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Unit 3 · Topic 4.2 · Vector calculus

Collision problems

This guide puts position vectors to work on moving particles. You will move between position, velocity and acceleration, turn a condition into an equation in $t$, and decide whether two particles collide or only cross paths. Nearly every recent paper has a question from this topic, and collision questions are the most common. Work through it in order, and tick off each skill once you can do it without help.

Skills in this topic

Only tick a skill when you could do it in an exam.

Differentiate a position vector to get velocity and acceleration, and find the speed. Part 1
Integrate an acceleration or a velocity vector, finding the constant vector from the starting conditions. Part 1
Show that a particle passes through a point, or use a given point to find an unknown constant. Part 2
Turn a condition, such as being at rest or position perpendicular to velocity, into an equation and solve for the time. Part 2
Decide whether two particles collide, and find where their paths cross. Part 3
Find the distance between two particles as a function of time, then minimise it or set it to a given value. Part 4

Part 1

Position, velocity and acceleration

The first question to ask is how many particles you are dealing with. With one particle, the questions move between its position, velocity and acceleration, or impose a condition on it. With two, they ask whether the particles collide, whether their paths cross, or how far apart they are. Parts 1 and 2 cover one particle and Parts 3 and 4 cover two.

To differentiate or integrate a vector, work on each component separately with the usual rules. Differentiate to move down from position to velocity to acceleration, and integrate to move back up.

The idea

What velocity and acceleration vectors look like

Differentiate down

$$\begin{aligned}\boldsymbol{r}(t)&=x(t)\hat{i}+y(t)\hat{j}+z(t)\hat{k}\\\boldsymbol{v}(t)&=\dot{x}(t)\hat{i}+\dot{y}(t)\hat{j}+\dot{z}(t)\hat{k}\\\boldsymbol{a}(t)&=\ddot{x}(t)\hat{i}+\ddot{y}(t)\hat{j}+\ddot{z}(t)\hat{k}\end{aligned}$$

A dot means one derivative with respect to $t$.

Speed

$$|\boldsymbol{v}(t)|=\sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2}$$

Velocity is a vector. Speed is its magnitude, so it is a number.

Integrate up

$$\begin{aligned}\boldsymbol{v}(t)&=\int\boldsymbol{a}(t)\,dt\\\boldsymbol{r}(t)&=\int\boldsymbol{v}(t)\,dt\end{aligned}$$

Each integral adds a constant vector, $\boldsymbol{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$. Find it from the position or velocity at a known time, usually $t=0$.

Practise

Exam questions: moving between position, velocity and acceleration

2025 Paper 1, Question 6

Multiple choice · Technology-free · [1 mark]

At time $t$, a particle travels with a velocity of $\boldsymbol{v}=\left(\dfrac{2}{1+t^2}\right)\hat{i}-2t\hat{j}$. Determine a general expression for the position vector, $\boldsymbol{r}$, of the particle during this motion.

Integrate each component. The $\hat{i}$ component is 2 times a standard integral.

$$\begin{aligned}\int\frac{2}{1+t^2}\,dt&=2\tan^{-1}(t)\\\int-2t\,dt&=-t^2\end{aligned}$$

(A) and (C) have $-2\hat{j}$, which is the derivative of $-2t$ rather than its integral. (C) and (D) halve the 2 instead of keeping it.

Answer
(B) $\boldsymbol{r}=2\tan^{-1}(t)\hat{i}-t^2\hat{j}+\boldsymbol{c}$

2021 Paper 1, Question 3

Multiple choice · Technology-free · [1 mark]

An object has a velocity $\boldsymbol{v}(t)=e^{-2t}\hat{i}+\left(\dfrac{1}{t}\right)\hat{k}$, where $t$ represents time $(t>0)$. The displacement $\boldsymbol{r}(t)$ of the object could be

$$\begin{aligned}\int e^{-2t}\,dt&=-\tfrac{1}{2}e^{-2t}\\\int\frac{1}{t}\,dt&=\ln(t)\end{aligned}$$

The second integral is $\ln(t)$ rather than $\ln|t|$ because $t>0$.

(A) and (B) multiply by $-2$, which is what differentiating does. (B) and (D) differentiate $\frac{1}{t}$ instead of integrating it. The words "could be" are there because the constant has been left out.

Answer
(C) $-\tfrac{1}{2}e^{-2t}\hat{i}+\ln(t)\hat{k}$

2023 Paper 1, Question 1

Multiple choice · Technology-free · [1 mark]

The position of a particle is given by $\boldsymbol{r}=(t+2)\hat{i}+t^2\hat{j}$ for $t\ge0$. Determine the corresponding Cartesian equation.

There are no trigonometric functions here, so make $t$ the subject of the $x$ equation and substitute.

$$\begin{aligned}x=t+2\ &\Rightarrow\ t=x-2\\y=t^2&=(x-2)^2\\&=x^2-4x+4\end{aligned}$$

This is the same skill as Topic 4.1.

Answer
(C) $y=x^2-4x+4$

2021 Paper 1, Question 14

Short response · Technology-free · [6 marks]

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

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

a)Determine the velocity of the object through the air, $\boldsymbol{v}(t)$, in metres per second.[2 marks]
b)Determine the number of seconds until the object reaches its maximum height.[2 marks]
c)Determine the maximum height that the object reaches, in metres.[2 marks]

Part a)

$$\boldsymbol{r}(t)=(40t-5t^2)\hat{j}$$

1 markCorrectly expands the given expression.

$$\boldsymbol{v}(t)=(40-10t)\hat{j}$$

1 markDetermines $\boldsymbol{v}(t)$.

The marking guide does not accept $40-10t$ on its own. The velocity is a vector, so keep the $\hat{j}$. The product rule is an equivalent method.

Part b)

At the maximum height the object is momentarily at rest, so $\boldsymbol{v}(t)=0\hat{j}$.

$$40-10t=0$$

1 markEstablishes an equation in terms of $t$.

$$t=4\text{ s}$$

1 markDetermines $t$.

Part c)

$$5\times4(8-4)$$

1 markSubstitutes the result from part b) into the expression for position.

$$=80\text{ m}$$

1 markDetermines the maximum height.
Answer
a) $\boldsymbol{v}(t)=(40-10t)\hat{j}$ b) 4 s c) 80 m

Part 2

One particle: points and conditions

Two kinds of question use a single position vector at a particular time. In the first, you substitute a point and solve, in either direction. In the second, the question states a condition in words and you turn it into an equation in $t$.

Show a particle passes through a point

  1. 1Set one component equal to the matching coordinate and solve for $t$.
  2. 2Substitute that $t$ into the other components.
  3. 3If every component matches, the particle passes through the point at that time.

Find an unknown constant from a point

  1. 1Use a component with no unknown constant to find $t$. Reject any time outside the domain.
  2. 2Substitute that $t$ into the component that contains the constant, and solve for it.

Turning words into an equation

The first two conditions have been tested on past papers. The others haven't been asked yet, but I can see them turning up.

Condition in wordsEquation
At rest, at its maximum height, or at a turning point$\boldsymbol{v}(t)=\boldsymbol{0}$
Position perpendicular to velocity$\boldsymbol{r}\cdot\boldsymbol{v}=0$
Velocity perpendicular to a vector $\boldsymbol{d}$$\boldsymbol{v}\cdot\boldsymbol{d}=0$
Position parallel to velocity$\boldsymbol{r}=k\boldsymbol{v}$
A given distance $k$ from the origin$|\boldsymbol{r}|=k$
A given speed $k$$|\boldsymbol{v}|=k$
Angle $\theta$ between position and velocity$\boldsymbol{r}\cdot\boldsymbol{v}=|\boldsymbol{r}||\boldsymbol{v}|\cos(\theta)$

For a vector to be zero, every component must be zero at the same time. In 2021 Paper 1, Question 14 in Part 1 there was only a $\hat{j}$ component, so one equation was enough.

Practise

Exam question: a condition on one particle

2024 Paper 2, Question 15 a) and b) also use a point on a path. They are in Part 4 with the rest of that question.

2022 Paper 1, Question 11 a) and b)

Short response · Technology-free · [3 of 6 marks]

The position vector of a particle, $\boldsymbol{r}_1$ (cm), over time, $t$ (s), is given by

$$\begin{aligned}\boldsymbol{r}_1(t)=&(2t+1)\hat{i}+(t+3)\hat{j}\\&-(2t-3)\hat{k}\end{aligned}$$

a)Determine the velocity vector of the particle.[1 mark]
b)Determine the time when the position vector of the particle is perpendicular to its velocity vector.[2 marks]

Part c) is in Part 3.

Part a)

$$\boldsymbol{v}_1(t)=2\hat{i}+\hat{j}-2\hat{k}$$

1 markCorrectly determines the velocity vector.

Part b)

Perpendicular means the scalar product is zero.

$$\begin{pmatrix}2t+1\\t+3\\-(2t-3)\end{pmatrix}\cdot\begin{pmatrix}2\\1\\-2\end{pmatrix}=0$$

$$\begin{aligned}&2(2t+1)+1(t+3)\\&\quad+2(2t-3)=0\end{aligned}$$

1 markEstablishes an equation in the form $x_1x_2+y_1y_2+z_1z_2=0$.

$$\begin{aligned}9t-1&=0\\t&=\tfrac{1}{9}\text{ s}\end{aligned}$$

1 markSolves the equation to determine the time.
Answer
a) $2\hat{i}+\hat{j}-2\hat{k}$ b) $t=\frac{1}{9}$ s

Part 3

Two particles: collisions and crossing paths

Two particles collide when they are in the same place at the same time. Their paths cross when they visit the same place, possibly at different times. Every collision happens where the paths cross, but most crossing points are not collisions.

The idea

Same place, same time

Game plan Do they collide?

  1. 1Use the same $t$ in both position vectors.
  2. 2Equate one pair of components, choosing the easier pair, and solve for $t$. Keep only times in the domain.
  3. 3Substitute each time into the other components. They collide only if every component agrees at that time.
  4. 4If they collide, substitute the time into either position vector to find the collision point.

Game plan Where do the paths cross?

  1. 1Give each particle its own parameter, $t_1$ and $t_2$.
  2. 2Equate the $\hat{i}$ components and the $\hat{j}$ components.
  3. 3Solve the two equations simultaneously, then check the third component in 3D.
  4. 4Substitute $t_1$ into the first position vector, or $t_2$ into the second, to find the crossing point.

Collision questions are very common on the QCAA papers. A question on crossing paths hasn't appeared yet, but it is in the syllabus, and the method matters for collisions too. In 3D, two components can agree at a time when the third does not, so always check all three.

Try it yourself

Collide or cross?

Particle A starts at $(-4,-2)$ and moves with velocity $2\hat{i}+\hat{j}$. Choose where B starts and how it moves. The working solves for where the paths cross, with a separate parameter for each particle, and then checks whether the two times agree. Move the time slider to watch both particles.

−6−6−4−4−2−2224466xy A B
path of A path of B paths cross collision

Position vectors

Where the paths cross

Practise

Exam questions: collisions

2025 Paper 1, Question 8

Multiple choice · Technology-free · [1 mark]

The position vectors of two objects over time, $t$, where $t\ge0$, are given by

$$\begin{aligned}\boldsymbol{r}_1(t)&=-2\hat{i}+t^2\hat{j}\\\boldsymbol{r}_2(t)&=at\hat{i}+4\hat{j}\quad(a\in\mathbb{R})\end{aligned}$$

Given that the two objects collide, the value of $a$ is

The $\hat{j}$ components have no unknown, so use them to find the time of the collision.

$$t^2=4\ \Rightarrow\ t=2\quad(t\ge0)$$

At that same time the $\hat{i}$ components must agree.

$$2a=-2\ \Rightarrow\ a=-1$$

(B) comes from using $t=-2$, which the domain rules out.

Answer
(C) $-1$

2020 Paper 2, Question 9

Multiple choice · Technology-active · [1 mark]

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

$$\begin{aligned}\boldsymbol{r}_P=&(t^2-4t)\hat{i}+(2t^2-t+3)\hat{j}\\&-(6-5t)\hat{k}\\\boldsymbol{r}_Q=&(-t^2+2t)\hat{i}+(3t+t^2)\hat{j}+t^2\hat{k}\end{aligned}$$

The objects will collide at

Equate each pair of components. Each equation is a quadratic with two solutions.

$$\begin{aligned}\hat{i}:&\ t^2-4t=-t^2+2t\\&\Rightarrow t=0\text{ or }3\\\hat{j}:&\ 2t^2-t+3=3t+t^2\\&\Rightarrow t=1\text{ or }3\\\hat{k}:&\ 5t-6=t^2\\&\Rightarrow t=2\text{ or }3\end{aligned}$$

Each of (A), (B) and (C) works in one component only. The only time that works in all three is $t=3$.

Answer
(D) $t=3$

2022 Paper 1, Question 11 c)

Short response · Technology-free · [3 of 6 marks]

From part a), the position vector of the first particle is $\boldsymbol{r}_1(t)=(2t+1)\hat{i}+(t+3)\hat{j}-(2t-3)\hat{k}$. The position vector of a second particle, $\boldsymbol{r}_2$ (cm), over time, $t$ (s), is given by

$$\begin{aligned}\boldsymbol{r}_2(t)=&(16-4t)\hat{i}-(3t-13)\hat{j}\\&+2\hat{k}\end{aligned}$$

c)Determine whether the two particles collide.[3 marks]

Let $\boldsymbol{r}_1(t)=\boldsymbol{r}_2(t)$ and equate the components.

$$\begin{aligned}2t+1&=16-4t&&(1)\\t+3&=-3t+13&&(2)\\-2t+3&=2&&(3)\end{aligned}$$

1 markCorrectly equates two pairs of components, one of which must be the $\hat{k}$ component, to form two equations.

From (1) and (2), $t=2.5$ s. From (3), $t=0.5$ s.

1 markSolves the equations to determine two different time values.

The times are not the same, so the particles do not collide.

1 markUses mathematical reasoning to make an appropriate conclusion about whether the particles collide.

The $\hat{i}$ and $\hat{j}$ components agree at $t=2.5$, which is why the marking guide insists on the $\hat{k}$ component. A conclusion with no reasoning doesn't earn the last mark.

Answer
The particles do not collide.

2023 Paper 1, Question 19

Short response · Technology-free · [6 marks]

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

$$\boldsymbol{r}_A=2\sqrt{3}t\hat{i}+3t\hat{j}+2t\hat{k},\quad t\ge0$$

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

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.

$$\begin{aligned}\boldsymbol{v}_A&=2\sqrt{3}\hat{i}+3\hat{j}+2\hat{k}\\|\boldsymbol{v}_A|&=\sqrt{12+9+4}=5\\|\boldsymbol{v}_B|&=5\sqrt{2}\text{ m s}^{-1}\end{aligned}$$

1 markCorrectly determines the value of $|\boldsymbol{v}_B|$.

B moves along the line from $P$ in the direction of $\overrightarrow{PQ}$. Use a parameter $l$ along that line. It is not a time, because B is released later than A.

$$\overrightarrow{PQ}=2\sqrt{3}\hat{i}+2\hat{j}+4\hat{k}$$

$$\begin{aligned}\boldsymbol{r}_B=&\left(3\sqrt{3}+2\sqrt{3}l\right)\hat{i}\\&+(6+2l)\hat{j}+4l\hat{k}\end{aligned}$$

1 markDetermines a vector, in terms of a parameter, for the position of object B as it moves.

Let the objects collide when A has travelled for $t_1$ seconds, and equate the components.

$$\begin{aligned}3\sqrt{3}+2\sqrt{3}l&=2\sqrt{3}t_1&&(1)\\6+2l&=3t_1&&(2)\\4l&=2t_1&&(3)\end{aligned}$$

1 markDetermines at least two simultaneous equations based on the collision of the objects.

From (3), $l=\frac{t_1}{2}$. Substitute into (1).

$$3\sqrt{3}+\sqrt{3}t_1=2\sqrt{3}t_1\ \Rightarrow\ t_1=3\text{ s}$$

1 markDetermines the time from release to collision for object A.

The collision point is $\boldsymbol{r}_A(3)=6\sqrt{3}\hat{i}+9\hat{j}+6\hat{k}$. B travels from $P$ to this point.

$$\begin{aligned}&\sqrt{\left(3\sqrt{3}\right)^2+3^2+6^2}\\&=\sqrt{72}=6\sqrt{2}\text{ m}\end{aligned}$$

1 markDetermines the distance that object B travels to reach the collision point.

B takes this long to reach the collision point:

$$\frac{6\sqrt{2}}{5\sqrt{2}}=1.2\text{ s}$$

A had been moving for 3 s, so the time between the releases is:

$$3-1.2=1.8\text{ s}$$

1 markDetermines the time between the release of the two objects.
Answer
1.8 s

Part 4

Two particles: the distance between them

The displacement of B relative to A is $\boldsymbol{r}_B(t)-\boldsymbol{r}_A(t)$. Its magnitude is the distance between the particles at time $t$, so it is a function of time you can work with. Past papers have asked for the shortest distance, and for the time when the distance equals a given value.

Distance between two particles

$$d(t)=\left|\boldsymbol{r}_B(t)-\boldsymbol{r}_A(t)\right|$$

Working with $d^2$ avoids the square root, and it has its minimum at the same time as $d$.

The idea

Watch the gap, not the paths

Watch it on an exam question

The shortest distance between two particles

Practise

Exam questions: separation

2024 Paper 2, Question 15

Short response · Technology-active · [8 marks]

The vectors representing the position (m) of particles A and B are given by

$$\begin{aligned}\boldsymbol{r}_A&=(4t-9)\hat{i}-2(5-t)\hat{j}-8\hat{k}\\\boldsymbol{r}_B&=(t^2+1)\hat{i}-3\hat{j}+(4-at^2)\hat{k}\end{aligned}$$

respectively, where $t$ is the time of motion for $0\le t\le10$ seconds.

a)Show that particle A passes through the point $P(5,-3,-8)$.[2 marks]
b)Given that particle B also passes through point $P$, determine the value of $a$.[2 marks]
c)Determine the vector that represents the displacement of particle B relative to particle A during the given time of motion. Express your answer in simplest form.[1 mark]
d)Use your result from Question 15c) to determine the shortest distance between particles A and B during the given time of motion.[3 marks]

Part a)

The $\hat{k}$ component of $\boldsymbol{r}_A$ is always $-8$, so check the other two.

$$4t-9=5\ \Rightarrow\ t=3.5\text{ s}$$

1 markCorrectly determines the time taken for particle A to reach $P$ using the $\hat{i}$ component.

$$-2(5-t)=-3\ \Rightarrow\ t=3.5\text{ s}$$

Both components give the same time, so particle A passes through $P$.

1 markCorrectly determines the time taken for particle A to reach $P$ using the $\hat{j}$ component.

The marking guide also accepts finding $t$ from one component, then substituting it to show $\boldsymbol{r}_A(3.5)=5\hat{i}-3\hat{j}-8\hat{k}$.

Part b)

$$t^2+1=5\ \Rightarrow\ t=2\text{ s}$$

Reject $t=-2$, because $0\le t\le10$.

1 markCorrectly determines the time for particle B to reach $P$.

$$\begin{aligned}4-a(2)^2&=-8\\a&=3\end{aligned}$$

1 markDetermines the value of $a$.

A is at $P$ when $t=3.5$ and B is at $P$ when $t=2$. $P$ is a point where the paths cross, but the particles do not collide there.

Part c)

$$\begin{aligned}\boldsymbol{r}_B-\boldsymbol{r}_A=&(t^2-4t+10)\hat{i}\\&+(7-2t)\hat{j}\\&+(12-3t^2)\hat{k}\text{ m}\end{aligned}$$

1 markDetermines the required displacement vector in simplest form.

Part d)

$$\begin{aligned}&|\boldsymbol{r}_B-\boldsymbol{r}_A|^2\\&=(t^2-4t+10)^2+(7-2t)^2\\&\quad+(12-3t^2)^2\\&=10t^4-8t^3-32t^2-108t+293\end{aligned}$$

1 markDetermines an expression that represents the distance, or the square of the distance, between the particles.

The particles are closest when this distance is a minimum.

1 markJustifies when the shortest distance between the particles occurs.

Using technology, the minimum on $0\le t\le10$ is at $t\approx2.03$ s.

$$\text{shortest distance}\approx6.69\text{ m}$$

1 markDetermines the shortest distance between the particles.
Answer
a) $t=3.5$ s from both components b) $a=3$ c) $(t^2-4t+10)\hat{i}+(7-2t)\hat{j}+(12-3t^2)\hat{k}$ d) 6.69 m

2021 Paper 1, Question 19

Short response · Technology-free · Complex unfamiliar · [7 marks]

This question sets the distance between two objects equal to 4 m. You need to integrate both velocities first, then use a compound angle formula to simplify the distance. It is worked in full, with an animation, in Topic 4.1, Part 4. There is also a full walkthrough of this question.

Game plan

Your game plan for this topic

For the whole topic on one page, see Game plan 4 · Motion in a plane in the External Game plan.

  1. 1Count the particles. With one, move between position, velocity and acceleration, or use a condition. With two, it is a collision, crossing paths or a distance.
  2. 2Differentiate down from $\boldsymbol{r}$ to $\boldsymbol{v}$ to $\boldsymbol{a}$. Integrate back up, and find the constant vector from the starting conditions.
  3. 3Turn the words into an equation in $t$, using the table in Part 2.
  4. 4Collision? Use one time for both particles, and check that every component agrees at that time.
  5. 5Crossing paths? Give each particle its own parameter, $t_1$ and $t_2$.
  6. 6Distance? Find $|\boldsymbol{r}_B-\boldsymbol{r}_A|$, then minimise it or set it equal to the given value.

Mistakes I see a lot

  • leaving out the constant of integration, or treating it as a number instead of a vector
  • dropping the unit vector from a velocity
  • checking only two components for a collision in 3D
  • using one parameter for both particles when finding where paths cross
  • keeping a negative time that the domain rules out
  • giving the velocity when the question asks for the speed

Practice

Practice questions

Ten questions of my own, worth 54 marks in total. Each one is labelled with the skill it practises, so if one goes wrong you know which part of the guide to go back to. Check whether each one is Technology-free or Technology-active before you start.

Question 1 of 10

Velocity; position perpendicular to velocity; collision in 3D · Technology-free · Simple familiar · [6 marks]

The position vector of a particle, $\boldsymbol{r}_1$ (cm), at time $t$ (s), is given by

$$\begin{aligned}\boldsymbol{r}_1(t)=&(3t-1)\,\hat{i}+(t+2)\,\hat{j}\\&+(4-2t)\,\hat{k}\end{aligned}$$

a)Determine the velocity vector of the particle.[1 mark]
b)Determine the time when the position vector of the particle is perpendicular to its velocity vector.[2 marks]

The position vector of a second particle, $\boldsymbol{r}_2$ (cm), at time $t$ (s), is given by

$$\begin{aligned}\boldsymbol{r}_2(t)=&(9-2t)\,\hat{i}+(3t-2)\,\hat{j}\\&+(t-2)\,\hat{k}\end{aligned}$$

c)Determine whether the two particles collide.[3 marks]

Part a)

$\boldsymbol{v}_1(t)=3\hat{i}+\hat{j}-2\hat{k}$

1 markCorrectly determines the velocity vector.

Part b)

$\boldsymbol{r}_1\cdot\boldsymbol{v}_1=3(3t-1)+(t+2)-2(4-2t)=14t-9$

1 markCorrectly determines $\boldsymbol{r}_1\cdot\boldsymbol{v}_1$.

$14t-9=0\Rightarrow t=\dfrac{9}{14}$ s

1 markDetermines the time.

Part c)

Equating $\hat{i}$ components: $3t-1=9-2t\Rightarrow t=2$

1 markEquates one pair of components to determine $t$.

At $t=2$: $\hat{j}$: $2+2=4$ and $3(2)-2=4$; $\hat{k}$: $4-4=0$ and $2-2=0$

1 markChecks the remaining components at the same time.

The particles collide at $t=2$ s, at $(5,4,0)$.

1 markStates the conclusion.
Answer
a) $3\hat{i}+\hat{j}-2\hat{k}$b) $t=\tfrac{9}{14}$ sc) they collide at $t=2$, at $(5,4,0)$

Question 2 of 10

Velocity; maximum height; speed · Technology-free · Simple familiar · [6 marks]

An object is projected vertically upwards from ground level. After $t$ seconds, its position vector, in metres, is modelled by

$$\boldsymbol{r}(t)=\left(30t-5t^{2}\right)\hat{j}$$

a)Determine the velocity of the object, $\boldsymbol{v}(t)$, in metres per second.[1 mark]
b)Determine the number of seconds until the object reaches its maximum height.[2 marks]
c)Determine the maximum height that the object reaches, in metres.[1 mark]
d)Determine the speed of the object when it returns to ground level.[2 marks]

Part a)

$\boldsymbol{v}(t)=(30-10t)\,\hat{j}$

1 markCorrectly determines the velocity.

Part b)

Maximum height when $\boldsymbol{v}(t)=\boldsymbol{0}$

1 markRecognises that the velocity is zero at maximum height.

$30-10t=0\Rightarrow t=3$ s

1 markDetermines the time.

Part c)

$\boldsymbol{r}(3)=(90-45)\,\hat{j}=45\,\hat{j}$; maximum height 45 m

1 markDetermines the maximum height.

Part d)

Ground level: $30t-5t^2=0\Rightarrow t=0$ or $t=6$; returns at $t=6$

1 markDetermines the time of return.

$\boldsymbol{v}(6)=-30\,\hat{j}$; speed $=30\ \text{m s}^{-1}$

1 markDetermines the speed.
Answer
a) $(30-10t)\hat{j}$b) 3 sc) 45 md) $30\text{ m s}^{-1}$

Question 3 of 10

Integrating a velocity with a constant vector; acceleration · Technology-free · Simple familiar · [5 marks]

A particle moves in a plane with velocity

$$\begin{aligned}\boldsymbol{v}(t)=&3e^{-t}\,\hat{i}\\&+\frac{2t}{t^{2}+1}\,\hat{j}\end{aligned}$$

for $t\ge0$.

At $t=0$ the position vector of the particle is $\boldsymbol{r}(0)=2\hat{i}+\hat{j}$.

a)Determine the position vector $\boldsymbol{r}(t)$.[3 marks]
b)Determine the acceleration vector $\boldsymbol{a}(t)$.[1 mark]
c)Determine the exact position of the particle when $t=1$.[1 mark]

Part a)

$\displaystyle \boldsymbol{r}(t)=\left(-3e^{-t}+c_1\right)\hat{i}+\left(\ln\left(t^2+1\right)+c_2\right)\hat{j}$

1 markCorrectly integrates both components.

$\boldsymbol{r}(0)=(-3+c_1)\hat{i}+c_2\hat{j}=2\hat{i}+\hat{j}\Rightarrow c_1=5,\ c_2=1$

1 markUses $\boldsymbol{r}(0)$ to determine the constants.

$\boldsymbol{r}(t)=\left(5-3e^{-t}\right)\hat{i}+\left(1+\ln\left(t^2+1\right)\right)\hat{j}$

1 markDetermines $\boldsymbol{r}(t)$.

Part b)

$\displaystyle \boldsymbol{a}(t)=-3e^{-t}\hat{i}+\frac{2\left(t^2+1\right)-2t(2t)}{\left(t^2+1\right)^2}\hat{j}=-3e^{-t}\hat{i}+\frac{2\left(1-t^2\right)}{\left(t^2+1\right)^2}\hat{j}$

1 markDetermines $\boldsymbol{a}(t)$.

Part c)

$\boldsymbol{r}(1)=\left(5-\dfrac3e\right)\hat{i}+\left(1+\ln2\right)\hat{j}$

1 markDetermines the exact position.
Answer
a) $\left(5-3e^{-t}\right)\hat{i}+\left(1+\ln\left(t^2+1\right)\right)\hat{j}$b) $-3e^{-t}\hat{i}+\dfrac{2\left(1-t^2\right)}{\left(t^2+1\right)^2}\hat{j}$c) $\left(5-\tfrac{3}{e},1+\ln(2)\right)$

Question 4 of 10

Collision test; crossing paths with two parameters · Technology-free · Simple familiar · [5 marks]

The positions of particles $A$ and $B$ at time $t\ge0$ are given by

$$\begin{aligned}\boldsymbol{r}_A(t)&=2t\,\hat{i}+(t+1)\,\hat{j}\\\boldsymbol{r}_B(t)&=(6-2t)\,\hat{i}+(5t-2)\,\hat{j}\end{aligned}$$

a)Show that the particles do not collide.[2 marks]
b)Show that the paths of the particles cross. Determine the point where the paths cross, and the time at which each particle passes through this point.[3 marks]

Part a)

Equating $\hat{i}$ components: $2t=6-2t\Rightarrow t=\tfrac32$

1 markEquates one pair of components to determine $t$.

At $t=\tfrac32$: $\hat{j}$ components are $\tfrac52$ and $\tfrac{11}{2}$, which are not equal.

The particles do not collide.

1 markShows the other components differ at that time and states the conclusion.

Part b)

Use separate times: $A$ at time $t$, $B$ at time $s$

$2t=6-2s$ and $t+1=5s-2$

1 markCorrectly forms two equations with separate time parameters.

From the first: $t=3-s$; substituting: $4-s=5s-2\Rightarrow s=1$, $t=2$

1 markSolves the equations simultaneously.

$\boldsymbol{r}_A(2)=4\hat{i}+3\hat{j}=\boldsymbol{r}_B(1)$

The paths cross at $(4,3)$. $A$ passes through it at $t=2$ and $B$ at $t=1$.

1 markDetermines the crossing point and both times.

Using the same $t$ for both particles in part b) tests for a collision, not a crossing, and gains no marks for part b).

Answer
a) no collisionb) the paths cross at $(4,3)$, with A there at $t=2$ and B at $t=1$

Question 5 of 10

Displacement of one object relative to another; shortest distance; distance equal to a value · Technology-active · Simple familiar · [5 marks]

The positions (m) of drones $A$ and $B$ at time $t$ seconds, $0\le t\le6$, are given by

$$\begin{aligned}\boldsymbol{r}_A(t)&=3t\,\hat{i}+4t\,\hat{j}\\\boldsymbol{r}_B(t)&=12\,\hat{i}+2t\,\hat{j}+5\,\hat{k}\end{aligned}$$

a)Determine the vector that represents the displacement of drone $B$ relative to drone $A$.[1 mark]
b)Use your result from Question 5a) to determine the shortest distance between the drones during the given time of motion, and the time at which it occurs. Give your answers rounded to two decimal places.[3 marks]
c)Determine the times when the drones are 10 m apart. Give your answers rounded to two decimal places.[1 mark]

Part a)

$\boldsymbol{r}_B-\boldsymbol{r}_A=(12-3t)\,\hat{i}-2t\,\hat{j}+5\,\hat{k}$

1 markCorrectly determines the relative displacement.

Part b)

$\left|\boldsymbol{r}_B-\boldsymbol{r}_A\right|=\sqrt{(12-3t)^2+4t^2+25}=\sqrt{13t^2-72t+169}$

1 markCorrectly expresses the distance as a function of $t$.

Using technology (or calculus on $13t^2-72t+169$): minimum at $\displaystyle t=\frac{36}{13}\approx2.77$ s

1 markDetermines the time of the minimum.

Shortest distance $\approx8.33$ m

1 markDetermines the shortest distance.

Part c)

$13t^2-72t+169=100$

Using technology: $t\approx1.23$ s and $t\approx4.31$ s

1 markDetermines both times.
Answer
a) $(12-3t)\hat{i}-2t\hat{j}+5\hat{k}$b) 8.33 m at $t\approx2.77$ sc) $t\approx1.23$ s and $t\approx4.31$ s

Question 6 of 10

Velocity; a perpendicular condition; speed; direction of motion · Technology-free · Simple familiar · [5 marks]

The position vector of a particle at time $t\ge0$ is

$$\begin{aligned}\boldsymbol{r}(t)=&\left(t^{2}-4t\right)\hat{i}\\&+(2t+1)\,\hat{j}\end{aligned}$$

a)Determine the velocity vector of the particle.[1 mark]
b)Determine the time when the velocity of the particle is perpendicular to the vector $\hat{i}-\hat{j}$.[1 mark]
c)Determine the times when the speed of the particle is $2\sqrt5$.[2 marks]
d)Determine the time when the particle is moving parallel to the $y$-axis.[1 mark]

Part a)

$\boldsymbol{v}(t)=(2t-4)\,\hat{i}+2\,\hat{j}$

1 markCorrectly determines the velocity.

Part b)

$\boldsymbol{v}\cdot\left(\hat{i}-\hat{j}\right)=(2t-4)-2=0\Rightarrow t=3$

1 markDetermines the time.

Part c)

$|\boldsymbol{v}|^2=(2t-4)^2+4=20$

1 markCorrectly forms an equation for the speed.

$(2t-4)^2=16\Rightarrow 2t-4=\pm4\Rightarrow t=0$ or $t=4$

1 markDetermines both times.

Part d)

Parallel to the $y$-axis when the $\hat{i}$ component of $\boldsymbol{v}$ is zero:

$2t-4=0\Rightarrow t=2$

1 markDetermines the time.
Answer
a) $(2t-4)\hat{i}+2\hat{j}$b) $t=3$c) $t=0$ and $t=4$d) $t=2$

Question 7 of 10

Collision with unknown constants · Technology-free · Simple familiar · [4 marks]

The positions of two objects at time $t\ge0$ are given by

$$\begin{aligned}\boldsymbol{r}_A(t)=&\left(t^{2}-1\right)\hat{i}+2t\,\hat{j}\\&+(3t-2)\,\hat{k}\end{aligned}$$

$$\begin{aligned}\boldsymbol{r}_B(t)=&(at+1)\,\hat{i}+(t+2)\,\hat{j}\\&+(bt-4)\,\hat{k}\end{aligned}$$

where $a,\,b\in\mathbb{R}$.

Given that the two objects collide, determine the values of $a$ and $b$, and the coordinates of the point of collision.

Equating $\hat{j}$ components: $2t=t+2\Rightarrow t=2$

1 markDetermines the time of collision from the component with no unknown constant.

$\hat{i}$: $2^2-1=2a+1\Rightarrow a=1$

1 markDetermines $a$.

$\hat{k}$: $3(2)-2=2b-4\Rightarrow b=4$

1 markDetermines $b$.

$\boldsymbol{r}_A(2)=3\hat{i}+4\hat{j}+4\hat{k}$; collision at $(3,4,4)$

1 markDetermines the point of collision.
Answer
$a=1$, $b=4$, collision at $(3,4,4)$

Question 8 of 10

Integrating acceleration twice; straight-line motion; at rest · Technology-free · Simple familiar · [6 marks]

A particle has acceleration $\boldsymbol{a}(t)=6t\left(2\hat{i}-\hat{j}+2\hat{k}\right)$, $t\ge0$. Its initial velocity is $\boldsymbol{v}(0)=-6\hat{i}+3\hat{j}-6\hat{k}$ and its initial position is $\boldsymbol{r}(0)=\hat{i}+\hat{k}$.

a)Determine the velocity vector $\boldsymbol{v}(t)$.[1 mark]
b)Determine the position vector $\boldsymbol{r}(t)$.[1 mark]
c)Explain why the particle moves along a straight line, and state the Cartesian equation of this line.[2 marks]
d)Determine the time when the particle is first at rest, and its distance from its initial position at that time.[2 marks]

Part a)

$\boldsymbol{v}(t)=3t^2\left(2\hat{i}-\hat{j}+2\hat{k}\right)+\boldsymbol{c}$, with $\boldsymbol{v}(0)=-3\left(2\hat{i}-\hat{j}+2\hat{k}\right)$

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

1 markCorrectly determines $\boldsymbol{v}(t)$.

Part b)

$\boldsymbol{r}(t)=\left(t^3-3t\right)\left(2\hat{i}-\hat{j}+2\hat{k}\right)+\hat{i}+\hat{k}$

1 markCorrectly determines $\boldsymbol{r}(t)$.

Part c)

$\boldsymbol{r}(t)=\left(\hat{i}+\hat{k}\right)+\lambda\left(2\hat{i}-\hat{j}+2\hat{k}\right)$ with $\lambda=t^3-3t$: a fixed point plus a scalar multiple of a fixed direction, so the particle moves along a straight line.

1 markExplains why the motion is along a straight line.

$\displaystyle \frac{x-1}{2}=\frac{y}{-1}=\frac{z-1}{2}$

1 markStates the Cartesian equation of the line.

Part d)

$\boldsymbol{v}(t)=\boldsymbol{0}\Rightarrow3t^2-3=0\Rightarrow t=1$ (as $t\ge0$)

1 markDetermines the time.

$\boldsymbol{r}(1)-\boldsymbol{r}(0)=-2\left(2\hat{i}-\hat{j}+2\hat{k}\right)$

Distance $=2\sqrt{4+1+4}=6$ units

1 markDetermines the distance.
Answer
a) $\boldsymbol{v}(t)=3(t^2-1)(2\hat{i}-\hat{j}+2\hat{k})$b) $\boldsymbol{r}(t)=\hat{i}+\hat{k}+(t^3-3t)(2\hat{i}-\hat{j}+2\hat{k})$c) $\dfrac{x-1}{2}=\dfrac{y}{-1}=\dfrac{z-1}{2}$d) $t=1$, 6 units from the start

Question 9 of 10

Collision with a later release; constant speed · Technology-free · Complex familiar · [6 marks]

Object $A$ is released from the origin with constant velocity, such that its position after $t$ seconds is given by

$$\boldsymbol{r}_A=2t\,\hat{i}+t\,\hat{j}+2t\,\hat{k}$$

for $t\ge0$.

At a later time, object $B$ is released from point $P(2,-8,4)$ and travels towards point $Q(11,10,10)$ with a constant speed of $7\ \text{m s}^{-1}$.

Given that objects $A$ and $B$ collide, determine the time between the release of the two objects. Assume all positions are given in metres.

$\overrightarrow{PQ}=9\hat{i}+18\hat{j}+6\hat{k}$, $\left|\overrightarrow{PQ}\right|=21$

Unit direction $=\tfrac17\left(3\hat{i}+6\hat{j}+2\hat{k}\right)$

1 markCorrectly determines the direction of $B$'s motion.

$\boldsymbol{v}_B=7\times\tfrac17\left(3\hat{i}+6\hat{j}+2\hat{k}\right)=3\hat{i}+6\hat{j}+2\hat{k}$

1 markDetermines $\boldsymbol{v}_B$.

Let $B$ be released at time $\tau$:

$\boldsymbol{r}_B(t)=\left(2\hat{i}-8\hat{j}+4\hat{k}\right)+(t-\tau)\left(3\hat{i}+6\hat{j}+2\hat{k}\right)$, $t\ge\tau$

1 markDetermines an expression for $B$'s position allowing for the later release.

Equating components:

$2t=2+3(t-\tau)$, $t=-8+6(t-\tau)$, $2t=4+2(t-\tau)$

1 markCorrectly forms the equations for a collision.

From the $\hat{k}$ equation: $\tau=2$; then $\hat{i}$: $t=4$

Check $\hat{j}$: $4=-8+6(2)$, which agrees.

1 markSolves the equations and checks consistency.

Object $B$ is released 2 s after object $A$.

1 markShows logical organisation, having fully attempted the question.

You can also write $\boldsymbol{r}_B=\overrightarrow{OP}+u\,\boldsymbol{v}_B$, where $u$ is the time since B was released. Then $t-u$ is the answer.

Answer
B is released 2 s after A.

Question 10 of 10

Integrating velocities; collision test; minimum distance · Technology-active · Complex familiar · [6 marks]

The velocities (m s$^{-1}$) of two objects at time $t$ seconds are

$$\begin{aligned}\boldsymbol{v}_A(t)&=4\,\hat{i}+2t\,\hat{j}\\\boldsymbol{v}_B(t)&=-2\,\hat{i}+\hat{j}\end{aligned}$$

Initially, object $A$ is at the origin and object $B$ is at $(12,3)$.

Determine whether the objects collide during $0\le t\le4$. If they do not, determine the minimum distance between them during this time and the time at which it occurs. Give your answers rounded to two decimal places.

$\boldsymbol{r}_A(t)=4t\,\hat{i}+t^2\,\hat{j}+\boldsymbol{c}$, $\boldsymbol{r}_A(0)=\boldsymbol{0}$, so $\boldsymbol{r}_A(t)=4t\,\hat{i}+t^2\,\hat{j}$

1 markCorrectly determines $\boldsymbol{r}_A(t)$.

$\boldsymbol{r}_B(t)=(12-2t)\,\hat{i}+(3+t)\,\hat{j}$

1 markCorrectly determines $\boldsymbol{r}_B(t)$.

$\hat{i}$: $4t=12-2t\Rightarrow t=2$; $\hat{j}$ at $t=2$: $4\ne5$. The objects do not collide.

1 markShows the objects do not collide.

$d(t)=\left|\boldsymbol{r}_B-\boldsymbol{r}_A\right|=\sqrt{(12-6t)^2+\left(3+t-t^2\right)^2}$

1 markCorrectly expresses the distance as a function of $t$.

Using technology: minimum of $d(t)$ on $0\le t\le4$ at $t\approx2.07$ s

1 markDetermines the time of the minimum.

Minimum distance $\approx0.89$ m at $t\approx2.07$ s

1 markShows logical organisation, having fully attempted the question.

The distance at $t=2$ is 1 m, but that is not the minimum. Assuming the minimum happens when one pair of components is equal loses the fifth mark.

Answer
no collisionminimum distance $\approx0.89$ m at $t\approx2.07$ s

Keep practising

Ready for something harder?

Complex unfamiliar questions on this topic join several steps together, like 2023 Paper 1, Question 19. Once these skills feel comfortable, try the vector questions in the collection.

All complex unfamiliar questions

Past questions and marking-guide steps are from the 2020–2025 QCAA Specialist Mathematics external assessments, © State of Queensland (QCAA) 2020–2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.

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