How many QCAA Specialist Mathematics questions cover Modelling motion?

AusGrader has 36 QCAA Specialist Mathematics questions on Modelling motion, all with instant AI grading and detailed marking feedback.

QCAA (QCE) Specialist Mathematics — Modelling motion Questions & Answers | AusGrader

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 $v = 9 + 8t - t^2 (\text{m s}^{-1})$ .
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

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

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

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.

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

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 $P$ 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 $v = 2$ .

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

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

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

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

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

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.

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

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

QCAA Specialist Mathematics Modelling motion

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