QCAA Specialist Mathematics Vector calculus

15 sample questions with marking guides and sample answers

Q17
2025
VCAA
Paper 2
1 mark
Q17
1 mark

The acceleration vector of a particle that starts from rest is given by
a∼(t)=4cos⁡(2t)i∼+10sin⁡(2t)j∼−6e−2tk∼\underset{\sim}{\text{a}}(t) = 4\cos(2t)\underset{\sim}{\text{i}} + 10\sin(2t)\underset{\sim}{\text{j}} - 6e^{-2t}\underset{\sim}{\text{k}}, where t≥0t \geq 0.

The velocity vector of the particle, v∼(t)\underset{\sim}{\text{v}}(t), is given by

A

v∼(t)=2sin⁡(2t)i∼−5cos⁡(2t)j∼+3e−2tk∼\underset{\sim}{\text{v}}(t) = 2\sin(2t)\underset{\sim}{\text{i}} - 5\cos(2t)\underset{\sim}{\text{j}} + 3e^{-2t}\underset{\sim}{\text{k}}

B

v∼(t)=2sin⁡(2t)i∼−5(cos⁡(2t)+1)j∼+3(e−2t+1)k∼\underset{\sim}{\text{v}}(t) = 2\sin(2t)\underset{\sim}{\text{i}} - 5(\cos(2t) + 1)\underset{\sim}{\text{j}} + 3(e^{-2t} + 1)\underset{\sim}{\text{k}}

C

v∼(t)=2sin⁡(2t)i∼−5(cos⁡(2t)−1)j∼+3(e−2t−1)k∼\underset{\sim}{\text{v}}(t) = 2\sin(2t)\underset{\sim}{\text{i}} - 5(\cos(2t) - 1)\underset{\sim}{\text{j}} + 3(e^{-2t} - 1)\underset{\sim}{\text{k}}

D

v∼(t)=−8sin⁡(2t)i∼+20cos⁡(2t)j∼+12e−2tk∼\underset{\sim}{\text{v}}(t) = -8\sin(2t)\underset{\sim}{\text{i}} + 20\cos(2t)\underset{\sim}{\text{j}} + 12e^{-2t}\underset{\sim}{\text{k}}

Reveal Answer
A

v∼(t)=2sin⁡(2t)i∼−5cos⁡(2t)j∼+3e−2tk∼\underset{\sim}{\text{v}}(t) = 2\sin(2t)\underset{\sim}{\text{i}} - 5\cos(2t)\underset{\sim}{\text{j}} + 3e^{-2t}\underset{\sim}{\text{k}}

This option represents the indefinite integral without accounting for the initial condition v∼(0)=0\underset{\sim}{\text{v}}(0) = 0. It fails to include the necessary constants of integration.

B

v∼(t)=2sin⁡(2t)i∼−5(cos⁡(2t)+1)j∼+3(e−2t+1)k∼\underset{\sim}{\text{v}}(t) = 2\sin(2t)\underset{\sim}{\text{i}} - 5(\cos(2t) + 1)\underset{\sim}{\text{j}} + 3(e^{-2t} + 1)\underset{\sim}{\text{k}}

This option has the wrong signs for the constants of integration. Evaluating this vector at t=0t=0 gives −10j∼+6k∼-10\underset{\sim}{\text{j}} + 6\underset{\sim}{\text{k}} instead of the required zero vector.

C

v∼(t)=2sin⁡(2t)i∼−5(cos⁡(2t)−1)j∼+3(e−2t−1)k∼\underset{\sim}{\text{v}}(t) = 2\sin(2t)\underset{\sim}{\text{i}} - 5(\cos(2t) - 1)\underset{\sim}{\text{j}} + 3(e^{-2t} - 1)\underset{\sim}{\text{k}}

Correct Answer

Integrating the acceleration vector gives v∼(t)=(2sin⁡(2t)+C1)i∼+(−5cos⁡(2t)+C2)j∼+(3e−2t+C3)k∼\underset{\sim}{\text{v}}(t) = (2\sin(2t) + C_1)\underset{\sim}{\text{i}} + (-5\cos(2t) + C_2)\underset{\sim}{\text{j}} + (3e^{-2t} + C_3)\underset{\sim}{\text{k}}. Applying the initial condition v∼(0)=0\underset{\sim}{\text{v}}(0) = 0 yields the correct constants: C1=0C_1=0, C2=5C_2=5, and C3=−3C_3=-3.

D

v∼(t)=−8sin⁡(2t)i∼+20cos⁡(2t)j∼+12e−2tk∼\underset{\sim}{\text{v}}(t) = -8\sin(2t)\underset{\sim}{\text{i}} + 20\cos(2t)\underset{\sim}{\text{j}} + 12e^{-2t}\underset{\sim}{\text{k}}

This option is the derivative of the acceleration vector (known as jerk), rather than the integral. Velocity is found by integrating acceleration, not differentiating it.

Q17
2023
QCAA
Paper 2
6 marks
Q17
6 marks

An object is projected upwards from ground level with an initial velocity of 15 m s−115 \text{ m s}^{-1} at an angle of 54∘54^\circ to the horizontal.

The object just passes over a drone hovering in the air. An observer is positioned directly below the drone and at a horizontal distance of 20 m20 \text{ m} from where the object is projected.

The observer commented that:

  • it took the object around 2 to 2.5 seconds after its projection to reach the drone
  • the object was still moving in an upwards direction as it passed the drone.

Assuming that air resistance is negligible, use a vector calculus approach to evaluate the reasonableness of the observer's comments.

Reveal Answer

Let i^\hat{i} and j^\hat{j} be the horizontal and vertical unit vectors respectively. Let tt represent the time in seconds after the projection of the object.
a(t)=−9.8j^a(t) = -9.8\hat{j}
v(t)=∫a(t)dt=−9.8tj^+cv(t) = \int a(t) dt = -9.8t\hat{j} + c
Given v(0)=15cos⁡(54∘)i^+15sin⁡(54∘)j^v(0) = 15\cos(54^\circ)\hat{i} + 15\sin(54^\circ)\hat{j}
v(t)=15cos⁡(54∘)i^+(15sin⁡(54∘)−9.8t)j^v(t) = 15\cos(54^\circ)\hat{i} + (15\sin(54^\circ) - 9.8t)\hat{j}
r(t)=∫v(t)dtr(t) = \int v(t) dt
=15cos⁡(54∘)ti^+(15sin⁡(54∘)t−4.9t2)j^+c= 15\cos(54^\circ)t\hat{i} + (15\sin(54^\circ)t - 4.9t^2)\hat{j} + c
Let origin be at the release point: r(0)=0i^+0j^r(0) = 0\hat{i} + 0\hat{j}
r(t)=15cos⁡(54∘)ti^+(15sin⁡(54∘)t−4.9t2)j^r(t) = 15\cos(54^\circ)t\hat{i} + (15\sin(54^\circ)t - 4.9t^2)\hat{j}
When rx=20⇒15cos⁡(54∘)t=20r_x = 20 \Rightarrow 15\cos(54^\circ)t = 20
Time object just passes drone: t=2.27st = 2.27\text{s}
Finding maximum value of ry:15sin⁡(54∘)t−4.9t2r_y: 15\sin(54^\circ)t - 4.9t^2
Using GDC
Time object reaches maximum height: t=1.24st = 1.24\text{s}

While the estimation of the time taken for the object to reach the drone is reasonable, the comment regarding the direction of the object as it passed the drone is not reasonable as it would have been moving in a downward direction at that time.

Marking Criteria
DescriptorMarks

correctly determines the velocity function of the object using vector calculus

1

determines displacement function of the object

1

determines time when the object just passes drone

1

determines time when the object reaches maximum height

1

uses mathematical justification to evaluate the reasonableness of both comments based on prior mathematical reasoning

1

shows logical organisation, communicating key steps

1
Q6
2025
QCAA
Paper 1
1 mark
Q6
1 mark

At time tt, a particle travels with a velocity of v=(21+t2)i^−2tj^v = \left(\frac{2}{1+t^2}\right)\hat{i} - 2t\hat{j}.

Determine a general expression for the position vector, rr, of the particle during this motion.

A

r=2tan⁡−1(t)i^−2j^+cr = 2\tan^{-1}(t)\hat{i} - 2\hat{j} + c

B

r=2tan⁡−1(t)i^−t2j^+cr = 2\tan^{-1}(t)\hat{i} - t^2\hat{j} + c

C

r=12tan⁡−1(t)i^−2j^+cr = \frac{1}{2}\tan^{-1}(t)\hat{i} - 2\hat{j} + c

D

r=12tan⁡−1(t)i^−t2j^+cr = \frac{1}{2}\tan^{-1}(t)\hat{i} - t^2\hat{j} + c

Reveal Answer
A

r=2tan⁡−1(t)i^−2j^+cr = 2\tan^{-1}(t)\hat{i} - 2\hat{j} + c

This is incorrect because the integral of the j^\hat{j} component, −2t-2t, is −t2-t^2, not −2-2.

B

r=2tan⁡−1(t)i^−t2j^+cr = 2\tan^{-1}(t)\hat{i} - t^2\hat{j} + c

Correct Answer

This is correct. The position vector is the integral of the velocity vector with respect to time. Integrating 21+t2\frac{2}{1+t^2} yields 2tan⁡−1(t)2\tan^{-1}(t), and integrating −2t-2t yields −t2-t^2.

C

r=12tan⁡−1(t)i^−2j^+cr = \frac{1}{2}\tan^{-1}(t)\hat{i} - 2\hat{j} + c

This is incorrect because both components are integrated improperly. The integral of 21+t2\frac{2}{1+t^2} is 2tan⁡−1(t)2\tan^{-1}(t), and the integral of −2t-2t is −t2-t^2.

D

r=12tan⁡−1(t)i^−t2j^+cr = \frac{1}{2}\tan^{-1}(t)\hat{i} - t^2\hat{j} + c

This is incorrect because the integral of the i^\hat{i} component, 21+t2\frac{2}{1+t^2}, is 2tan⁡−1(t)2\tan^{-1}(t), not 12tan⁡−1(t)\frac{1}{2}\tan^{-1}(t).

Q19
2021
QCAA
Paper 1
7 marks
Q19
7 marks

The velocity vectors of two objects A and B (in m s−1^{-1}) at time tt (in s) are given respectively by

vA=6sin⁡(3t)i^+6cos⁡(3t)j^\boldsymbol{v}_{\mathrm{A}} = 6\sin(3t)\hat{\boldsymbol{i}} + 6\cos(3t)\hat{\boldsymbol{j}}
vB=cos⁡(t)i^−sin⁡(t)j^\boldsymbol{v}_{\mathrm{B}} = \cos(t)\hat{\boldsymbol{i}} - \sin(t)\hat{\boldsymbol{j}}

Objects A and B are initially at (−2,0,2)(-2, 0, 2) and (0,1,−1)(0, 1, -1) respectively. Determine the position of Object A when it is 4 metres away from Object B for the first time.

Reveal Answer

vA=6sin⁡(3t)i+6cos⁡(3t)jv_A = 6\sin(3t)i + 6\cos(3t)j
vB=cos⁡(t)i−sin⁡(t)jv_B = \cos(t)i - \sin(t)j

rA=∫vAdt=−2cos⁡(3t)i+2sin⁡(3t)j+cAr_A = \int v_A dt = -2\cos(3t)i + 2\sin(3t)j + c_A
When t=0t = 0
−2i+2k=−2cos⁡(0)i+2sin⁡(0)j+cA⇒cA=2k-2i + 2k = -2\cos(0)i + 2\sin(0)j + c_A \Rightarrow c_A = 2k
∴rA=−2cos⁡(3t)i+2sin⁡(3t)j+2k\therefore r_A = -2\cos(3t)i + 2\sin(3t)j + 2k

rB=∫vBdt=sin⁡(t)i+cos⁡(t)j+cBr_B = \int v_B dt = \sin(t)i + \cos(t)j + c_B
When t=0t = 0
j−k=sin⁡(0)i+cos⁡(0)j+cB⇒cB=−kj - k = \sin(0)i + \cos(0)j + c_B \Rightarrow c_B = -k
∴rB=sin⁡(t)i+cos⁡(t)j−k\therefore r_B = \sin(t)i + \cos(t)j - k

rB−rAr_B - r_A
=(sin⁡(t)i+cos⁡(t)j−k)…= (\sin(t)i + \cos(t)j - k) \dots
⋯−(−2cos⁡(3t)i+2sin⁡(3t)j+2k)\dots - (-2\cos(3t)i + 2\sin(3t)j + 2k)
=(sin⁡(t)+2cos⁡(3t))i+(cos⁡(t)−2sin⁡(3t))j−3k= (\sin(t) + 2\cos(3t))i + (\cos(t) - 2\sin(3t))j - 3k

∣rB−rA∣|r_B - r_A|
=sin⁡2(t)+4sin⁡(t)cos⁡(3t)+4cos⁡2(3t)+…= \sqrt{\sin^2(t) + 4\sin(t)\cos(3t) + 4\cos^2(3t) + \dots}
…cos⁡2(t)−4cos⁡(t)sin⁡(3t)+4sin⁡2(3t)+9‾\overline{\dots \cos^2(t) - 4\cos(t)\sin(3t) + 4\sin^2(3t) + 9}
=14−4(sin⁡(3t)cos⁡(t)−cos⁡(3t)sin⁡(t))= \sqrt{14 - 4(\sin(3t)\cos(t) - \cos(3t)\sin(t))}
=14−4sin⁡(2t)= \sqrt{14 - 4\sin(2t)}

Given ∣rB−rA∣=4|r_B - r_A| = 4
14−4sin⁡(2t)=4\sqrt{14 - 4\sin(2t)} = 4
sin⁡(2t)=−12\sin(2t) = -\frac{1}{2}
2t=7π62t = \frac{7\pi}{6}
t=7π12t = \frac{7\pi}{12} s (first positive solution)

Position of A
rA=−2cos⁡(3t)i+2sin⁡(3t)j+2kr_A = -2\cos(3t)i + 2\sin(3t)j + 2k
=−2cos⁡(7π4)i+2sin⁡(7π4)j+2k= -2\cos(\frac{7\pi}{4})i + 2\sin(\frac{7\pi}{4})j + 2k
=−2i−2j+2k= -\sqrt{2}i - \sqrt{2}j + 2k (m)

Marking Criteria
DescriptorMarks

correctly determines the expression for the position of Object A

1

correctly determines the expression for the position of Object B

1

determines an expression to represent the relative position of Objects A and B

1

determines an expression to represent the distance (or square of the distance) between the objects

1

uses a trigonometric identity to determine an expression in terms of a single trigonometric function that represents the distance (or square of the distance) between the objects

1

determines the first time that Object A is 4 metres away from Object B

1

determines position of Object A

1
Q16
2025
VCAA
Paper 2
1 mark
Q16
1 mark

The position vector of a particle at time tt is given by r∼(t)=ne−2ti∼−t2j∼\underset{\sim}{\text{r}}(t) = ne^{-2t}\underset{\sim}{\text{i}} - t^2\underset{\sim}{\text{j}}, where nn is a positive constant.

For what value of nn is the particle's acceleration perpendicular to its velocity when t=12t = \frac{1}{2}?

A

2e2e

B

e0.52\frac{e^{0.5}}{2}

C

e2\frac{e}{2}

D

e22\frac{e}{2\sqrt{2}}

Reveal Answer
A

2e2e

Incorrect. This value results from an algebraic error when solving the dot product equation −8n2e−2+2=0-8n^2e^{-2} + 2 = 0, perhaps by incorrectly isolating nn.

B

e0.52\frac{e^{0.5}}{2}

Incorrect. This value likely comes from incorrectly evaluating the derivatives at a different time or making an exponent error during differentiation.

C

e2\frac{e}{2}

Correct Answer

Correct. The velocity is v(t)=−2ne−2ti−2tj\mathbf{v}(t) = -2ne^{-2t}\mathbf{i} - 2t\mathbf{j} and acceleration is a(t)=4ne−2ti−2j\mathbf{a}(t) = 4ne^{-2t}\mathbf{i} - 2\mathbf{j}. Setting their dot product at t=1/2t=1/2 to zero gives −8n2e−2+2=0-8n^2e^{-2} + 2 = 0, which yields n=e2n = \frac{e}{2}.

D

e22\frac{e}{2\sqrt{2}}

Incorrect. This answer stems from a miscalculation in the dot product, such as setting 8n2e−2=18n^2e^{-2} = 1 instead of 22, leading to n=e22n = \frac{e}{2\sqrt{2}}.

Q18
2025
QCAA
Paper 1
6 marks
Q18
6 marks

An object is projected at an acute angle of θ\theta below the horizontal, with an initial speed of 30 m s−130 \text{ m s}^{-1} from a position 90 m90 \text{ m} above ground level.

The object hits the ground 90 m90 \text{ m} horizontally from its projection point.

Use vector calculus to determine θ\theta in its simplest form.

Assume that the magnitude of mean acceleration due to gravity on Earth is 10 m s−210 \text{ m s}^{-2} and that there is no air resistance.

Reveal Answer

Let i^\hat{\boldsymbol{i}} and j^\hat{\boldsymbol{j}} be the horizontal and vertical unit vectors and assume downwards as the positive direction.

Let tt represent the time (s) after projection.

a(t)=10j^\boldsymbol{a}(t) = 10\hat{\boldsymbol{j}}

v(t)=∫a(t)dt=10tj^+c\boldsymbol{v}(t) = \int \boldsymbol{a}(t) dt = 10t\hat{\boldsymbol{j}} + \boldsymbol{c}

Given v(0)=30cos⁡(θ)i^+30sin⁡(θ)j^\boldsymbol{v}(0) = 30\cos(\theta)\hat{\boldsymbol{i}} + 30\sin(\theta)\hat{\boldsymbol{j}}

v(t)=30cos⁡(θ)i^+(30sin⁡(θ)+10t)j^\boldsymbol{v}(t) = 30\cos(\theta)\hat{\boldsymbol{i}} + (30\sin(\theta) + 10t)\hat{\boldsymbol{j}}

r(t)=∫v(t)dt=30cos⁡(θ)ti^+(30sin⁡(θ)t+5t2)j^+c\begin{align*} \boldsymbol{r}(t) &= \int \boldsymbol{v}(t) dt\\ &= 30\cos(\theta)t\hat{\boldsymbol{i}} + \left(30\sin(\theta)t + 5t^2\right)\hat{\boldsymbol{j}} + \boldsymbol{c} \end{align*}

Let origin be at the release point: r(0)=0i^+0j^\boldsymbol{r}(0) = 0\hat{\boldsymbol{i}} + 0\hat{\boldsymbol{j}}

r(t)=30cos⁡(θ)ti^+(30sin⁡(θ)t+5t2)j^\boldsymbol{r}(t) = 30\cos(\theta)t\hat{\boldsymbol{i}} + \left(30\sin(\theta)t + 5t^2\right)\hat{\boldsymbol{j}}

Consider the object's position as it hits the ground
30cos⁡(θ)t=90…(1)30\cos(\theta)t = 90 \quad \dots (1)
30sin⁡(θ)t+5t2=90…(2)30\sin(\theta)t + 5t^2 = 90 \quad \dots (2)

From (1), t=3cos⁡(θ)t = \frac{3}{\cos(\theta)}

Substituting into (2):

30sin⁡(θ)(3cos⁡(θ))+5(3cos⁡(θ))2=9090tan⁡(θ)+45sec⁡2(θ)=902tan⁡(θ)+(1+tan⁡2(θ))=2tan⁡2(θ)+2tan⁡(θ)−1=0\begin{align*} 30\sin(\theta)\left(\frac{3}{\cos(\theta)}\right) + 5\left(\frac{3}{\cos(\theta)}\right)^2 &= 90\\ 90\tan(\theta) + 45\sec^2(\theta) &= 90\\ 2\tan(\theta) + \left(1 + \tan^2(\theta)\right) &= 2\\ \tan^2(\theta) + 2\tan(\theta) - 1 &= 0 \end{align*}

Using the quadratic formula

tan⁡(θ)=−2±4−4×1×−12=−2±222=−1±2\begin{align*} \tan(\theta) &= \frac{-2 \pm \sqrt{4 - 4 \times 1 \times -1}}{2}\\ &= \frac{-2 \pm 2\sqrt{2}}{2}\\ &= -1 \pm \sqrt{2} \end{align*}

As θ\theta is acute, θ=tan⁡−1(−1+2)\theta = \tan^{-1}\left(-1 + \sqrt{2}\right).

Marking Criteria
DescriptorMarks

correctly determines the position vector of the object showing evidence of the use of vector calculus

1

determines two simultaneous equations in terms of θ\theta and tt

1

uses simultaneous equations to determine equation in terms of θ\theta

1

determines equation in terms of tan⁡(θ)\tan(\theta)

1

determines both possible values of tan⁡(θ)\tan(\theta)

1

determines θ\theta by evaluating the reasonableness of the solution

1
Q8
2025
QCAA
Paper 1
1 mark
Q8
1 mark

The position vectors of two objects over time, tt, where t≥0t \ge 0, are given by

r1(t)=−2i^+t2j^r_1(t) = -2\hat{i} + t^2\hat{j} r2(t)=ati^+4j^(where a∈R)r_2(t) = at\hat{i} + 4\hat{j} \quad (\text{where } a \in R)

Given that the two objects collide, the value of aa is

A

2

B

1

C

-1

D

-2

Reveal Answer
A

2

If a=2a = 2, equating the i^\hat{i} components (−2=2t-2 = 2t) would mean the collision happens at t=−1t = -1, but time must be non-negative (t≥0t \ge 0).

B

1

If a=1a = 1, the second object's x-position at the collision time (t=2t = 2) would be 22, which does not match the first object's constant x-position of −2-2.

C

-1

Correct Answer

For a collision, the j^\hat{j} components must be equal (t2=4  ⟹  t=2t^2 = 4 \implies t = 2, since t≥0t \ge 0). Equating the i^\hat{i} components at t=2t = 2 gives −2=2a-2 = 2a, which simplifies to a=−1a = -1.

D

-2

If a=−2a = -2, the second object's x-position at the collision time (t=2t = 2) would be −4-4, missing the first object which is always at x=−2x = -2.

Q3
2021
QCAA
Paper 1
1 mark
Q3
1 mark

An object has a velocity v(t)=e−2ti^+(1t)k^v(t) = e^{-2t}\hat{i} + \left(\frac{1}{t}\right)\hat{k}, where tt represents time (t>0t > 0).

The displacement r(t)r(t) of the object could be

A

−2e−2ti^+ln⁡(t)k^-2e^{-2t}\hat{i} + \ln(t)\hat{k}

B

−2e−2ti^−1t2k^-2e^{-2t}\hat{i} - \frac{1}{t^2}\hat{k}

C

−12e−2ti^+ln⁡(t)k^-\frac{1}{2}e^{-2t}\hat{i} + \ln(t)\hat{k}

D

−12e−2ti^−1t2k^-\frac{1}{2}e^{-2t}\hat{i} - \frac{1}{t^2}\hat{k}

Reveal Answer
A

−2e−2ti^+ln⁡(t)k^-2e^{-2t}\hat{i} + \ln(t)\hat{k}

This option incorrectly differentiates the i^\hat{i} component instead of integrating it. The integral of e−2te^{-2t} is −12e−2t-\frac{1}{2}e^{-2t}, not −2e−2t-2e^{-2t}.

B

−2e−2ti^−1t2k^-2e^{-2t}\hat{i} - \frac{1}{t^2}\hat{k}

This option represents the acceleration (derivative of velocity) rather than displacement. Both components were differentiated: ddt(e−2t)=−2e−2t\frac{d}{dt}(e^{-2t}) = -2e^{-2t} and ddt(t−1)=−t−2\frac{d}{dt}(t^{-1}) = -t^{-2}.

C

−12e−2ti^+ln⁡(t)k^-\frac{1}{2}e^{-2t}\hat{i} + \ln(t)\hat{k}

Correct Answer

Displacement is found by integrating the velocity function with respect to time: ∫e−2tdt=−12e−2t\int e^{-2t} dt = -\frac{1}{2}e^{-2t} and ∫1tdt=ln⁡(t)\int \frac{1}{t} dt = \ln(t).

D

−12e−2ti^−1t2k^-\frac{1}{2}e^{-2t}\hat{i} - \frac{1}{t^2}\hat{k}

The k^\hat{k} component is incorrect because it represents the derivative of 1t\frac{1}{t} (which is −1t2-\frac{1}{t^2}) rather than the integral (which is ln⁡(t)\ln(t)).

Q16
2024
VCAA
Paper 2
1 mark
Q16
1 mark

Particle 1 has position vector r∼1(t)=cos⁡(t)i∼+sin⁡(t)j∼+sin⁡(2t)k∼\underset{\sim}{\mathrm{r}}_1(t) = \cos(t)\underset{\sim}{\mathrm{i}} + \sin(t)\underset{\sim}{\mathrm{j}} + \sqrt{\sin(2t)}\underset{\sim}{\mathrm{k}} and Particle 2 has position vector r∼2(t)=sin⁡(t)i∼+cos⁡(t)j∼+sin⁡(2t)k∼\underset{\sim}{\mathrm{r}}_2(t) = \sin(t)\underset{\sim}{\mathrm{i}} + \cos(t)\underset{\sim}{\mathrm{j}} + \sqrt{\sin(2t)}\underset{\sim}{\mathrm{k}}, where tt is measured in seconds and t∈(0,π2]t \in \left(0, \frac{\pi}{2}\right].

The number of times the velocity of Particle 1 is perpendicular to the position vector r∼2(t)\underset{\sim}{\mathrm{r}}_2(t) during the first π2\frac{\pi}{2} seconds is

A

11

B

22

C

33

D

44

Reveal Answer
A

11

Correct Answer

The dot product of the velocity of Particle 1 and the position of Particle 2 simplifies to 2cos⁡(2t)2\cos(2t). Setting this to zero for perpendicularity gives exactly one solution, t=π4t = \frac{\pi}{4}, in the interval (0,π2](0, \frac{\pi}{2}].

B

22

This is incorrect because the equation 2cos⁡(2t)=02\cos(2t) = 0 only has one solution in the given interval. Two solutions would occur if the interval was extended to (0,π](0, \pi].

C

33

This is incorrect because the dot product 2cos⁡(2t)=02\cos(2t) = 0 only yields one valid time t=π4t = \frac{\pi}{4} within the restricted domain of (0,π2](0, \frac{\pi}{2}].

D

44

This is incorrect because the vectors are only perpendicular once in the given interval. Four solutions would require a much larger time interval, such as (0,2π](0, 2\pi].

Q9
2020
QCAA
Paper 2
1 mark
Q9
1 mark

Two objects, P and Q, move in three-dimensional space such that their positions, r\mathbf{r}, over time, tt, are described by the following vectors until they collide.

rP=(t2−4t)i+(2t2−t+3)j−(6−5t)k\mathbf{r}_P = (t^2 - 4t)\mathbf{i} + (2t^2 - t + 3)\mathbf{j} - (6 - 5t)\mathbf{k}
rQ=(−t2+2t)i+(3t+t2)j+t2k\mathbf{r}_Q = (-t^2 + 2t)\mathbf{i} + (3t + t^2)\mathbf{j} + t^2\mathbf{k}

The objects will collide at

A

t=0t = 0

B

t=1t = 1

C

t=2t = 2

D

t=3t = 3

Reveal Answer
A

t=0t = 0

At t=0t=0, only the i\mathbf{i}-components are equal (00). The objects are at different positions because the j\mathbf{j} and k\mathbf{k} components do not match (3≠03 \neq 0 and −6≠0-6 \neq 0).

B

t=1t = 1

At t=1t=1, only the j\mathbf{j}-components are equal (44). The i\mathbf{i} and k\mathbf{k} components differ, meaning the objects have not collided.

C

t=2t = 2

At t=2t=2, only the k\mathbf{k}-components are equal (44). The i\mathbf{i} and j\mathbf{j} components differ, so the objects are not at the same point in space.

D

t=3t = 3

Correct Answer

For a collision, rP=rQ\mathbf{r}_P = \mathbf{r}_Q for all components simultaneously. Solving the equations for each component (2t2−6t=02t^2-6t=0, t2−4t+3=0t^2-4t+3=0, and t2−5t+6=0t^2-5t+6=0) reveals that t=3t=3 is the only common solution.

Q19
2023
QCAA
Paper 1
6 marks
Q19
6 marks

Object A is released from the origin with constant velocity, vA\boldsymbol{v}_A, such that its position after tt seconds is given by

rA=23ti+3tj+2tk,t≥0\boldsymbol{r}_A = 2\sqrt{3}t\boldsymbol{i} + 3t\boldsymbol{j} + 2t\boldsymbol{k}, t \ge 0.

At a later time, object B is released from point P(33,6,0)P(3\sqrt{3}, 6, 0) and travels towards point Q(53,8,4)Q(5\sqrt{3}, 8, 4) with constant velocity, vB\boldsymbol{v}_B, such that ∣vB∣=2∣vA∣|\boldsymbol{v}_B| = \sqrt{2}|\boldsymbol{v}_A|.

Given that objects A and B collide, determine the time between the release of the two objects.

Assume all positions are given in metres and all velocities are given in metres per second.

Reveal Answer

Method 1

rA=23ti^+3tj^+2tk^,t≥0r_A = 2\sqrt{3}t\hat{i} + 3t\hat{j} + 2t\hat{k}, t \ge 0
vA=23i^+3j^+2k^v_A = 2\sqrt{3}\hat{i} + 3\hat{j} + 2\hat{k}

∣vA∣=(23)2+32+22=5|v_A| = \sqrt{(2\sqrt{3})^2 + 3^2 + 2^2} = 5
∴∣vB∣=52 ms−1\therefore |v_B| = 5\sqrt{2} \text{ ms}^{-1}

Given vBv_B is constant, the position of object B from the origin as it moves can be represented along the line rB=b+ld,l∈Rr_B = b + ld, l \in \mathbb{R}, where
b=33i^+6j^b = 3\sqrt{3}\hat{i} + 6\hat{j}
d=53i^+8j^+4k^−(33i^+6j^)d = 5\sqrt{3}\hat{i} + 8\hat{j} + 4\hat{k} - (3\sqrt{3}\hat{i} + 6\hat{j})
=23i^+2j^+4k^= 2\sqrt{3}\hat{i} + 2\hat{j} + 4\hat{k}

rB=(33+23l)i^+(6+2l)j^+4lk^r_B = (3\sqrt{3} + 2\sqrt{3}l)\hat{i} + (6+2l)\hat{j} + 4l\hat{k}

Let the objects collide when t=t1t = t_1
rA(t1)=23t1i^+3t1j^+2t1k^r_A(t_1) = 2\sqrt{3}t_1\hat{i} + 3t_1\hat{j} + 2t_1\hat{k}

Collision occurs when rA=rBr_A = r_B

Equating i^,j^\hat{i}, \hat{j} and k^\hat{k} components:
33+23l=23t13\sqrt{3} + 2\sqrt{3}l = 2\sqrt{3}t_1 ... (1)
6+2l=3t16 + 2l = 3t_1 ... (2)
4l=2t14l = 2t_1 ... (3)

From (3), l=t12l = \frac{t_1}{2}
Substituting into (1):
33+23(t12)=23t13\sqrt{3} + 2\sqrt{3}\left(\frac{t_1}{2}\right) = 2\sqrt{3}t_1
t1=3 st_1 = 3 \text{ s}

The collision occurs 3 seconds after object A is released.
Collision point is
23(3)i^+3(3)j^+2(3)k^2\sqrt{3}(3)\hat{i} + 3(3)\hat{j} + 2(3)\hat{k}
=63i^+9j^+6k^= 6\sqrt{3}\hat{i} + 9\hat{j} + 6\hat{k}

Distance object B travels from P to collision point is
(63−33)2+(6−9)2+62=62\sqrt{(6\sqrt{3}-3\sqrt{3})^2 + (6-9)^2 + 6^2} = 6\sqrt{2}

Time taken for object B to reach collision point is
ds=6252\frac{d}{s} = \frac{6\sqrt{2}}{5\sqrt{2}}
=1.2 s= 1.2 \text{ s}

Time between the release of the two objects is
3−1.2=1.8 s3 - 1.2 = 1.8 \text{ s}

Marking Criteria
DescriptorMarks

correctly determines the value of ∣vB∣|v_B|

1

determines a vector in terms of a parameter representing the position of object B from the origin as it moves

1

determines at least two simultaneous equations based on the collision of the objects

1

determines time from release to collision for object A

1

determines distance that object B travels to reach collision point

1

determines time between the release of the two objects

1
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 (m s−2)\left(\text{m s}^{-2}\right) of the falling instrument can be modelled by 9.8−0.1v9.8 - 0.1v, where vv is its velocity (m s−1)\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 m s−120 \text{ m s}^{-1}. However, if it hits with a speed above 50 m s−150 \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=9.8−0.1va = 9.8 - 0.1v
vdvdx=9.8−0.1vv \frac{dv}{dx} = 9.8 - 0.1v
∫v9.8−0.1vdv=∫dx\int \frac{v}{9.8 - 0.1v} dv = \int dx
∫vv−98dv=∫−110dx\int \frac{v}{v - 98} dv = \int \frac{-1}{10} dx
∫1+98v−98dv=∫−110dx\int 1 + \frac{98}{v - 98} dv = \int \frac{-1}{10} dx
v+98ln⁡∣v−98∣=−x10+cv + 98 \ln|v - 98| = \frac{-x}{10} + c

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

v+98ln⁡∣v−98∣=−x10+98ln⁡(98)v + 98 \ln|v - 98| = \frac{-x}{10} + 98 \ln(98)
Let the distance to the ocean surface be hh metres.
Consider time of drop for each required velocity
When v=20 m s−1,x=hv = 20 \text{ m s}^{-1}, x = h
20+98ln⁡∣−78∣=−h10+98ln⁡(98)20 + 98 \ln|-78| = \frac{-h}{10} + 98 \ln(98)
h=23.7 mh = 23.7 \text{ m}

When v=50 m s−1,x=hv = 50 \text{ m s}^{-1}, x = h
50+98ln⁡∣−48∣=98ln⁡(98)−h1050 + 98 \ln|-48| = 98 \ln(98) - \frac{h}{10}
h=199.5 mh = 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
DescriptorMarks

correctly establishes a differential equation in terms of vv and xx

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
Q14
2021
QCAA
Paper 1
6 marks
Q14

An object is projected vertically upwards from ground level. After the object has been in motion for tt seconds, its position vector through the air, in metres, is modelled by

r(t)=5t(8−t)j^\boldsymbol{r}(t) = 5t(8-t)\boldsymbol{\hat{j}}

Q14a
2 marks

Determine the velocity of the object through the air, v(t)\boldsymbol{v}(t), in metres per second.

Reveal Answer

r(t)=5t(8−t)jr(t) = 5t(8 - t)j
=(40t−5t2)j= (40t - 5t^2)j
v(t)=(40−10t)jv(t) = (40 - 10t)j

Marking Criteria
DescriptorMarks

correctly expands the given expression

1

determines v(t)v(t)

1
Q14b
2 marks

Determine the number of seconds until the object reaches its maximum height.

Reveal Answer

Maximum height occurs when v(t)=0jv(t) = 0j
40−10t=040 - 10t = 0
10t=4010t = 40
t=4t = 4 s

Marking Criteria
DescriptorMarks

establishes equation in terms of tt

1

determines tt

1
Q14c
2 marks

Determine the maximum height that the object reaches, in metres.

Reveal Answer

Maximum height =5×4(8−4)= 5 \times 4(8 - 4)
=20×4= 20 \times 4
=80= 80 m

Marking Criteria
DescriptorMarks

substitutes result from 14b) into expression for position

1

determines maximum height

1
Q17
2024
QCAA
Paper 2
6 marks
Q17
6 marks

An object moves with a constant speed of vv in a circular path.

The position vector of the object is given by

r=rcos⁡(ωt)i^+rsin⁡(ωt)j^r = r\cos(\omega t)\hat{i} + r\sin(\omega t)\hat{j}

where

  • rr is the radius (metres) of the circle
  • ω\omega is the angular velocity (radians per second)
  • tt is the time (seconds) of motion for t≥0t \ge 0.

Use vector calculus to prove that the magnitude of the acceleration of the object is ∣a∣=v2r|a| = \frac{v^2}{r}.

Reveal Answer

Given r=rcos⁡(ωt)i^+rsin⁡(ωt)j^r = r\cos(\omega t)\hat{i} + r\sin(\omega t)\hat{j}

v=drdt=−rωsin⁡(ωt)i^+rωcos⁡(ωt)j^v = \frac{dr}{dt} = -r\omega\sin(\omega t)\hat{i} + r\omega\cos(\omega t)\hat{j}

a=dvdt=−rω2cos⁡(ωt)i^−rω2sin⁡(ωt)j^a = \frac{dv}{dt} = -r\omega^2\cos(\omega t)\hat{i} - r\omega^2\sin(\omega t)\hat{j}

v=∣v∣v = |v|

=(−rωsin⁡(ωt))2+(rωcos⁡(ωt))2= \sqrt{(-r\omega\sin(\omega t))^2 + (r\omega\cos(\omega t))^2}

=r2ω2(sin⁡2(ωt)+cos⁡2(ωt))= \sqrt{r^2\omega^2(\sin^2(\omega t) + \cos^2(\omega t))}

=rω= r\omega

∣a∣=(−rω2cos⁡(ωt))2+(−rω2sin⁡(ωt))2|a| = \sqrt{(-r\omega^2\cos(\omega t))^2 + (-r\omega^2\sin(\omega t))^2}

∣a∣=r2ω4(cos⁡2(ωt)+sin⁡2(ωt))|a| = \sqrt{r^2\omega^4(\cos^2(\omega t) + \sin^2(\omega t))}

=rω2=r2ω2r= r\omega^2 = \frac{r^2\omega^2}{r}

=v2r= \frac{v^2}{r}

Marking Criteria
DescriptorMarks

correctly determines the velocity vector

1

determines an acceleration vector

1

determines a simplified expression for the modulus of v

1

determines an expression for the modulus of a

1

correctly completes the proof based on prior evidence

1

shows logical organisation of a fully attempted proof, communicating key steps

1
Q13
2022
VCAA
Paper 2
1 mark
Q13
1 mark

The acceleration of a body moving in a plane is given by r¨∼(t)=sin⁡(t)i∼+2cos⁡(t)j∼\underset{\sim}{\ddot{r}}(t) = \sin(t)\underset{\sim}{i} + 2\cos(t)\underset{\sim}{j}, where t≥0t \ge 0.

Given that r˙∼(0)=2i∼+j∼\underset{\sim}{\dot{r}}(0) = 2\underset{\sim}{i} + \underset{\sim}{j}, the velocity of the body at time tt, r˙∼(t)\underset{\sim}{\dot{r}}(t), is given by

A

−cos⁡(t)i∼+2sin⁡(t)j∼-\cos(t)\underset{\sim}{i} + 2\sin(t)\underset{\sim}{j}

B

(3−cos⁡(t))i∼+(2sin⁡(t)+1)j∼(3 - \cos(t))\underset{\sim}{i} + (2\sin(t) + 1)\underset{\sim}{j}

C

(1+cos⁡(t))i∼+(2sin⁡(t)+1)j∼(1 + \cos(t))\underset{\sim}{i} + (2\sin(t) + 1)\underset{\sim}{j}

D

(2+sin⁡(t))i∼+(2cos⁡(t)−1)j∼(2 + \sin(t))\underset{\sim}{i} + (2\cos(t) - 1)\underset{\sim}{j}

E

(1+cos⁡(t))i∼+(1−2sin⁡(t))j∼(1 + \cos(t))\underset{\sim}{i} + (1 - 2\sin(t))\underset{\sim}{j}

Reveal Answer
A

−cos⁡(t)i∼+2sin⁡(t)j∼-\cos(t)\underset{\sim}{i} + 2\sin(t)\underset{\sim}{j}

This option represents the antiderivative of the acceleration but fails to include the constant of integration C∼\underset{\sim}{C} required by the initial velocity.

B

(3−cos⁡(t))i∼+(2sin⁡(t)+1)j∼(3 - \cos(t))\underset{\sim}{i} + (2\sin(t) + 1)\underset{\sim}{j}

Correct Answer

Integrating the acceleration gives −cos⁡(t)i∼+2sin⁡(t)j∼+C∼-\cos(t)\underset{\sim}{i} + 2\sin(t)\underset{\sim}{j} + \underset{\sim}{C}. Applying the initial condition r˙∼(0)=2i∼+j∼\underset{\sim}{\dot{r}}(0) = 2\underset{\sim}{i} + \underset{\sim}{j} yields C∼=3i∼+j∼\underset{\sim}{C} = 3\underset{\sim}{i} + \underset{\sim}{j}, which gives this correct velocity vector.

C

(1+cos⁡(t))i∼+(2sin⁡(t)+1)j∼(1 + \cos(t))\underset{\sim}{i} + (2\sin(t) + 1)\underset{\sim}{j}

This option incorrectly integrates sin⁡(t)\sin(t) as cos⁡(t)\cos(t) instead of −cos⁡(t)-\cos(t), which leads to an incorrect constant of integration for the i∼\underset{\sim}{i} component.

D

(2+sin⁡(t))i∼+(2cos⁡(t)−1)j∼(2 + \sin(t))\underset{\sim}{i} + (2\cos(t) - 1)\underset{\sim}{j}

This option incorrectly adds the initial velocity components directly to the acceleration components instead of properly integrating the acceleration function.

E

(1+cos⁡(t))i∼+(1−2sin⁡(t))j∼(1 + \cos(t))\underset{\sim}{i} + (1 - 2\sin(t))\underset{\sim}{j}

This option uses incorrect signs during integration, mistakenly integrating sin⁡(t)\sin(t) to cos⁡(t)\cos(t) and 2cos⁡(t)2\cos(t) to −2sin⁡(t)-2\sin(t).

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