Mechanics 203-SN1-RE • CEGEP, Montreal

Revision sheet: vector kinematics in two and three dimensions (203-SN1-RE)

Vector kinematics is the chapter where mechanics students lose marks without understanding why: the calculation is right, but it is done on the wrong object, a magnitude instead of a component, an angle measured from the wrong axis, a uniformly accelerated motion equation applied to an acceleration that varies.

This sheet does not copy the course. It gives what is really lost on the exam, the move that avoids it, and a complete example where each step says why it exists. The corrected exercises on the same chapter make you practise each of these moves.

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The thread of the chapter

A vector is handled component by component, and each component has its own story: you calculate in components, you return to magnitude and direction only to answer, and it is the sign of $\vec{v}\cdot\vec{a}$ , never that of a component, that tells whether an object goes faster or slower.

This chapter is part of 203-SN1-RE, Mechanics

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Position vector
Unit vectors
Average velocity
Average acceleration
Instantaneous velocity
Uniformly accelerated motion
Components of a vector
Direction of a vector
Dot product
Tangential acceleration
Normal acceleration
Horizontal launch

The essentials

Work in components, conclude in magnitude

• $\vec{r}=x\,\hat{\imath}+y\,\hat{\jmath}+z\,\hat{k}$ : three straight-line motions that share the same time $t$ . Every operation is done line by line, and the magnitude $\sqrt{x^{2}+y^{2}+z^{2}}$ is calculated only at the end, to answer.
• Instantaneous: $\vec{v}=\dfrac{d\vec{r}}{dt}$ and $\vec{a}=\dfrac{d\vec{v}}{dt}$ , differentiating each component. Average: $\vec{v}_{\mathrm{avg}}=\dfrac{\Delta\vec{r}}{\Delta t}$ and $\vec{a}_{\mathrm{avg}}=\dfrac{\Delta\vec{v}}{\Delta t}$ , which only need two states.
• Uniformly accelerated motion, $\vec{a}$ CONSTANT: $\vec{v}=\vec{v}_{0}+\vec{a}\,t$ and $\vec{r}=\vec{r}_{0}+\vec{v}_{0}\,t+\frac{1}{2}\vec{a}\,t^{2}$ . A single component of $\vec{a}$ that depends on time is enough to rule these equations out.
• Direction: $\theta=\arctan\dfrac{v_{y}}{v_{x}}$ , then CHECK the quadrant on the signs of the components, because the calculator only returns an angle between $-90^{\circ}$ and $90^{\circ}$ .

Adding two vectors means adding their components. Adding their magnitudes is correct only if they are parallel and point the same way, and otherwise gives a result that is too large.

The angle between velocity and acceleration

• $\vec{v}\cdot\vec{a}>0$ : the speed increases. $\vec{v}\cdot\vec{a}<0$ : it decreases. $\vec{v}\cdot\vec{a}=0$ : it is unchanged at that instant.
• The direction of $\vec{v}$ changes only if $\vec{a}$ has a component PERPENDICULAR to $\vec{v}$ ; in 2D, you test it with $v_{x}a_{y}-v_{y}a_{x}\neq 0$ .
• Tangential acceleration $a_{t}=\dfrac{\vec{v}\cdot\vec{a}}{|\vec{v}|}$ , which changes the speed; normal acceleration $a_{n}=\sqrt{|\vec{a}|^{2}-a_{t}^{2}}$ , which bends the path.
• Projectile: $v_{x}$ constant, $\vec{a}=-g\,\hat{\jmath}$ throughout the flight. At the top, only $v_{y}$ vanishes.

On the left

\vec{a}

follows

\vec{v}

: the speed increases without turning. In the center it is perpendicular to it: the velocity turns without changing its magnitude. On the right it opposes it.

The sign of ONE component of $\vec{a}$ says nothing about speed: an object with $a_{x}<0$ can go faster and faster if the other components win out.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Reading a compass angle from the wrong axis

2 marks, and every question that uses this displacement

What not to write

“ $3.00$ km at $20^{\circ}$ west of north: $\Delta x=-3.00\cos 20^{\circ}=-2.82$ km.”

What to write

“The angle starts from NORTH: $\Delta x=-3.00\sin 20^{\circ}=-1.03$ km and $\Delta y=3.00\cos 20^{\circ}=2.82$ km.”

You start from north and turn 20° toward the west: the component on the starting axis takes the cosine,

\Delta y=2.82

km, the other takes the sine,

\Delta x=-1.03

km.

Why: The cosine goes on the axis you start from when measuring the angle. In “west of north”, you start from north, so the cosine goes on $y$ . A ten-second sketch is enough to see that a walk “almost due north” cannot have its large component in $x$ .

2. Adding the magnitudes of two non-parallel vectors

2 to 3 marks, and a final speed wrong by nearly 15%

What not to write

“Velocity of $12.0$ m/s, gust of $3.20$ m/s² for $2.50$ s: $v=12.0+8.00=20.0$ m/s.”

What to write

“ $\vec{v}_{0}$ at $45^{\circ}$ and $\vec{a}\,\Delta t$ at $-15^{\circ}$ : $\vec{v}=(16.2, 6.41)$ m/s, that is $17.4$ m/s.”

\vec{a}\,\Delta t

(brown) is placed at the tip of

\vec{v}_{0}

: the sum (black) is 17.4 m/s at 21.6° north of east, less than

12.0+8.00=20.0

m/s, because the two vectors make 60°.

Why: $\vec{v}=\vec{v}_{0}+\vec{a}\,\Delta t$ is a sum of VECTORS. The magnitudes add only if the two vectors are parallel and point the same way; otherwise part of $\vec{a}\,\Delta t$ goes into turning, and the result is always smaller than the sum of the magnitudes.

3. Forgetting the z-component in a magnitude

1 mark, and an error that does not show because the result stays plausible

What not to write

“ $\vec{v}_{\mathrm{avg}}=(2.0, 2.0, -0.75)$ m/s, so $|\vec{v}_{\mathrm{avg}}|=\sqrt{2^{2}+2^{2}}=2.83$ m/s.”

What to write

“ $|\vec{v}_{\mathrm{avg}}|=\sqrt{2.0^{2}+2.0^{2}+0.75^{2}}=2.93$ m/s.”

Why: The magnitude takes ALL the squares. Working in the plane out of habit makes you forget the third component, especially when it is small; count the components again before writing the root.

4. Writing that the velocity is zero at the top of a projectile

2 marks, and everything after it if you use it for an energy or a range

What not to write

“At the top, $\vec{v}=\vec{0}$ and $\vec{a}=\vec{0}$ .”

What to write

“At the top, $v_{y}=0$ : $\vec{v}=v_{0x}\,\hat{\imath}$ and $\vec{a}=-g\,\hat{\jmath}$ .”

At the top, the velocity is horizontal and equals

v_{0x}

, the acceleration is vertical and equals

g

: neither is zero, and they are perpendicular.

Why: Only the vertical component vanishes; the horizontal velocity has never changed. And the acceleration stays $g$ downward throughout the flight: if it vanished at the top, the projectile would hang there.

5. Using the average of the velocities outside uniformly accelerated motion

2 marks, and a wrong conclusion about the nature of the motion

What not to write

“ $\vec{v}_{\mathrm{avg}}=\dfrac{\vec{v}_{1}+\vec{v}_{2}}{2}=(0, 5.0, 0)$ m/s.”

What to write

“ $\vec{v}_{\mathrm{avg}}=\dfrac{\vec{r}_{2}-\vec{r}_{1}}{\Delta t}=(2.0, 2.0, -0.75)$ m/s.”

Why: $\dfrac{\vec{v}_{1}+\vec{v}_{2}}{2}$ equals the average velocity only if the acceleration is constant. The definition $\dfrac{\Delta\vec{r}}{\Delta t}$ always holds; and if the two calculations differ, that is the proof the acceleration varied.

6. Concluding that an object slows down from the sign of a component

2 marks, on a question that asks for a single word

What not to write

“ $a_{x}=-1.0$ m/s² is negative, so the helicopter slows down.”

What to write

“ $\vec{v}\cdot\vec{a}=-2.0+18+5.76=21.8>0$ : the speed increases, $a_{t}=3.22$ m/s².”

Why: Speeding up depends on the angle between the two VECTORS, which only the dot product measures. A negative component of $\vec{a}$ can be more than offset by the others.

7. Confusing the minimum coordinate with the minimum speed

2 to 3 marks, and a wrong instant reused in the following questions

What not to write

“ $y$ is a minimum at $t=2.0$ s, so the speed is a minimum there: $1.0$ m/s.”

What to write

“ $|\vec{v}|^{2}=5t^{2}-22t+25$ is a minimum at $t=2.2$ s: $|\vec{v}|_{\mathrm{min}}=0.89$ m/s.”

Why: A minimum $y$ cancels $v_{y}$ only. The speed is a minimum when $\vec{v}\cdot\vec{a}=0$ , which depends on BOTH components. The two instants coincide only if $\vec{a}$ is vertical, as for a projectile.

8. Putting the whole initial speed in the maximum height

2 marks, and a height wrong by more than half

What not to write

“ $v_{0}=25$ m/s at $53.1^{\circ}$ : $h_{\mathrm{max}}=\dfrac{25^{2}}{2\times 9.80}=31.9$ m.”

What to write

“ $v_{0y}=25\sin 53.1^{\circ}=20$ m/s: $h_{\mathrm{max}}=\dfrac{20^{2}}{19.6}=20.4$ m.”

Why: Only the vertical component fights gravity; the horizontal component carries on unchanged. $\dfrac{v_{0}^{2}}{2g}$ is the height of a VERTICAL kick.

Which method to choose

Which tool, depending on what the problem gives

What the problem provides, and what the question asks for

If a position vector $\vec{r}(t)$ as a function of time → DIFFERENTIATE each component to get $\vec{v}(t)$ , then again for $\vec{a}(t)$

Example: $x=4.0t-0.50t^{2}$ gives $v_{x}=4.0-t$ and $a_{x}=-1.0$ m/s²
If two states, positions and velocities at two instants → AVERAGES: $\dfrac{\Delta\vec{r}}{\Delta t}$ and $\dfrac{\Delta\vec{v}}{\Delta t}$ , nothing else

Example: $\Delta\vec{r}=(8, 8, -3)$ m in $4.0$ s: $\vec{v}_{\mathrm{avg}}=(2, 2, -0.75)$ m/s
If $\vec{r}_{0}$ , $\vec{v}_{0}$ and a CONSTANT acceleration → Constant-acceleration equations, one per component, then solve on the component that carries the condition

Example: $y=6-4t+t^{2}$ is a minimum when $v_{y}=-4+2t=0$ , at $t=2.0$ s
If an acceleration that depends on time → NO constant-acceleration equations: differentiate or integrate component by component

Example: $a_{z}=-1.2t$ : none of the constant-acceleration equations applies
If “does it go faster or slower?” → sign of $\vec{v}\cdot\vec{a}$ , then $a_{t}=\dfrac{\vec{v}\cdot\vec{a}}{|\vec{v}|}$ if you want a number

Example: $\vec{v}=(3, 4)$ , $\vec{a}=(0, -10)$ : $-40<0$ , $a_{t}=-8$ m/s²
If a coordinate extremum, a top, a crossing of an axis → set the relevant COMPONENT to zero, of velocity or of position, without touching the others

Example: top of a projectile: $v_{y}=0$ ; crossing the $y$ -axis: $x=0$

A horizontal launch is constant-acceleration motion with $v_{0y}=0$ : the fall time depends only on the height, $t=\sqrt{\dfrac{2h}{g}}$ , and the horizontal velocity plays no role.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Solving a uniformly accelerated motion with vectors

When to use it: The problem gives an initial position, an initial velocity and a constant acceleration, in components or in magnitude and direction.

1 Choose and DRAW the axes, with their positive direction; convert each vector given as magnitude and direction into signed components.
2 Write the two or three lines $x(t)$ , $y(t)$ , $z(t)$ and the velocity lines, without calculating yet.
3 Translate the condition of the question onto ONE component: $v_{y}=0$ , $x=0$ , $y=h$ .
4 Solve for $t$ , discard the negative or earlier root while saying so, then substitute $t$ into the other lines.
5 Answer in magnitude and direction only if the question asks for it, with the unit.

Concluding sentence

“Since the acceleration is constant, each component follows the constant-acceleration equation: $x(t)=x_{0}+v_{0x}t+\frac{1}{2}a_{x}t^{2}$ . The root $t=-0.61$ s is before the start, so it is rejected.”

The trap: Forgetting to justify that the acceleration is constant. Without this sentence, using the constant-acceleration equations is not justified, and several markers take off the method mark even when the calculation is right.

Marking: Typically 1 mark for the signed components, 1 for the equations, 1 for the condition, 1 for the solution with the root discarded, 1 for the answer with its unit.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The quadrant of the angle

After an $\arctan$ , look at the signs of the components. Two negative components give the same $\arctan$ as two positive ones, but a vector pointing the opposite way.

$(-2.44, -2.32)$ km: $43.6^{\circ}$ SOUTH of WEST, not north of east.
The magnitude larger than each component

The magnitude of a vector is at least the absolute value of its largest component, and at most the sum of the absolute values. Outside that interval, there is an error.

$\vec{v}=(2.0, 6.0, -2.4)$ m/s: the magnitude must lie between $6.0$ and $10.4$ ; $6.76$ passes.
The vector sum smaller than the sum of the magnitudes

The magnitude of a sum of two non-parallel vectors is always STRICTLY less than the sum of their magnitudes. An equal result gives away added magnitudes.

$12.0$ and $8.00$ m/s at $60^{\circ}$ to each other: $17.4<20.0$ m/s.
Return to the same height

Without friction, a projectile that lands at its starting height has the same speed as at the start, and the descent lasts as long as the ascent.

Ball kicked at $25$ m/s from the ground: it lands at $25$ m/s, after $2\times 2.04=4.08$ s.

The typical problem, taken apart

The first crossing of the x-axis, with an acceleration that is not gravity

A particle starts from $\vec{r}_{0}=(1.0\,\hat{\imath}+4.0\,\hat{\jmath})$ m with $\vec{v}_{0}=(6.0\,\hat{\imath}-3.0\,\hat{\jmath})$ m/s and a constant acceleration $\vec{a}=(-1.0\,\hat{\imath}+1.0\,\hat{\jmath})$ m/s².

Where and when does it cross the $x$ -axis for the first time? What is its velocity then, and is it going faster or slower at that instant?

The starting configuration:

\vec{v}_{0}

points down and to the right,

\vec{a}

up and to the left. Their obtuse angle already announces a slowdown.

Step 1

$x(t)=1.0+6.0t-0.50t^{2}$ and $y(t)=4.0-3.0t+0.50t^{2}$ ; $v_{x}=6.0-t$ and $v_{y}=-3.0+t$ .

Why

The acceleration is constant: write the constant-acceleration equations line by line BEFORE any calculation. This is the method mark, and it is where a forgotten $\frac{1}{2}$ shows.

Step 2

$x$ -axis: $y=0$ , that is $t^{2}-6t+8=0$ , so $t=2.0$ s or $t=4.0$ s.

Why

The condition is on ONE component, the position in $y$ . Both roots are positive: the particle crosses the axis going down, then crosses it again coming back up. The question asks for the FIRST.

Step 3

$x(2.0)=1+12-2=11$ m: the particle passes through $(11, 0)$ m.

Why

You substitute the instant found into the OTHER line. Time is the only link between the components.

Step 4

$\vec{v}(2.0)=(4.0, -1.0)$ m/s, of magnitude $\sqrt{17}=4.12$ m/s.

Why

The magnitude comes at the end, to answer. The vector also says that the particle is still going down, $v_{y}<0$ , which confirms that this is indeed the first crossing.

Step 5

$\vec{v}\cdot\vec{a}=(4)(-1)+(-1)(1)=-5<0$ : the particle slows down, $a_{t}=\dfrac{-5}{4.12}=-1.21$ m/s².

Why

Speed is judged by the dot product, not by the sign of $a_{y}$ , which is positive here and would wrongly suggest speeding up.

The conclusion, written out

“The particle crosses the $x$ -axis for the first time at $t=2.0$ s, at $(11, 0)$ m, with velocity $(4.0\,\hat{\imath}-1.0\,\hat{\jmath})$ m/s of magnitude $4.12$ m/s; since $\vec{v}\cdot\vec{a}<0$ , it slows down at that instant.”

The classic mistake on this problem: Taking the root $t=4.0$ s, which is the second crossing, or concluding that the particle speeds up because $a_{y}>0$ .

Learn by heart

• Calculate in components, answer in magnitude and direction.
• The cosine goes on the axis you START FROM when measuring the angle.
• Magnitude in 3D: all three squares, $z$ included.
• $\vec{v}_{\mathrm{avg}}=\dfrac{\Delta\vec{r}}{\Delta t}$ always; $\dfrac{\vec{v}_{i}+\vec{v}_{f}}{2}$ only if $\vec{a}$ is constant.
• Constant acceleration: $\vec{r}=\vec{r}_{0}+\vec{v}_{0}t+\frac{1}{2}\vec{a}t^{2}$ , one line per component.
• Faster or slower: sign of $\vec{v}\cdot\vec{a}$ .
• The direction changes only if $\vec{a}$ has a component perpendicular to $\vec{v}$ .
• At the top of a projectile: $\vec{v}=v_{0x}\,\hat{\imath}$ and $\vec{a}=-g\,\hat{\jmath}$ .
• Horizontal launch: the height alone sets the fall time.
• $h_{\mathrm{max}}=\dfrac{v_{0y}^{2}}{2g}$ , with the VERTICAL component.

Frequently asked questions

How can you tell whether an object speeds up or slows down when velocity and acceleration are vectors?

Calculate the dot product of the velocity and the acceleration by multiplying the components two by two and adding. If it is positive, the speed increases; if it is negative, it decreases; if it is zero, the speed is unchanged at that instant. The sign of a single component of the acceleration is not enough to conclude.

When can you use the constant-acceleration equations in two or three dimensions?

Only when the acceleration vector is constant, that is, when each of its components is constant. You then write one equation per axis, with the same time variable. If even one component of the acceleration depends on time, these equations are wrong and you must differentiate or integrate component by component.

How do you read an angle such as 20 degrees west of north?

Start from the direction named last, here north, and turn 20 degrees toward the first one, west. The cosine of the angle therefore gives the northward component and the sine the westward component, with a minus sign if the x-axis points east. A quick sketch prevents almost every error.

Is the velocity of a projectile zero at the top of its path?

No. At the top, only the vertical component of the velocity is zero; the horizontal component is the same as at the start, since nothing changes it without air resistance. The acceleration, for its part, is always g downward. The velocity is zero at the top only for a perfectly vertical throw.

Practise it

Corrected exercises: Vector kinematics in two and three dimensions, 203-SN1-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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Previous sheet Kinematics in one and two dimensions Next sheet Dynamics, Newton's laws and friction

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-phys1-vector-kinematics. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: vector kinematics in two and three dimensions (203-SN1-RE)

The essentials

Work in components, conclude in magnitude

The angle between velocity and acceleration

The mistakes that cost marks

1. Reading a compass angle from the wrong axis

2. Adding the magnitudes of two non-parallel vectors

3. Forgetting the z-component in a magnitude

4. Writing that the velocity is zero at the top of a projectile

5. Using the average of the velocities outside uniformly accelerated motion

6. Concluding that an object slows down from the sign of a component

7. Confusing the minimum coordinate with the minimum speed

8. Putting the whole initial speed in the maximum height

Which method to choose

Which tool, depending on what the problem gives

How the answer is expected to be written

Solving a uniformly accelerated motion with vectors

Check before you hand in

The typical problem, taken apart

The first crossing of the x-axis, with an acceleration that is not gravity

Learn by heart

Frequently asked questions

How can you tell whether an object speeds up or slows down when velocity and acceleration are vectors?

When can you use the constant-acceleration equations in two or three dimensions?

How do you read an angle such as 20 degrees west of north?

Is the velocity of a projectile zero at the top of its path?

Corrected exercises: Vector kinematics in two and three dimensions, 203-SN1-RE

See also

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