Vector modelling: displacement, velocity and forces

Vector modelling turns a situation with several directions into one component calculation. The central principle is:

$$\boxed{\text{resultant vector}=\text{vector sum of the individual effects}.}$$

The arithmetic is usually straightforward. The difficult work is deciding what each vector means, choosing positive directions, keeping units consistent and interpreting the answer in context.

Prerequisites

You should be able to:

• add vectors and multiply them by scalars;
• find a vector’s magnitude and direction;
• form a displacement using end minus start;
• use position vectors in two and three dimensions;
• distinguish vectors from scalars.

Review vector arithmetic, magnitude and direction and position vectors if necessary.

A reliable modelling routine

For almost every vector model, use the following sequence.

1. Define axes and state their positive directions.
2. Translate every directed quantity into components.
3. Convert all quantities to compatible units.
4. Combine vectors according to the situation.
5. Find a magnitude or direction only if it answers the question.
6. Interpret the result, including its units and any modelling assumptions.

For horizontal maps, a common convention is

$$\mathbf i=\text{one unit east},\qquad \mathbf j=\text{one unit north}.$$

Then $3\mathbf i-2\mathbf j$ means $3$ units east and $2$ units south.

Components depend on the chosen axes

The same physical vector has different components under different coordinate systems. Components are not meaningful until the axes are defined.

Displacement models

Displacement records overall change in position. If a journey consists of displacements $\mathbf d_1,\mathbf d_2,\ldots,\mathbf d_n$, its resultant displacement is

$$\boxed{\mathbf d=\mathbf d_1+\mathbf d_2+\cdots+\mathbf d_n.}$$

Distance travelled is different:

$$\boxed{\text{distance}=|\mathbf d_1|+|\mathbf d_2|+\cdots+|\mathbf d_n|.}$$

In general, $|\mathbf d_1+\mathbf d_2|\ne |\mathbf d_1|+|\mathbf d_2|$. The left side is the straight line displacement from start to finish. The right side is the total length travelled.

Worked example 1: a two stage walk

East is the positive $x$ direction and north is the positive $y$ direction. A walker travels $6$ km east and then $8$ km north. Find the resultant displacement and the distance travelled.

The two displacements are

$$\mathbf d_1=\begin{pmatrix}6\\0\end{pmatrix}, \qquad \mathbf d_2=\begin{pmatrix}0\\8\end{pmatrix}.$$

Hence

$$\mathbf d=\mathbf d_1+\mathbf d_2 =\begin{pmatrix}6\\8\end{pmatrix}\text{ km}.$$

Its magnitude is

$$|\mathbf d|=\sqrt{6^2+8^2}=10\text{ km}.$$

If $\theta$ is measured north of east, then

$$\tan\theta=\frac86, \qquad \theta=53.1^\circ.$$

The resultant displacement is $\boxed{10\text{ km at }53.1^\circ\text{ north of east}}$.

The distance travelled is

$$6+8=\boxed{14\text{ km}}.$$

Worked example 2: returning to a known point

A survey drone makes three displacements:

$$\mathbf a=\begin{pmatrix}4\\-1\\2\end{pmatrix}\text{ m}, \quad \mathbf b=\begin{pmatrix}-3\\5\\1\end{pmatrix}\text{ m}, \quad \mathbf c=\begin{pmatrix}2\\0\\-4\end{pmatrix}\text{ m}.$$

Find the displacement $\mathbf r$ needed to return to its starting point.

For a closed journey, the total displacement is the zero vector:

$$\mathbf a+\mathbf b+\mathbf c+\mathbf r=\mathbf 0.$$

Therefore

$$\begin{aligned} \mathbf r &=-(\mathbf a+\mathbf b+\mathbf c)\\ &=-\begin{pmatrix}4-3+2\\-1+5+0\\2+1-4\end{pmatrix}\\ &=\boxed{\begin{pmatrix}-3\\-4\\1\end{pmatrix}\text{ m}}. \end{aligned}$$

A useful check is

$$\begin{pmatrix}3\\4\\-1\end{pmatrix} +\begin{pmatrix}-3\\-4\\1\end{pmatrix} =\mathbf 0.$$

Constant velocity models

Velocity is the rate of change of position and includes direction. For constant velocity $\mathbf v$ over time $t$,

$$\boxed{\mathbf d=t\mathbf v.}$$

If an object starts with position vector $\mathbf r_0$, then after time $t$ its position vector is

$$\boxed{\mathbf r(t)=\mathbf r_0+t\mathbf v.}$$

This is the vector form of position equals initial position plus displacement.

The dimensions provide a check:

$$(\mathrm{s})(\mathrm{m\,s^{-1}})=\mathrm m.$$

Worked example 3: position after a given time

Relative to a harbour, a boat starts at

$$\mathbf r_0=\begin{pmatrix}-2\\5\end{pmatrix}\text{ km}$$

and travels with constant velocity

$$\mathbf v=\begin{pmatrix}12\\-5\end{pmatrix}\text{ km h}^{-1}.$$

Find its position after $20$ minutes and its speed.

First convert the time:

$$20\text{ minutes}=\frac13\text{ hour}.$$

Its displacement is

$$t\mathbf v =\frac13\begin{pmatrix}12\\-5\end{pmatrix} =\begin{pmatrix}4\\-5/3\end{pmatrix}\text{ km}.$$

Therefore

$$\begin{aligned} \mathbf r\left(\frac13\right) &=\begin{pmatrix}-2\\5\end{pmatrix} +\begin{pmatrix}4\\-5/3\end{pmatrix}\\ &=\boxed{\begin{pmatrix}2\\10/3\end{pmatrix}\text{ km}}. \end{aligned}$$

Speed is the magnitude of velocity:

$$|\mathbf v|=\sqrt{12^2+(-5)^2}=\boxed{13\text{ km h}^{-1}}.$$

The negative velocity component means motion south. It does not make the speed negative.

Worked example 4: when does an object reach a line?

A particle has position

$$\mathbf r(t)=\begin{pmatrix}3\\-4\end{pmatrix} +t\begin{pmatrix}2\\5\end{pmatrix}, \qquad t\geq 0,$$

where position is measured in metres and time in seconds. When does it cross the line $y=11$, and where does this happen?

The $y$ component is

$$y=-4+5t.$$

Set this equal to $11$:

$$-4+5t=11 \quad\Longrightarrow\quad t=3.$$

At $t=3$,

$$x=3+2(3)=9.$$

The particle crosses the line after $\boxed{3\text{ s}}$ at $\boxed{(9,11)}$.

Use only the component you need

To find when a particle reaches a horizontal line, begin with its vertical component. Substitute the resulting time into the other component afterwards.

Combining velocities

When one motion occurs within another moving system, velocities add. If a person walks with velocity $\mathbf v_{P/T}$ relative to a train, and the train has velocity $\mathbf v_{T/G}$ relative to the ground, then

$$\boxed{\mathbf v_{P/G}=\mathbf v_{P/T}+\mathbf v_{T/G}.}$$

The subscripts make the reference frame explicit. Reading $P/T$ as “person relative to train” helps prevent reversed subtraction.

Worked example 5: aircraft and wind

An aircraft’s velocity relative to the air is $180\mathbf i+240\mathbf j$ km h$^{-1}$. The wind velocity is $40\mathbf i-20\mathbf j$ km h$^{-1}$, relative to the ground. Find the aircraft’s ground velocity, ground speed and bearing.

Add the velocities:

$$\begin{aligned} \mathbf v_{\text{ground}} &=\mathbf v_{\text{aircraft/air}}+\mathbf v_{\text{air/ground}}\\ &=\begin{pmatrix}180\\240\end{pmatrix} +\begin{pmatrix}40\\-20\end{pmatrix}\\ &=\boxed{\begin{pmatrix}220\\220\end{pmatrix}\text{ km h}^{-1}}. \end{aligned}$$

Thus the ground speed is

$$\sqrt{220^2+220^2}=220\sqrt2\approx\boxed{311\text{ km h}^{-1}}.$$

The east and north components are equal and positive, so the direction is north east. Bearings are measured clockwise from north, giving a bearing of $\boxed{045^\circ}$.

Worked example 6: choosing a heading to cancel drift

A river flows east at $3\text{ m s}^{-1}$. A boat moves at $5\text{ m s}^{-1}$ relative to the water. It must travel due north relative to the bank. Find the boat’s required velocity relative to the water and its resulting speed relative to the bank.

Take east and north as positive component directions. Let the boat’s velocity relative to the water be

$$\mathbf v_{B/W}=\begin{pmatrix}x\\y\end{pmatrix}.$$

The water’s velocity relative to the bank is

$$\mathbf v_{W/G}=\begin{pmatrix}3\\0\end{pmatrix}.$$

For the resultant to be due north, its east component must be zero:

$$x+3=0, \qquad x=-3.$$

The boat’s speed relative to the water is $5$, so

$$(-3)^2+y^2=5^2.$$

Since the boat must head north, $y>0$:

$$y=4.$$

Therefore

$$\boxed{\mathbf v_{B/W}=\begin{pmatrix}-3\\4\end{pmatrix}\text{ m s}^{-1}}.$$

Its velocity relative to the bank is

$$\begin{pmatrix}-3\\4\end{pmatrix} +\begin{pmatrix}3\\0\end{pmatrix} =\begin{pmatrix}0\\4\end{pmatrix}\text{ m s}^{-1},$$

so its speed relative to the bank is $\boxed{4\text{ m s}^{-1}}$.

The boat points partly west even though its actual path is north. Heading and track are not the same when there is wind or current.

Relative velocity

To describe how $A$ appears to move to an observer travelling with $B$, subtract $B$‘s velocity:

$$\boxed{\mathbf v_{A/B}=\mathbf v_A-\mathbf v_B.}$$

This is the velocity version of end minus start. If $\mathbf v_{A/B}=\mathbf 0$, the objects have no relative motion, even if both are moving relative to the ground.

Worked example 7: relative motion of two vehicles

Vehicle $A$ has velocity

$$\mathbf v_A=\begin{pmatrix}18\\6\end{pmatrix}\text{ m s}^{-1},$$

and vehicle $B$ has velocity

$$\mathbf v_B=\begin{pmatrix}12\\-2\end{pmatrix}\text{ m s}^{-1}.$$

Find the velocity and speed of $A$ relative to $B$.

$$\begin{aligned} \mathbf v_{A/B} &=\mathbf v_A-\mathbf v_B\\ &=\begin{pmatrix}18\\6\end{pmatrix} -\begin{pmatrix}12\\-2\end{pmatrix}\\ &=\boxed{\begin{pmatrix}6\\8\end{pmatrix}\text{ m s}^{-1}}. \end{aligned}$$

The relative speed is

$$|\mathbf v_{A/B}|=\sqrt{6^2+8^2}=\boxed{10\text{ m s}^{-1}}.$$

Notice that $\mathbf v_{B/A}=(-6,-8)^T$. Reversing the observer reverses the relative velocity, but not the relative speed.

Resultant forces

Forces are vectors. If several forces act on a particle, their resultant is

$$\boxed{\mathbf R=\sum \mathbf F.}$$

If $\mathbf R=\mathbf 0$, the forces are balanced. This means zero acceleration, not necessarily zero velocity.

Worked example 8: resultant of three forces

Three coplanar forces act on a particle:

$$\mathbf F_1=\begin{pmatrix}7\\2\end{pmatrix}\text{ N}, \quad \mathbf F_2=\begin{pmatrix}-4\\5\end{pmatrix}\text{ N}, \quad \mathbf F_3=\begin{pmatrix}1\\-9\end{pmatrix}\text{ N}.$$

Their resultant is

$$\mathbf R =\begin{pmatrix}7-4+1\\2+5-9\end{pmatrix} =\boxed{\begin{pmatrix}4\\-2\end{pmatrix}\text{ N}}.$$

Its magnitude is

$$|\mathbf R|=\sqrt{4^2+(-2)^2}=\boxed{2\sqrt5\text{ N}}.$$

It acts at an angle

$$\tan^{-1}\left(\frac24\right)=26.6^\circ$$

south of east.

Worked example 9: finding an equilibrant

Two forces $3\mathbf i-4\mathbf j$ N and $-8\mathbf i+\mathbf j$ N act on a particle. Find the single additional force $\mathbf E$ that would produce equilibrium.

For equilibrium,

$$\mathbf F_1+\mathbf F_2+\mathbf E=\mathbf 0.$$

The existing resultant is

$$\mathbf F_1+\mathbf F_2 =(3-8)\mathbf i+(-4+1)\mathbf j =-5\mathbf i-3\mathbf j.$$

Therefore

$$\boxed{\mathbf E=5\mathbf i+3\mathbf j\text{ N}}.$$

This force is called the equilibrant. It has the same magnitude as the existing resultant but the opposite direction.

For force diagrams, resolving angled forces and applying Newton’s laws, continue to forces and free body diagrams and resolving forces.

Modelling assumptions and limitations

A vector calculation may be exact for the model while only approximating reality. Common assumptions include:

• velocity is constant over the stated time interval;
• wind or current is uniform and constant;
• the map is treated as a flat plane;
• an object is represented by a particle, so its size and rotation are ignored;
• listed forces are the only significant forces;
• measurements and initial positions are exact.

An assumption matters because it justifies an equation. For example, $\mathbf d=t\mathbf v$ requires constant velocity. If velocity changes, multiplying one instantaneous velocity by the whole time does not generally give displacement.

Worked example 10: evaluating a model

A cyclist’s displacement over $30$ seconds is predicted using a constant measured velocity

$$\mathbf v=\begin{pmatrix}6\\2\end{pmatrix}\text{ m s}^{-1}.$$

The model gives

$$\mathbf d=30\mathbf v =\begin{pmatrix}180\\60\end{pmatrix}\text{ m}.$$

State two reasons why the actual displacement might differ.

Possible answers include:

1. The cyclist may accelerate, brake or turn, so velocity is not constant.
2. The measured velocity may contain error.
3. Wind, gradient or traffic may alter the motion.

Saying only “the model is unrealistic” is too vague. Identify an assumption and explain how reality may violate it.

Common misconceptions

Adding speeds instead of velocities

Speeds have no direction. Two velocities of magnitude $5$ need not combine to give speed $10$. For example,

$$\begin{pmatrix}5\\0\end{pmatrix} +\begin{pmatrix}-5\\0\end{pmatrix} =\mathbf 0.$$

Confusing position and displacement

If $\mathbf r_0$ is not zero, then $t\mathbf v$ is displacement, not final position. Use

$$\mathbf r(t)=\mathbf r_0+t\mathbf v.$$

Mixing time units

A velocity in km h$^{-1}$ must be multiplied by time in hours. Convert minutes or seconds before multiplying.

Taking a magnitude too early

In general,

$$|\mathbf a+\mathbf b|\ne |\mathbf a|+|\mathbf b|.$$

Add components first, then find the resultant magnitude.

Using $\tan^{-1}(y/x)$ without checking the quadrant

The ratio $y/x$ does not uniquely identify a direction. Inspect the signs of both components, sketch the vector, then state the angle clearly as a bearing or relative to an axis.

Self check

1. A walker travels with displacements $\begin{pmatrix}5\\-2\end{pmatrix}$ km and $\begin{pmatrix}-1\\7\end{pmatrix}$ km. Find the resultant displacement and its magnitude.

2. A particle starts at $\begin{pmatrix}4\\-3\end{pmatrix}$ m and moves with constant velocity $\begin{pmatrix}-2\\5\end{pmatrix}$ m s$^{-1}$. Find its position after $6$ seconds.

3. A plane has air velocity $\begin{pmatrix}100\\160\end{pmatrix}$ km h$^{-1}$ and the wind velocity is $\begin{pmatrix}-20\\30\end{pmatrix}$ km h$^{-1}$. Find its ground velocity and ground speed.

4. Objects $A$ and $B$ have velocities $4\mathbf i+7\mathbf j$ and $-2\mathbf i+3\mathbf j$ m s$^{-1}$. Find the velocity of $A$ relative to $B$.

5. Forces $\begin{pmatrix}6\\-1\end{pmatrix}$ N and $\begin{pmatrix}-2\\4\end{pmatrix}$ N act on a particle. Find the equilibrant.

6. Explain why the equation $\mathbf d=t\mathbf v$ may fail for a car travelling through a town.

Answers

1. The resultant is

   $$\begin{pmatrix}5\\-2\end{pmatrix} +\begin{pmatrix}-1\\7\end{pmatrix} =\begin{pmatrix}4\\5\end{pmatrix}\text{ km},$$

   with magnitude $\sqrt{4^2+5^2}=\boxed{\sqrt{41}\text{ km}}$.

$$\mathbf r(6)=\begin{pmatrix}4\\-3\end{pmatrix} +6\begin{pmatrix}-2\\5\end{pmatrix} =\boxed{\begin{pmatrix}-8\\27\end{pmatrix}\text{ m}}.$$

3. The ground velocity is

   $$\begin{pmatrix}100\\160\end{pmatrix} +\begin{pmatrix}-20\\30\end{pmatrix} =\boxed{\begin{pmatrix}80\\190\end{pmatrix}\text{ km h}^{-1}}.$$

   Its speed is $\sqrt{80^2+190^2}=\boxed{10\sqrt{425}\text{ km h}^{-1}}$, approximately $206\text{ km h}^{-1}$.

$$\mathbf v_{A/B}=(4\mathbf i+7\mathbf j)-(-2\mathbf i+3\mathbf j) =\boxed{6\mathbf i+4\mathbf j\text{ m s}^{-1}}.$$

5. The existing resultant is $\begin{pmatrix}4\\3\end{pmatrix}$ N, so the equilibrant is $\boxed{\begin{pmatrix}-4\\-3\end{pmatrix}\text{ N}}$.

6. The car’s velocity changes when it accelerates, brakes or turns. A single constant velocity therefore may not represent the whole time interval.

Exam checklist

Before finishing a vector modelling problem, ask:

• Have I defined the component directions?
• Are all units compatible?
• Did I add vectors before taking the magnitude?
• Have I distinguished position, displacement, distance, velocity and speed?
• Does my angle lie in the correct quadrant?
• Have I answered with a quantity, units and contextual meaning?
• If asked to assess the model, have I linked a limitation to an assumption?

Next steps

Continue to two dimensional motion for vector kinematics with changing velocity, and resolving forces for forces given by angles rather than components. To strengthen the wider modelling process, study mathematical modelling and interpreting context.