{\pi}{30}\right)\sin\left(\frac{\pi t}{30}\right)$

$\therefore \text{Max } v = A\left(\frac{\pi}{30}\right) = \frac{\pi}{2} \implies \therefore A = 15$

Hence $x(t) = 15\cos\left(\frac{\pi t}{30}\right)$

Marking Criteria

Descriptor	Marks
determines $n$ correctly	1
differentiates and forms the correct expression for the maximum speed	1
determines $A$ correctly	1

Q12b

3 marks

Determine the horizontal acceleration, correct to the nearest $0.001\text{ m/s}^2$ , when the horizontal displacement is 10 metres.

Reveal Answer

Condition for S.H.M. is $a = -\left(\frac{\pi}{30}\right)^2 x$

$\therefore$ When $x = 10 \quad a = -\left(\frac{\pi^2}{900}\right)(10) = -0.10966 ... \text{ metres/sec}^2$

i.e. acceleration is $-0.110 \text{ m/sec}^2$ (3 d.p.)

Marking Criteria

Descriptor	Marks
applies the condition for S.H.M. correctly	1
substitutes $x = 10$ correctly	1
determines the acceleration correct to $0.001 \text{ m/sec}^2$	1

Q13

2025

VCAA

Paper 2

1 mark

Q13

1 mark

From an open window, a person projects a ball vertically up using an outstretched arm so the ball does not strike any part of the building. The point of projection of the ball is $50 \text{ m}$ above the ground and its velocity of projection is $20 \text{ m s}^{-1}$ .

The time, in seconds, it takes for the ball to reach the tray of a truck that is $1 \text{ m}$ above the ground directly below the point of projection is closest to

A

$1.72$

B

$5.80$

C

$5.83$

D

$1.75$

Reveal Answer

A

$1.72$

This is the magnitude of the negative root of the kinematic equation, which represents the time if the trajectory was extended backwards before projection.

B

$5.80$

Correct Answer

Using the kinematic equation $\Delta y = ut + \frac{1}{2}at^2$ with a displacement of $\Delta y = -49 \text{ m}$ (since the tray is $1 \text{ m}$ above the ground), $u = 20 \text{ m s}^{-1}$ , and $a = -9.8 \text{ m s}^{-2}$ yields $t \approx 5.80 \text{ s}$ .

C

$5.83$

This is the time it would take for the ball to reach the ground ( $\Delta y = -50 \text{ m}$ ), failing to account for the truck tray being $1 \text{ m}$ above the ground.

D

$1.75$

This is the magnitude of the negative root if the displacement was incorrectly set to $-50 \text{ m}$ (reaching the ground instead of the truck tray).

Q12

2024

VCAA

Paper 2

1 mark

Q12

1 mark

The position, $x$ metres, of a particle moving in a straight line from a fixed origin $O$ at time, $t$ seconds, is given by $x = e^{(k - 1)t}$ , where $k > 1$ .

The acceleration of the particle, in $\text{m s}^{-2}$ , when $x = k + 1$ is

A

$k^2 - 1$

B

$(k^2 - 1)(k + 1)$

C

$(k^2 - 1)(k - 1)$

D

$(k - 1)^2$

Reveal Answer

A

$k^2 - 1$

Incorrect. This represents the velocity of the particle when $x = k + 1$ , found by taking the first derivative $v = (k - 1)x$ , rather than the acceleration.

B

$(k^2 - 1)(k + 1)$

Incorrect. This expression does not match the acceleration formula $a = (k - 1)^2 x$ . It incorrectly multiplies the velocity by $x$ instead of taking the second derivative.

C

$(k^2 - 1)(k - 1)$

Correct Answer

Correct. The acceleration is the second derivative of position, $a = (k - 1)^2 x$ . Substituting $x = k + 1$ gives $a = (k - 1)^2(k + 1) = (k - 1)(k^2 - 1).

D

$(k - 1)^2$

Incorrect. This is the coefficient of $x$ in the acceleration equation $a = (k - 1)^2 x$ , but it fails to substitute the given value $x = k + 1$ .

Q17

2020

QCAA

Paper 2

7 marks

Q17

7 marks

An object is released from rest at a height of 100 m above the ground.
The motion of the vertical descent of the object is modelled by

$v\frac{dv}{dx} = 9.8 - 0.004v^2 \quad (v \ge 0)$

where $v$ is the velocity (m s $^{-1}$ ) and $x$ is the displacement from the ground (m).
Determine the velocity of the object when it strikes the ground.

Reveal Answer

$v\frac{dv}{dx} = 9.8 - 0.004v^2, v > 0$
$\int \frac{v}{9.8 - 0.004v^2} dv = \int dx$
$\frac{-1}{0.008} \int \frac{-0.008v}{9.8 - 0.004v^2} dv = \int dx$
$-125 \ln|9.8 - 0.004v^2| = x + c$

Given $v = 0$ when $x = -100$
$-125 \ln|9.8| = -100 + c$
$c \approx -185.298$

$-125 \ln|9.8 - 0.004v^2| = x - 185.298$
Determining $v$ when $x = 0$
$-125 \ln|9.8 - 0.004v^2| = -185.298$

Using graph facility of GDC
$v \approx -36.7 \text{ ms}^{-1}$ or $v \approx 36.7 \text{ ms}^{-1}$
As $v > 0$ , the negative solution is rejected
$\therefore v \approx 36.7 \text{ ms}^{-1}$

Marking Criteria

Descriptor	Marks
correctly uses separation of variables	1
correctly develops the general solution of the differential equation	1
correctly uses the given position of the origin	1
uses the given condition to determine value for c	1
substitutes the displacement at impact to form an equation in terms of v	1
determines one reasonable solution of v	1
shows logical organisation communicating key steps	1

Q17

2023

QCAA

Paper 1

7 marks

Q17

7 marks

An object of mass 2 kg is moving with a constant velocity ( $\text{m s}^{-1}$ ) of $\boldsymbol{v} = 3\boldsymbol{i} + \boldsymbol{k}$ .

At an instant, two forces (N), $\boldsymbol{F}_1 = 5t\boldsymbol{j} - 3\boldsymbol{k}$ and $\boldsymbol{F}_2 = -t\boldsymbol{j} + \boldsymbol{k}$ , act simultaneously on the object for $t$ seconds, where $0 \le t \le 2$ .

Determine the magnitude of the momentum of the object when $t = 1$ .

Reveal Answer

Method 1

$F_{net} = F_1 + F_2 = (5t\hat{j} - 3\hat{k}) + (-t\hat{j} + \hat{k})$
$= 4t\hat{j} - 2\hat{k}, 0 \le t \le 2$

$= ma$
$4t\hat{j} - 2\hat{k} = 2a$
$a = 2t\hat{j} - \hat{k}$

$v = \int a dt$
$= t^2\hat{j} - t\hat{k} + c$

At $t=0, v = 3\hat{i} + \hat{k}$
$\therefore c = 3\hat{i} + \hat{k}$

$v = t^2\hat{j} - t\hat{k} + 3\hat{i} + \hat{k}$
$= 3\hat{i} + t^2\hat{j} + (1-t)\hat{k}$

Momentum $= mv$
$= 2(3\hat{i} + t^2\hat{j} + (1-t)\hat{k})$

At $t=1, mv = 2(3\hat{i} + \hat{j})$

$|mv| = 2\sqrt{3^2 + 1^2} = 2\sqrt{10} \text{ kg ms}^{-1}$

Marking Criteria

Descriptor	Marks
correctly determines an expression for the net force acting on the object	1
determines an expression involving the object's acceleration after the forces act	1
determines a general solution representing the object's velocity after the forces act	1
determines a particular solution representing the velocity of the object after the forces act	1
determines an expression for the momentum of the object at $t=1$	1
determines a magnitude of the momentum of the object at $t=1$	1
shows logical organisation, communicating key steps to at least the start of determining the momentum of the object	1

Q1

2020

QCAA

Paper 2

1 mark

Q1

1 mark

The position $x$ (m) at time $t$ (s) of a 7 kg particle moving in a straight line is given by

$x = 3t^3 - 5t^2 + 2t - 4 \text{ for } 0 \le t \le 10$

Determine the time when the particle has a momentum of 620 kg m s $^{-1}$ .

A

1.73 s

B

2.60 s

C

3.66 s

D

3.71 s

Reveal Answer

A

1.73 s

This time value is incorrect. It may result from a calculation error or solving an incorrect kinematic equation.

B

2.60 s

This is an incorrect answer. Substituting $t = 2.60$ s into the derived momentum equation yields a value significantly lower than 620 kg m s $^{-1}$ .

C

3.66 s

This value is close but incorrect. Using $t = 3.66$ s results in a momentum of approximately 602 kg m s $^{-1}$ , which is less than the required value.

D

3.71 s

Correct Answer

First, find the velocity function by differentiating the position: $v(t) = \frac{dx}{dt} = 9t^2 - 10t + 2$ . Then, use the momentum formula $p = mv$ to set up $620 = 7(9t^2 - 10t + 2)$ . Solving this quadratic equation for $t$ yields $t \approx 3.71$ s.

Q16

2021

VCAA

Paper 2

1 mark

Q16

1 mark

An object of mass $m$ kilograms slides down a smooth slope that is inclined at an angle of $\theta^\circ$ to the horizontal, where $0^\circ < \theta^\circ < 45^\circ$ . The acceleration of the object down the slope is $a \text{ ms}^{-2}$ , $a > 0$ .

If the angle of inclination of the slope is doubled to $2\theta^\circ$ , then the acceleration of the object down the slope, in $\text{ms}^{-2}$ , is

A

$2a$

B

$\frac{2a}{g}\sqrt{g^2 - a^2}$

C

$\frac{2a^2 - g^2}{g}$

D

$\frac{a}{g}\sqrt{g^2 - a^2}$

E

$2a\sqrt{g^2 - a^2}$

Reveal Answer

A

$2a$

This incorrectly assumes that $\sin(2\theta) = 2\sin\theta$ , which is a common trigonometric misconception.

B

$\frac{2a}{g}\sqrt{g^2 - a^2}$

Correct Answer

The new acceleration is $g\sin(2\theta) = 2g\sin\theta\cos\theta$ . Substituting $\sin\theta = \frac{a}{g}$ and $\cos\theta = \frac{\sqrt{g^2-a^2}}{g}$ yields $\frac{2a}{g}\sqrt{g^2 - a^2}$ .

C

$\frac{2a^2 - g^2}{g}$

This expression is equivalent to $-g\cos(2\theta)$ , which does not represent the acceleration $g\sin(2\theta)$ down the slope.

D

$\frac{a}{g}\sqrt{g^2 - a^2}$

This expression evaluates to $\frac{1}{2}g\sin(2\theta)$ , which is exactly half of the correct new acceleration.

E

$2a\sqrt{g^2 - a^2}$

This expression is missing a factor of $\frac{1}{g}$ that arises when substituting $\cos\theta = \frac{\sqrt{g^2-a^2}}{g}$ into the double angle formula.

Q14

2021

VCAA

Paper 2

1 mark

Q14

1 mark

A body of mass 5 kg is acted on by a net force of magnitude $F$ newtons. This force causes the body to move so that its velocity, $v \text{ ms}^{-1}$ , along a straight line of motion is given by $v = 3 + 2x$ , where $x$ metres is the position of the body at time $t$ seconds.

When $x = 2$ , $F$ is equal to

A

10

B

14

C

35

D

70

E

175

Reveal Answer

A

10

This incorrectly calculates $m \frac{dv}{dx} = 5 \times 2 = 10$ . Acceleration is given by the chain rule $a = v \frac{dv}{dx}$ , not just $\frac{dv}{dx}$ .

B

14

This is the acceleration of the body ( $a = 14 \text{ ms}^{-2}$ ), but finding the net force requires multiplying this acceleration by the mass ( $F = ma$ ).

C

35

This calculates the momentum of the body ( $p = mv = 5 \times 7 = 35$ ), rather than the force ( $F = ma$ ).

D

70

Correct Answer

Acceleration is found using the chain rule: $a = v \frac{dv}{dx} = (3+2x)(2)$ . At $x=2$ , $a = 7 \times 2 = 14 \text{ ms}^{-2}$ . Using Newton's second law, $F = ma = 5 \times 14 = 70$ N.

E

175

This likely results from incorrectly multiplying the momentum by the mass ( $35 \times 5 = 175$ ). The correct formula for force is $F = m \left(v \frac{dv}{dx}\right)$ .

Q8

2022

VCAA

Paper 1

4 marks

Q8

4 marks

A body moves in a straight line so that when its displacement from a fixed origin $O$ is $x$ metres, its acceleration, $a$ , is $-4x \text{ ms}^{-2}$ . The body accelerates from rest and its velocity, $v$ , is equal to $-2 \text{ ms}^{-1}$ as it passes through the origin. The body then comes to rest again.

Find $v$ in terms of $x$ for this interval.

Reveal Answer

$\frac{d}{dx}\left(\frac{1}{2}v^2\right) = -4x$
$\Rightarrow \frac{1}{2}v^2 = -2x^2 + c$

When $x = 0$ , $v = -2$ and so $c = 2$ .

Therefore

$v^2 = -4x^2 + 4$
$\Rightarrow v = -\sqrt{-4x^2 + 4}$
$= -2\sqrt{1 - x^2}$

choosing the negative square root as $v = -2$ when $x = 0$ .

Marking Criteria

Descriptor	Marks
Equates an appropriate acceleration equivalent to $-4x$ , e.g., $\frac{d}{dx}\left(\frac{1}{2}v^2\right) = -4x$	1
Integrates correctly to obtain an expression for $v^2$ or $\frac{1}{2}v^2$ with a constant of integration	1
Substitutes $x = 0$ and $v = -2$ to find the constant of integration	1
Correctly expresses $v$ in terms of $x$ , choosing the negative square root, e.g., $v = -2\sqrt{1 - x^2}$	1

Q17

2024

QCAA

Paper 1

6 marks

Q17

6 marks

The acceleration ( $\text{m s}^{-2}$ ) of an object that moves in a straight line in an easterly direction over time $t$ for $0 \le t \le \frac{\pi}{6}$ seconds is given by $a = 2(1+v^2)$ , where $v$ is its velocity ( $\text{m s}^{-1}$ ).

The object is initially at rest at a position that is $\ln(\sqrt{2})$ metres west of the origin.

A student uses this information to calculate that the object is positioned at the origin when $t = \frac{\pi}{6}$ seconds.

Evaluate the reasonableness of the student's calculation.

Reveal Answer

Method 1

$a = 2(1+v^2)$
$\frac{dv}{dt} = 2(1+v^2)$

$\int \frac{1}{1+v^2} dv = \int 2 dt$
$\tan^{-1}(v) = 2t + c$

At $t=0, v=0: \tan^{-1}(0) = 0 + c$
$\therefore c = 0$

$\therefore \tan^{-1}(v) = 2t$
$v = \tan(2t)$
$x = \int v dt$
$= \int \tan(2t) dt$
$= -\frac{1}{2} \ln(\cos(2t)) + c$
$= -\frac{1}{2} \ln(\cos(2t)) + c$

Given $x = -\ln(\sqrt{2})$ when $t=0$
$-\ln(\sqrt{2}) = -\frac{1}{2} \ln(\cos(0)) + c$
$c = -\ln(\sqrt{2})$

$x = -\frac{1}{2} \ln(\cos(2t)) - \ln(\sqrt{2})$

When $t = \frac{\pi}{6}$
$x = -\frac{1}{2} \ln(\cos(\frac{\pi}{3})) - \ln(\sqrt{2})$
$= -\frac{1}{2} \ln(\frac{1}{2}) - \ln(\sqrt{2})$

$x = \ln((\frac{1}{2})^{-\frac{1}{2}}) - \ln(\sqrt{2})$
$= \ln(\sqrt{2}) - \ln(\sqrt{2})$
$= 0$
So the calculation is reasonable.

Marking Criteria

Descriptor	Marks
correctly determines a general solution to the differential equation in terms of v and time	1
determines an appropriate constant of integration	1
determines a general solution for displacement in terms of time	1
determines an appropriate constant of integration	1
determines a result for displacement when $t = \frac{\pi}{6}$ without a trigonometric term	1
provides an appropriate statement of reasonableness based on mathematical reasoning	1

Q14

2020

SCSA

Paper 2

5 marks

Q14

A particle travels in a straight line so that its velocity $v$ cm per second and displacement $x$ cm are related by the equation:

$v = -0.2x$

Q14a

2 marks

Determine the acceleration $a$ in terms of its displacement $x$ .

Reveal Answer

$\text{Acceleration } a = \frac{dv}{dt} = \frac{dv}{dx} \times \frac{dx}{dt} = (-0.2) \times (-0.2x) = 0.04x$

Marking Criteria

Descriptor	Marks
uses the chain rule correctly to relate $\frac{dv}{dt}$ in terms of $v$	1
obtains $a$ correctly in terms of $x$	1

Q14b

1 mark

Does the particle's motion constitute simple harmonic motion? Justify your answer.

Reveal Answer

$\text{Since } a = 0.04x \text{ then we could write } a = (0.2)^2 x$

$\text{But the condition for S.H.M. is that } a = -n^2 x \text{ .}$

$\text{Hence the motion does NOT follow simple harmonic motion.}$

Marking Criteria

Descriptor	Marks
justifies why the motion is NOT simple harmonic	1

Q14c

2 marks

It is known that the initial displacement of the particle is $x = 4$ cm.

Determine, correct to the nearest 0.01 second, when the particle has a displacement of 2 cm.

Reveal Answer

$\text{We have } v = \frac{dx}{dt} = -0.2x$

\begin{align*} \therefore \int \frac{1}{x} dx &= \int -0.2 dt \quad \text{using separation of variables}\\ \text{i.e. } \ln|x| &= -0.2t + c_1\\ \text{i.e. } x &= e^{-0.2t + c} = k \times e^{-0.2t} \end{align*}

$\text{Using } x(0) = 4 \text{, then } 4 = k\left(e^0\right) \implies \text{ Hence } k = 4$

$\text{i.e. } x = e^{-0.2t + c} = 4e^{-0.2t}$

$\text{Solving } x(t) = 2 \text{ yields } 2 = 4e^{-0.2t}$

$\text{i.e. } t = 5\ln 2 = 3.47 \text{ seconds (correct to 0.01 sec)}$

Marking Criteria

Descriptor	Marks
determines the function $x(t)$ correctly	1
solves for $t$ correct to 0.01 seconds	1

Q20

2020

VCAA

Paper 2

1 mark

Q20

1 mark

An object of mass 2 kg is suspended from a spring balance that is inside a lift travelling downwards.

If the reading on the spring balance is 30 N, the acceleration of the lift is

A

$5.2\text{ ms}^{-2}$ upwards.

B

$5.2\text{ ms}^{-2}$ downwards.

C

$9.8\text{ ms}^{-2}$ downwards.

D

$10.4\text{ ms}^{-2}$ upwards.

E

$10.4\text{ ms}^{-2}$ downwards.

Reveal Answer

A

$5.2\text{ ms}^{-2}$ upwards.

Correct Answer

The apparent weight (30 N) is greater than the true weight ( $2 \times 9.8 = 19.6$ N), so the net force is $10.4$ N upwards. Using $F=ma$ , the acceleration is $10.4 / 2 = 5.2 \text{ ms}^{-2}$ upwards.

B

$5.2\text{ ms}^{-2}$ downwards.

A downward acceleration of $5.2 \text{ ms}^{-2}$ would mean the net force is downwards, resulting in a spring reading less than the true weight ( $19.6 - 10.4 = 9.2$ N).

C

$9.8\text{ ms}^{-2}$ downwards.

A downward acceleration of $9.8 \text{ ms}^{-2}$ means the lift is in free fall, which would result in a spring balance reading of 0 N.

D

$10.4\text{ ms}^{-2}$ upwards.

This value represents the net force ( $10.4$ N) acting on the object, not the acceleration. You must divide the net force by the mass (2 kg) to find the acceleration.

E

$10.4\text{ ms}^{-2}$ downwards.

This value represents the magnitude of the net force ( $10.4$ N), not the acceleration. Additionally, the direction is incorrect because an apparent weight greater than the true weight requires an upward acceleration.

Q9

2024

VCAA

Paper 1

4 marks

Q9

A car is travelling along a straight, flat road. The velocity, $v$ km h $^{-1}$ , of the car and its position, $x$ kilometres, are measured from the position on the road where $x = 0$ .

The velocity $v$ and the position $x$ of the car are related by $v^2 = 1600 + \frac{672}{\pi}\arccos\left(\frac{x}{20}\right)$ , where $-15 \le x \le 15$ and $v \ge 0$ .

A speed detection device is positioned to detect the speed of a car as it passes the position $x = 0$ . The speed limit on the road is 40 km h $^{-1}$ .

The speed detection device will be activated if the car is travelling at 10% or more above the speed limit.

Q9a

1 mark

Determine, with evidence, whether the speed detection device will be activated.

Reveal Answer

$v \geq 44$ for the speed detection device to be activated.

At $x = 0$ , $v^2 = 1600 + \frac{672}{\pi}\arccos\left(\frac{0}{20}\right) = 1936 = 44^2$

Therefore, the speed detection device is activated.

Marking Criteria

Descriptor	Marks
Shows that at $x = 0$ , $v = 44$ (or $v^2 = 1936$ ) and concludes that the device is activated	1

Q9b

3 marks

Find the acceleration of the car, in km h $^{-2}$ , when $x = 12$ .

Give your answer in the form $\frac{k}{\pi}$ , where $k \in Z$ .

Reveal Answer

$a = \frac{d}{dx}\left(\frac{1}{2}v^2\right) = \frac{1}{2}\frac{d}{dx}\left(1600 + \frac{672}{\pi}\arccos\left(\frac{x}{20}\right)\right) \text{ or } \frac{d}{dx}\left(800 + \frac{336}{\pi}\arccos\left(\frac{x}{20}\right)\right)$

$a = \frac{-336}{\pi\sqrt{400 - x^2}} \text{ or } \frac{336}{\pi} \cdot \frac{1}{20} \cdot \frac{-1}{\sqrt{1 - \frac{x^2}{400}}} = \frac{336}{\pi} \cdot \frac{1}{20} \cdot \frac{-1}{\sqrt{1 - \left(\frac{x}{20}\right)^2}}$

When $x = 12$ , $a = \frac{-21}{\pi}$

Marking Criteria

Descriptor	Marks
Calculates the correct value for $a$	1
Differentiates correctly to find an expression for $a$	1
Uses $a = \frac{d}{dx}\left(\frac{1}{2}v^2\right)$	1

Q12

2025

VCAA

Paper 2

1 mark

Q12

1 mark

A particle moves along a straight line with constant acceleration. It passes through a point $A$ with velocity $u \text{ m s}^{-1}$ and then through a point $B$ with velocity $v \text{ m s}^{-1}$ .

The velocity of the particle at the midpoint of the line segment $AB$ is given by

A

$\frac{u + v}{2}$

B

$u + \frac{u + v}{2}$

C

$\frac{u^2 + v^2}{2}$

D

$\sqrt{\frac{u^2 + v^2}{2}}$

Reveal Answer

A

$\frac{u + v}{2}$

This formula represents the average velocity with respect to time, not the instantaneous velocity at the spatial midpoint.

B

$u + \frac{u + v}{2}$

This expression does not represent the velocity at the midpoint; it incorrectly adds the initial velocity to the time-average velocity.

C

$\frac{u^2 + v^2}{2}$

This is the square of the velocity at the midpoint ( $v_m^2$ ). It is missing the final square root step required to find the actual velocity.

D

$\sqrt{\frac{u^2 + v^2}{2}}$

Correct Answer

Using the kinematic equation $v^2 = u^2 + 2as$ , the velocity at the midpoint (distance $s/2$ ) is found by solving $v_m^2 = u^2 + 2a(s/2)$ . Substituting $2as = v^2 - u^2$ yields $v_m = \sqrt{\frac{u^2 + v^2}{2}}$ .

Q19

2022

QCAA

Paper 2

7 marks

Q19

7 marks

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.

Reveal Answer

Assume downwards as the positive direction
$a = 9.8 - 0.1v$
$v \frac{dv}{dx} = 9.8 - 0.1v$
$\int \frac{v}{9.8 - 0.1v} dv = \int dx$
$\int \frac{v}{v - 98} dv = \int \frac{-1}{10} dx$
$\int 1 + \frac{98}{v - 98} dv = \int \frac{-1}{10} dx$
$v + 98 \ln|v - 98| = \frac{-x}{10} + c$

Assume the origin is at the point of release.
Given $v(0) = 0 \Rightarrow c = 98 \ln(98)$

$v + 98 \ln|v - 98| = \frac{-x}{10} + 98 \ln(98)$
Let the distance to the ocean surface be $h$ metres.
Consider time of drop for each required velocity
When $v = 20 \text{ m s}^{-1}, x = h$
$20 + 98 \ln|-78| = \frac{-h}{10} + 98 \ln(98)$
$h = 23.7 \text{ m}$

When $v = 50 \text{ m s}^{-1}, x = h$
$50 + 98 \ln|-48| = 98 \ln(98) - \frac{h}{10}$
$h = 199.5 \text{ m}$

The range of the drone's flying height above the ocean surface should be between 23.7 m and 199.5 m.

Marking Criteria

Descriptor	Marks
correctly establishes a differential equation in terms of $v$ and $x$	1
determines general solution of the differential equation	1
determines a value for the constant	1
determines a model for the velocity in terms of its displacement and initial height	1
determines displacement of the drop for the minimum acceptable speed	1
determines displacement of the drop for the maximum acceptable speed	1
communicates range of the drone’s flying height including units	1

QCAA Specialist Mathematics Modelling motion

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QCAA Specialist Mathematics Modelling motion

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