Vectors in Two Dimensions & Relative Motion

In L01 you added vectors along one line: give each a sign, then add the signed numbers. But what if two vectors point in directions that are not the same line, like 30 m east and 40 m north? You cannot reach "70" anything, because the two motions are at right angles. Walking east never undoes or adds to walking north; they build a corner. The single vector that gets you from start to finish is the diagonal across that corner.

When two vectors are perpendicular (at 90° to each other), they form the two short sides of a right-angled triangle, and their resultant is the hypotenuse. We find the resultant in two steps: its magnitude from Pythagoras, and its direction from trigonometry.

For example, a hiker walks 30 m east then 40 m north. The resultant displacement has magnitude $R = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50$ m. Its direction, measured as an angle north of east, is $\theta = \tan^{-1}\!\left(\frac{40}{30}\right) \approx 53°$. So the single displacement is 50 m at 53° north of east (a bearing of about 037°).

Two perpendicular vectors form a right-angled triangle. The resultant's magnitude is $R=\sqrt{R_x^2+R_y^2}$ (Pythagoras); its direction is $\theta=\tan^{-1}(R_y/R_x)$ (trig), stated as an angle from a reference line or a bearing. Perpendicular vectors are NEVER added arithmetically.

Pause, copy both formulas (Pythagoras for magnitude, tan for direction) and the 30-40-50 worked example into your book before moving on.

We just saw how to combine two perpendicular vectors into one resultant. That raises a question: the velocities we combine, measured by whom? A pilot and a ground observer report different speeds for the same plane. This card answers it → every velocity is measured relative to a chosen observer, and to switch observers we subtract vectors.

Sit on a train rolling at 80 km/h and watch a second train on the next track also doing 80 km/h the same way. Relative to you, it is not moving at all, it hangs level with your window. Relative to the platform, both trains scream past at 80 km/h. Neither answer is wrong. Velocity is never "absolute"; it is always measured against some observer.

The velocity of A relative to B is what an observer travelling with B would measure for A. It is found by vector subtraction:

Along a single line this is just signed arithmetic (from L01). Take east as positive. Car A drives east at 100 km/h ($+100$) and car B drives east at 60 km/h ($+60$). The velocity of A relative to B is $\vec{v}_{A} - \vec{v}_{B} = (+100) - (+60) = +40$ km/h, A pulls away from B at 40 km/h. If instead B drives west at 60 km/h ($-60$), then A relative to B is $(+100) - (-60) = +160$ km/h: they close on each other far faster.

Velocity is always relative to an observer. The velocity of A relative to B is $\vec{v}_{A\,\text{rel}\,B}=\vec{v}_A-\vec{v}_B$. Along one line, use signed values: same direction subtracts magnitudes, opposite directions add them (subtracting a negative).

Pause, copy the relative-velocity formula and both car examples (+40 km/h same way, +160 km/h head-on) into your book before moving on.

We just saw that relative velocity along a line is signed subtraction. That raises a question: what happens when the two velocities are not on the same line, like a plane heading north while the wind blows east? This card answers it → we add them as perpendicular vectors, exactly the triangle from Card 1.

An aircraft moves through the air, but the air itself can be moving over the ground. The plane's air velocity (its heading and airspeed) is what the pilot controls. The wind velocity is how the air moves over the ground. The plane's ground velocity its real path over the paddocks, is the vector sum of the two.

Take the Wagga plane: airspeed 200 km/h due north, wind 80 km/h from the west (so blowing toward the east). The two velocities are perpendicular, so the ground velocity is the hypotenuse of a triangle with sides 200 (north) and 80 (east):

$v_{\text{ground}} = \sqrt{200^2 + 80^2} = \sqrt{40000 + 6400} = \sqrt{46400} \approx 215$ km/h. The drift angle east of north is $\theta = \tan^{-1}\!\left(\frac{80}{200}\right) \approx 22°$. So the plane actually tracks across the ground at about 215 km/h on a bearing of 022°, even though its nose points due north (000°).

Ground velocity = air velocity + wind velocity (vector sum). When the wind is a crosswind (perpendicular to heading), use the triangle: $v_{\text{ground}}=\sqrt{v_{\text{air}}^2+v_{\text{wind}}^2}$ for the speed, and $\tan\theta=v_{\text{wind}}/v_{\text{air}}$ for the drift angle. The nose direction and the ground track are NOT the same.

Pause, copy the ground-velocity equation and the Wagga worked example (215 km/h on bearing 022°) into your book before moving on.

We just saw a plane's ground velocity built from air + wind. That raises a question: does the sideways push always change how long the crossing takes? This card answers it → for a boat pointed straight across a river, the current is perpendicular, so it shifts the landing point downstream without changing the crossing time at all.

A boat is steered straight across a river while the current sweeps it downstream. The boat's velocity relative to the water (across) and the current's velocity (downstream) are perpendicular, so the boat's velocity relative to the bank is, once again, the hypotenuse of a right-angled triangle.

The crucial insight is that the two perpendicular components act independently. The across-river component is set only by the boat's own speed and the river's width; the current adds nothing across the river, only along it. So:

Worked example: a river is 60 m wide. A boat heads straight across at 3.0 m/s while the current flows at 4.0 m/s downstream. The crossing time is $t = \frac{60}{3.0} = 20$ s, the current is irrelevant to time. In those 20 s the current carries the boat $4.0 \times 20 = 80$ m downstream. The boat's velocity relative to the bank is $\sqrt{3.0^2 + 4.0^2} = 5.0$ m/s, directed at $\tan^{-1}\!\left(\frac{4.0}{3.0}\right) \approx 53°$ downstream of straight-across.

A boat pointed straight across a river: the across-river and downstream velocities are perpendicular and independent. Crossing time $t=\dfrac{\text{width}}{v_{\text{across}}}$, the current does NOT change it. Downstream drift $=v_{\text{current}}\times t$; velocity relative to bank $=\sqrt{v_{\text{across}}^2+v_{\text{current}}^2}$.

Pause, copy the crossing-time rule and the 60 m / 3-4-5 worked example (20 s, 80 m drift, 5.0 m/s) into your book before moving on.

We just saw the crosswind and river cases solved one at a time. That raises a question: is there a single repeatable method that handles any 2D relative-motion problem? This card answers it → yes, draw the triangle, use Pythagoras for the magnitude, use tan for the direction, then state a bearing.

Crosswinds and river crossings are the same problem wearing different clothes: two perpendicular velocities adding to a resultant. A single five-step method handles both, and earns full marks because it forces you to show the vector diagram the marker is looking for.

Every 2D relative-motion problem: (1) draw the tip-to-tail vector triangle, (2) find the right angle, (3) magnitude with Pythagoras, (4) direction with $\tan^{-1}$, (5) state magnitude AND direction (as a bearing). The labelled diagram is worth marks even if the arithmetic slips.

Pause, copy the five-step method into your book and re-draw the crosswind triangle with all three vectors labelled.

We just saw the five-step method for combining two perpendicular vectors into one resultant. That raises a question: what if you are only given the single resultant, a velocity or a force at some angle, and need to know its horizontal and vertical parts? This card answers it → resolution is combination run in reverse.

NESA syllabus (IQ2): "analyse vectors in one and two dimensions... to resolve a vector into two perpendicular components." Cards 1 to 5 always started with two known perpendicular components and combined them into one vector. Resolution starts at the other end: you are given ONE vector at a known angle, and you split it back into its two perpendicular components.

Any vector $R$ acting at an angle $\theta$ to a reference direction (usually the horizontal) can be resolved into a horizontal component and a vertical component using the same right-angled triangle, this time solved with cosine and sine instead of Pythagoras and inverse tan.

Worked example: a ball is launched with a velocity of 50 m/s at 40° above the horizontal. Its horizontal component is $v_x = 50\cos 40° \approx 38$ m/s, and its vertical component is $v_y = 50\sin 40° \approx 32$ m/s. Check with Pythagoras (Card 1 in reverse): $\sqrt{38^2+32^2} \approx \sqrt{1444+1024} = \sqrt{2468} \approx 50$ m/s, the original magnitude, confirming the split is correct.

Resolution is combination in reverse: instead of combining $R_x$ and $R_y$ into $R$, split a single vector $R$ at angle $\theta$ into $R_x=R\cos\theta$ and $R_y=R\sin\theta$. Which trig ratio gives which component depends on which side $\theta$ is measured from, check the diagram.

Pause, copy both component formulas and the 50 m/s at 40° worked example (38 m/s horizontal, 32 m/s vertical) into your book before moving on.

We just saw how to resolve a single vector into components. That raises a question: Card 2 gave us relative VELOCITY, but where is one moving object relative to another, right now? This card answers it → relative position uses exactly the same vector-subtraction logic as relative velocity, applied to positions instead.

NESA syllabus (IQ2): "describe and analyse... the relative positions... of one object relative to another object on a plane." Every position can be written as a position vector, a displacement from some chosen origin (a wharf, an airfield, a lighthouse). Once two objects each have a position vector measured from the same origin, the position of one relative to the other is found the same way relative velocity was in Card 2, by subtraction.

So one hour after take-off, aircraft P is 500 km from aircraft Q, on a bearing of about 323° (37° west of north). Notice the method is identical to Card 1's triangle and Card 2's subtraction, only the quantity being subtracted has changed from a velocity to a position.

Relative position uses the same subtraction as relative velocity: $\vec{r}_{A\,\text{rel}\,B}=\vec{r}_A-\vec{r}_B$, where each position vector is measured from a common origin. Solve with the same triangle, Pythagoras for magnitude, tan for direction, stated as a bearing.

Pause, copy the relative-position formula and the two-aircraft worked example (500 km, bearing 323°) into your book before moving on.