How many QCAA Specialist Mathematics Alternative Sequence questions cover Modelling motion?

AusGrader has 17 QCAA Specialist Mathematics Alternative Sequence questions on Modelling motion, all with instant AI grading and detailed marking feedback.

QCAA Specialist Mathematics Alternative Sequence — Modelling motion Questions & Answers

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 value for the constant	1
determines model for the velocity in terms of displacement	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

Q3

2022

QCAA

Paper 1

1 mark

Q3

1 mark

A particle travels in a straight line over time, $t$ , with a constant acceleration, $a(t)$ .

Which function could represent the particle's displacement, $x(t)$ ?

A

$x(t) = t^3$

B

$x(t) = t^2$

C

$x(t) = \frac{1}{t}$

D

$x(t) = \sqrt{t}$

Reveal Answer

A

$x(t) = t^3$

If $x(t) = t^3$ , the acceleration is the second derivative, $a(t) = 6t$ . This represents an acceleration that changes with time, not a constant acceleration.

B

$x(t) = t^2$

Correct Answer

Acceleration is the second derivative of displacement. For $x(t) = t^2$ , the first derivative (velocity) is $2t$ and the second derivative (acceleration) is $2$ , which is a constant.

C

$x(t) = \frac{1}{t}$

If $x(t) = \frac{1}{t}$ , the second derivative is $a(t) = \frac{2}{t^3}$ , which is not a constant value.

D

$x(t) = \sqrt{t}$

If $x(t) = \sqrt{t}$ , the second derivative is $a(t) = -\frac{1}{4t^{3/2}}$ , which means the acceleration is not constant.

Q14

2022

QCAA

Paper 2

5 marks

Q14

An object is moving in a straight line with an acceleration represented by the differential equation $\frac{dv}{dt} = -\left(4 + v^2\right)$ , where $v$ is the object's velocity $\left(\text{m s}^{-1}\right)$ over time, $t\text{ (s)}$ , where $t \ge 0$ , until it comes to rest.

Q14a

3 marks

Determine the general solution of the differential equation.

Reveal Answer

$\frac{dv}{dt} = -(4 + v^2)$
$\frac{1}{4 + v^2} \frac{dv}{dt} = -1$
$\int \frac{1}{4 + v^2} dv = \int -1 dt$
$\frac{1}{2} \int \frac{2}{4 + v^2} dv = \int -1 dt$
$\frac{1}{2} \tan^{-1} \left(\frac{v}{2}\right) = -t + c$

Marking Criteria

Descriptor	Marks
correctly establishes a suitable integration result based on the separation of variables technique	1
determines one side of the general solution in terms of $v$	1
determines the other side of the general solution in terms of $t$	1

Q14b

2 marks

The initial velocity of the object is $1.5 \text{ m s}^{-1}$ .

Determine the time when the particle comes to rest.

Reveal Answer

Given $v(0) = 1.5$
$\frac{1}{2} \tan^{-1} \left(\frac{1.5}{2}\right) = c$
$c \approx 0.32$

When $v = 0$ :
$\frac{1}{2} \tan^{-1}(0) \approx -t + 0.32$
$t \approx 0.32 \text{ s}$
The particle comes to rest after 0.32 s.

Marking Criteria

Descriptor	Marks
determines an expression that represents the integration constant	1
determines the time when the particle is at rest	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 ( $\text{m s}^{-1}$ ) 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

\text{Given } v = 0 \text{ when } x = -100

-125\ln|9.8| = -100 + c

c \approx -185.298

-125\ln|9.8 - 0.004v^2| = x - 185.298

\text{Determining } v \text{ when } x = 0

-125\ln|9.8 - 0.004v^2| = -185.298

\text{Using graph facility of GDC}

v \approx -36.7 \text{ ms}^{-1} \text{ or } v \approx 36.7 \text{ ms}^{-1}

\text{As } v > 0 \text{, 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

Q15

2020

QCAA

Paper 1

4 marks

Q15

The motion of an object that moves in a straight line is given by $v(x) = \cos^{-1}(2x)$ where $v$ is the velocity ( $\text{m s}^{-1}$ ) and $x$ is the displacement (m) from the origin.

Q15a

2 marks

Determine $a(x)$ where $a$ is the acceleration ( $\text{m s}^{-2}$ ) of the object.

Reveal Answer

$v(x)=\cos^{-1}(2x)$
$a=v\frac{dv}{dx}$
$\frac{dv}{dx}=\frac{-1}{\sqrt{0.25-x^2}}$
$a=\frac{-\cos^{-1}(2x)}{\sqrt{0.25-x^2}}$

Marking Criteria

Descriptor	Marks
correctly determines $\frac{dv}{dx}$	1
determines an expression for acceleration as a function of displacement	1

Q15b

2 marks

Use the result from 15a) to determine $a(0)$ , given $-2\pi \le a(0) \le 0$ . Express your answer in simplest form.

Reveal Answer

When $x=0$
$a=\frac{-\cos^{-1}(0)}{0.5}$
$\cos^{-1}(0)=\frac{\pi}{2}$
$a(0)= -\frac{\left(\frac{\pi}{2}\right)}{0.5}$
$=-\pi\ (\text{m s}^{-2})$

Marking Criteria

Descriptor	Marks
determines a correct exact value for the inverse trigonometric expression on the numerator	1
determines a reasonable solution for $a(0)$ based on the given range ( $-2\pi\le a(0)\le 0$ )	1

Q4

2020

QCAA

Paper 2

1 mark

Q4

1 mark

A particle is moving with simple harmonic motion described by the equation $x = 1.32 \cos\left(\frac{\pi t}{2}\right)$ where $x$ (m) is the displacement of the particle from a central position over time $t$ (s), $t \ge 0$ .

The maximum speed of the particle is

A

$2.07 \text{ m s}^{-1}$

B

$4.15 \text{ m s}^{-1}$

C

$4.30 \text{ m s}^{-1}$

D

$5.28 \text{ m s}^{-1}$

Reveal Answer

A

$2.07 \text{ m s}^{-1}$

Correct Answer

The maximum speed of a particle in simple harmonic motion is given by $v_{max} = A\omega$ . With amplitude $A = 1.32$ and angular frequency $\omega = \frac{\pi}{2}$ , the maximum speed is $1.32 \times \frac{\pi}{2} \approx 2.07 \text{ m s}^{-1}$ .

B

$4.15 \text{ m s}^{-1}$

This incorrect answer comes from multiplying the amplitude by $\pi$ instead of $\frac{\pi}{2}$ , effectively using the wrong angular frequency.

C

$4.30 \text{ m s}^{-1}$

This is an incorrect calculation. The maximum speed must be found using the derivative of the displacement function, which yields $v_{max} = A\omega$ .

D

$5.28 \text{ m s}^{-1}$

This value is $1.32 \times 4$ , which incorrectly multiplies the amplitude by a factor of 4 rather than the angular frequency $\omega = \frac{\pi}{2}$ .

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

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

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

A

1.73 s

B

2.60 s

C

3.66 s

D

3.71 s

Reveal Answer

A

1.73 s

Incorrect. This value does not satisfy the momentum equation. You must find the velocity function by differentiating the position function, then solve $m v(t) = 620$ .

B

2.60 s

Incorrect. This result likely comes from a sign error in the quadratic formula, mistakenly using $-b = -70$ instead of $+70$ when solving $63t^2 - 70t - 606 = 0$ .

C

3.66 s

Incorrect. This result comes from an algebraic error in the quadratic formula, specifically forgetting to square the $b$ term in the discriminant ( $b^2 - 4ac$ ).

D

3.71 s

Correct Answer

Correct. Velocity is the derivative of position, $v(t) = 9t^2 - 10t + 2$ . Setting momentum $p = mv = 7(9t^2 - 10t + 2) = 620$ and solving the resulting quadratic equation $63t^2 - 70t - 606 = 0$ yields $t \approx 3.71 \text{ s}$ .

Q6

2022

QCAA

Paper 2

1 mark

Q6

1 mark

A 4 kg object moves in a straight line over time, $t\text{ (s)}$ , where $0 \le t \le 5$ with velocity $v = 9 + 8t - t^2 \left(\text{m s}^{-1}\right)$ .

Determine the momentum of the object when $t = 3$ .

A

$24 \text{ kg m s}^{-1}$

B

$27 \text{ kg m s}^{-1}$

C

$96 \text{ kg m s}^{-1}$

D

$100 \text{ kg m s}^{-1}$

Reveal Answer

A

$24 \text{ kg m s}^{-1}$

This is the velocity of the object at $t = 3$ , not the momentum. You must multiply the velocity by the mass to find the momentum.

B

$27 \text{ kg m s}^{-1}$

This is an incorrect calculation. To find the momentum, first calculate the velocity at $t = 3$ , then multiply by the mass of $4 \text{ kg}$ .

C

$96 \text{ kg m s}^{-1}$

Correct Answer

The velocity at $t = 3$ is $v = 9 + 8(3) - 3^2 = 24 \text{ m s}^{-1}$ . Multiplying this velocity by the mass ( $4 \text{ kg}$ ) gives the correct momentum of $96 \text{ kg m s}^{-1}$ .

D

$100 \text{ kg m s}^{-1}$

This is the momentum of the object at $t = 4$ , not $t = 3$ . Ensure you substitute the correct time into the velocity equation.

Q17

2021

QCAA

Paper 2

7 marks

Q17

7 marks

An object with a mass of 2 kg is released from rest at the top of a 1 metre long frictionless plane inclined at $30^\circ$ to the horizontal.

A force of $\boldsymbol{P}$ newtons acting parallel to the plane opposes the motion of the object as it travels down the plane.

When the object is $x$ metres from the top of the plane, its velocity is $v\text{ m s}^{-1}$ .

Given $|\boldsymbol{P}| = \frac{4}{\sqrt{4-x^2}}$ , determine $x$ when $v = 2$ .

Reveal Answer

Method 1
Resolving net forces along the plane
$F_{\text{net}} = 2g \sin(30^\circ) - \frac{4}{\sqrt{4 - x^2}}$

$F_{\text{net}} = ma$
$g - \frac{4}{\sqrt{4 - x^2}} = 2a$

$g - \frac{4}{\sqrt{4 - x^2}} = 2v \frac{dv}{dx}$

$v \frac{dv}{dx} = 4.9 - \frac{2}{\sqrt{4 - x^2}}$

$\int v \ dv = \int 4.9 - \frac{2}{\sqrt{4 - x^2}} \ dx$

$\frac{v^2}{2} = 4.9x - 2 \sin^{-1} \left(\frac{x}{2}\right) + c$

Given $v = 0$ when $x = 0$
$0 = 0 - 2 \sin^{-1}(0) + c$
$c = 0$
$\therefore v^2 = 9.8x - 4 \sin^{-1} \left(\frac{x}{2}\right)$

When $v = 2$
$4 = 9.8x - 4 \sin^{-1} \left(\frac{x}{2}\right)$

Solving for $x$ using GDC
$x = 0.51 \text{ m}$

Marking Criteria

Descriptor	Marks
correctly determines the net forces along the plane	1
determines equation for acceleration along the plane	1
determines differential equation in terms of velocity and displacement	1
determines general solution to a differential equation	1
determines value of arbitrary constant	1
establishes equation to solve for $x$ when $v = 2$	1
determines $x$	1

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