Visighta

Vector Kinematics in Two and Three Dimensions

3.5 Relative Motion and Moving Reference Frames

Crossing the River

You are on a boat crossing a river. You point the bow straight at the opposite bank and push the throttle forward. The engine drives you through the water exactly where you aimed --- perpendicular to the banks.

But when you look up a minute later, you are not heading straight across. The current has been carrying you downstream the entire time. You end up landing far from the point you aimed at, on a diagonal path you never intended.

Here is the strange part: from the perspective of the water itself, you did go straight across. Your velocity relative to the water pointed exactly where you aimed. But your velocity relative to the ground --- the thing that determines where you actually land --- pointed somewhere else entirely.

Which velocity is the "real" one?

Both. And that is the point. Velocity is always relative to something, and different reference frames give different velocities for the same object. The question is not which frame is correct, but which frame answers the question you are asking.

Before you read on: A plane flies due north at 200 km/h through air that is blowing due east at 50 km/h. In which direction does the plane actually move relative to the ground?

Sketch the two velocity vectors (plane relative to air, and air relative to ground) and try to draw the resultant before continuing.

[Interactive: Predict-Then-Reveal. Students draw or select a direction on a compass rose. After submitting, the correct answer is shown: the plane moves slightly east of north. The ground velocity vector is the diagonal of the rectangle formed by 200 km/h north and 50 km/h east, pointing at about 14 degrees east of north with a magnitude of approximately 206 km/h.]

If you expected "north," you were thinking about the plane's heading --- its direction relative to the air. But the ground does not care about the air. The ground sees the plane drifting northeast, carried partly by the wind. The plane's velocity relative to the ground combines two contributions: the plane's motion through the air, and the air's motion over the ground.

This section is about how to combine those contributions correctly.

The Key Question

How do velocities combine when the observer and the object are both moving?

You have encountered reference frames before. Section 1.3 introduced the idea that the description of motion depends on who is watching --- the choice of reference frame. At the time, that was a conceptual idea. Now we make it computational.

The answer turns out to be remarkably simple: velocity addition is vector addition. But the simplicity hides a trap. Adding velocities correctly requires keeping careful track of which velocity is measured relative to which frame. A subscript notation will save us from confusion.

Exploring Relative Motion

[Interactive: River Crossing Simulator. A top-down view of a river flowing from left to right. The student controls two things: (1) the boat's speed through the water (a slider, 1--10 m/s), and (2) the boat's heading angle relative to the bank (a dial, 0--180 degrees, where 90 degrees points straight across). The river current speed is adjustable (a slider, 0--5 m/s). The simulation shows three things simultaneously: the boat's velocity relative to the water (a blue arrow), the river current velocity (a green arrow), and the boat's velocity relative to the ground (a red arrow, drawn as the vector sum of the other two). As the boat crosses, its path is traced on the screen. A target marker sits on the opposite bank.]

Challenge 1: Set the boat heading to 90 degrees (straight across) and observe the path. Does the boat reach the opposite bank directly across from where it started?

Challenge 2: Adjust the heading until the boat reaches the target directly across. What angle did you need? Why is it not 90 degrees?

Challenge 3: Now try to find the heading that gets you across in the minimum time. Is it the same heading that gets you directly across?

Challenge 4: Increase the river current until it is faster than the boat. Can you still reach the target directly across? What happens?

If you spent time with that simulator, you discovered several things. Pointing straight across gets you across quickly but you drift downstream. Angling upstream compensates for the drift, but the crossing takes longer because some of your velocity is now fighting the current rather than carrying you across. And if the current is faster than your boat, you cannot aim upstream enough to cancel the drift --- no heading will get you straight across.

These are not separate facts to memorize. They all follow from one idea: the velocity you see from the ground is the vector sum of the velocity through the water and the velocity of the water itself.

Velocity Addition: The Concept

Let's name three things:

  • A = the boat
  • B = the water (the river)
  • C = the ground (the riverbank)

The boat moves through the water. The water moves over the ground. We want the boat's motion relative to the ground.

Here is the rule, stated in words first:

The velocity of A relative to C equals the velocity of A relative to B plus the velocity of B relative to C.

In symbols:

$$\vec{v}{A/C} = \vec{v}$$} + \vec{v}_{B/C

Read the subscripts like a chain: A relative to B, plus B relative to C, gives A relative to C. The intermediate frame (B) appears as both a denominator and a numerator, and it "cancels" in the chain, leaving A relative to C.

This is not a new physical law. It is the same vector addition you have been using since Section 3.1. The only novelty is the bookkeeping --- the subscript notation that keeps track of which frame each velocity belongs to.

The Subscript Chain: Why It Works

The subscript notation $\vec{v}_{A/B}$ means "the velocity of A as measured by an observer in frame B." Think of the slash as "relative to" or "as seen from."

The chain rule works because of what relative position means. Position of A relative to C is:

$$\vec{r}{A/C} = \vec{r}$$} + \vec{r}_{B/C

This says: to get from C to A, first go from C to B, then from B to A. It is just vector addition of displacements. Now take the time derivative of both sides:

$$\frac{d}{dt}\vec{r}{A/C} = \frac{d}{dt}\vec{r}$$} + \frac{d}{dt}\vec{r}_{B/C

$$\vec{v}{A/C} = \vec{v}$$} + \vec{v}_{B/C

The velocity addition formula is a consequence of the position addition formula, which is a consequence of basic vector geometry. There is nothing to memorize --- only something to understand.

A useful fact: reversing the subscripts reverses the vector.

$$\vec{v}{B/A} = -\vec{v}$$

If you see the boat moving east at 5 m/s from the riverbank, then from the boat, the riverbank moves west at 5 m/s. Same speed, opposite direction.

Worked Example: The Plane in the Wind

A pilot wants to fly due north. The plane's airspeed (speed relative to the air) is 200 km/h. A wind blows from west to east at 50 km/h. What is the plane's velocity relative to the ground?

Identify the frames:

  • A = plane
  • B = air
  • C = ground

Write the known velocities:

  • $\vec{v}_{A/B}$ = 200 km/h north (plane's velocity relative to the air)
  • $\vec{v}_{B/C}$ = 50 km/h east (air's velocity relative to the ground --- this is the wind)

Apply the formula:

$$\vec{v}{A/C} = \vec{v}$$} + \vec{v}_{B/C

Using components with north as $+y$ and east as $+x$:

$$\vec{v}_{A/C} = (0, 200) + (50, 0) = (50, 200) \text{ km/h}$$

Find the magnitude:

$$|\vec{v}_{A/C}| = \sqrt{50^2 + 200^2} = \sqrt{2500 + 40000} = \sqrt{42500} \approx 206 \text{ km/h}$$

Find the direction:

$$\theta = \arctan\left(\frac{50}{200}\right) = \arctan(0.25) \approx 14° \text{ east of north}$$

The plane ends up moving at about 206 km/h in a direction 14 degrees east of north --- even though the pilot aimed due north. The wind carried the plane sideways.

Before you read on: If the pilot actually wants to arrive at a city due north, what heading should the pilot choose? The answer is not "north." Think about what heading would make $\vec{v}_{A/C}$ point north, and then try to set it up before reading the next example.

Worked Example: Correcting for the Wind

Same setup: airspeed 200 km/h, wind 50 km/h due east. But now the pilot wants the ground velocity to point due north. What heading should the pilot use?

This time, $\vec{v}{A/C}$ must point north --- its east-west component must be zero. We need to choose $\vec{v}$ so that it cancels the wind's eastward push.

Let the heading be an angle $\alpha$ west of north. Then:

$$\vec{v}_{A/B} = (-200\sin\alpha, \, 200\cos\alpha)$$

The wind is still $\vec{v}_{B/C} = (50, 0)$. The ground velocity is:

$$\vec{v}_{A/C} = (-200\sin\alpha + 50, \, 200\cos\alpha)$$

Set the east-west component to zero:

$$-200\sin\alpha + 50 = 0 \implies \sin\alpha = \frac{50}{200} = 0.25 \implies \alpha \approx 14.5°$$

The pilot must aim about 14.5 degrees west of north. The ground speed is then:

$$v_{A/C} = 200\cos(14.5°) \approx 194 \text{ km/h}$$

Notice: correcting for the wind costs speed. The ground speed dropped from 200 to 194 km/h. Some of the plane's airspeed is now spent fighting the wind, leaving less for northward progress.

Two Classic River Problems

The riverboat scenario has two natural questions, and they have different answers. Understanding why they differ is the key insight.

Setup: A river is $d = 200$ m wide. The boat's speed through the water is $v_b = 5$ m/s. The river current is $v_r = 3$ m/s.

Question 1: Minimum crossing time

Before you read on: To cross the river in the shortest possible time, should you (a) aim straight across, (b) aim upstream, or (c) aim downstream?

Check your answer **(a) Aim straight across.** Crossing time depends only on how fast you move across the river --- the perpendicular component of your velocity. Aiming straight across puts all of your boat speed into the perpendicular direction. Any other heading diverts some of your speed along the river, reducing the perpendicular component and taking longer. Crossing time: $t = \frac{d}{v_b} = \frac{200}{5} = 40$ s. You will drift downstream by $v_r \times t = 3 \times 40 = 120$ m. But you will be across in the minimum possible time.

Question 2: Minimum drift (arriving directly across)

Before you read on: To arrive directly across from your starting point, should you (a) aim straight across, (b) aim upstream at some angle, or (c) it depends on the current speed?

Check your answer **(b) Aim upstream.** You need the ground velocity to have zero component along the river. That means the upstream component of your boat velocity must exactly cancel the river current. Let $\alpha$ be the angle upstream from straight across. Then: $$v_b \sin\alpha = v_r \implies \sin\alpha = \frac{v_r}{v_b} = \frac{3}{5} \implies \alpha \approx 37°$$ Your perpendicular speed is $v_b \cos\alpha = 5 \times \frac{4}{5} = 4$ m/s. Crossing time: $t = \frac{200}{4} = 50$ s. This is longer than the minimum crossing time (40 s), but you arrive exactly across. Note: this only works if $v_b > v_r$. If the current is faster than your boat, no heading can cancel the drift, and you cannot arrive directly across.

The contrast between these two problems illustrates the core tension: you can minimize time or minimize drift, but generally not both. The heading that crosses fastest accepts drift. The heading that eliminates drift accepts a slower crossing. This is a consequence of the vector nature of velocity --- you are allocating a fixed vector magnitude among two perpendicular directions, and optimizing one costs the other.

The Misconception: Adding Speeds Instead of Velocities

Here is an error students make frequently enough that it deserves its own discussion.

The mistake: "The boat goes 5 m/s and the river goes 3 m/s, so the ground speed is 8 m/s."

This is wrong in general. It happens to be correct in one special case: when the boat and the current move in the same direction. If the boat heads downstream, the speeds add: $5 + 3 = 8$ m/s.

But if the boat heads straight across, the ground speed is $\sqrt{5^2 + 3^2} \approx 5.8$ m/s. And if the boat heads upstream, the ground speed is $5 - 3 = 2$ m/s.

Scalar speed addition works only for motion along the same line. In every other case, you must add velocity vectors, which means using components and the Pythagorean theorem (or the full vector addition formula). This is a direct echo of the chapter's hero concept: components matter, and direction matters.

A Brief Historical Note

The problem of relative motion has practical roots. Sailors navigating rivers, pilots flying through winds, and travelers on moving ships have needed velocity addition for centuries. Galileo described a version of the problem in his Dialogue Concerning the Two Chief World Systems (1632), arguing that motion below decks on a smoothly sailing ship is indistinguishable from motion on land --- an early statement of what we now call Galilean relativity. The velocity addition formula $\vec{v}{A/C} = \vec{v}$ is sometimes called the Galilean velocity transformation. It works beautifully at everyday speeds. At speeds approaching the speed of light, Einstein showed that it needs modification --- but that is a story for a different course.} + \vec{v}_{B/C

Connection

Section 1.3 introduced reference frames as a conceptual tool: the description of motion depends on who is watching. Now you have seen the computational machinery behind that idea. Switching reference frames is not mysterious. It is vector addition. The subscript chain $\vec{v}{A/B} + \vec{v}$ is the entire technique.} = \vec{v}_{A/C

And notice how this connects to the hero concept of the chapter. When the boat crosses the river, the across-the-river component and the along-the-river component behave independently. The current does not affect how fast you cross; it only affects how far you drift. This is component independence at work, applied to reference-frame problems.

Practice

Layer 1: Concrete

A swimmer can swim at 2.0 m/s in still water. She swims in a river that flows east at 1.5 m/s. She aims her body due north (perpendicular to the current).

(a) What is her velocity relative to the ground? Give magnitude and direction.

(b) If the river is 80 m wide (north-south), how long does it take her to cross?

(c) How far downstream does she end up?

Check your answer **(a)** Her velocity relative to the water is $\vec{v}_{S/W} = (0, 2.0)$ m/s (north). The water's velocity relative to the ground is $\vec{v}_{W/G} = (1.5, 0)$ m/s (east). $$\vec{v}_{S/G} = (0 + 1.5, \, 2.0 + 0) = (1.5, 2.0) \text{ m/s}$$ Magnitude: $\sqrt{1.5^2 + 2.0^2} = \sqrt{2.25 + 4.0} = \sqrt{6.25} = 2.5$ m/s. Direction: $\theta = \arctan(1.5/2.0) = \arctan(0.75) \approx 36.9°$ east of north. **(b)** The northward component of her ground velocity is 2.0 m/s. Crossing time: $t = 80/2.0 = 40$ s. **(c)** Downstream drift: $1.5 \times 40 = 60$ m east.

Layer 2: Pattern

In each river scenario below, the river is 100 m wide, the boat speed in water is $v_b$, and the current is $v_r$. For each case, find (i) the minimum crossing time and the drift, and (ii) the heading for zero drift and the crossing time at that heading.

(a) $v_b = 4$ m/s, $v_r = 2$ m/s

(b) $v_b = 4$ m/s, $v_r = 4$ m/s

(c) $v_b = 4$ m/s, $v_r = 6$ m/s

Check your answer **(a)** $v_b = 4$, $v_r = 2$. (i) Minimum crossing time: aim straight across. $t = 100/4 = 25$ s. Drift: $2 \times 25 = 50$ m. (ii) Zero drift: $\sin\alpha = 2/4 = 0.5$, so $\alpha = 30°$ upstream. Perpendicular speed: $4\cos 30° \approx 3.46$ m/s. Time: $100/3.46 \approx 28.9$ s. **(b)** $v_b = 4$, $v_r = 4$. (i) Minimum crossing time: aim straight across. $t = 100/4 = 25$ s. Drift: $4 \times 25 = 100$ m. (ii) Zero drift: $\sin\alpha = 4/4 = 1$, so $\alpha = 90°$ --- you would aim directly upstream. Perpendicular speed: $4\cos 90° = 0$ m/s. You never cross. At the boundary case $v_r = v_b$, zero drift requires all your speed to fight the current, leaving nothing for crossing. **(c)** $v_b = 4$, $v_r = 6$. (i) Minimum crossing time: aim straight across. $t = 100/4 = 25$ s. Drift: $6 \times 25 = 150$ m. (ii) Zero drift: impossible. $\sin\alpha = 6/4 = 1.5 > 1$. No angle works. When the current is faster than the boat, you cannot cancel the drift. **Pattern:** As $v_r/v_b$ increases from 0 to 1, the zero-drift heading swings from straight across to straight upstream. Beyond 1, zero drift is impossible. The minimum crossing time is always $d/v_b$, regardless of the current.

Layer 3: Structure

The subscript chain says $\vec{v}{A/B} + \vec{v}$. The "inner" subscripts (B) match and "cancel."} = \vec{v}_{A/C

(a) What would $\vec{v}{A/B} + \vec{v}$ represent? Can you simplify it using the subscript chain? Why or why not?

(b) A bird flies relative to the air, the air moves relative to a train, and the train moves relative to the ground. Write the chain that gives the bird's velocity relative to the ground.

(c) Why does the chain work for any number of intermediate frames, not just one?

Check your answer **(a)** The inner subscripts are B and C --- they do not match, so the chain does not simplify directly. To use it, you would need to rewrite $\vec{v}_{C/A} = -\vec{v}_{A/C}$, giving $\vec{v}_{A/B} - \vec{v}_{A/C}$. Using the chain rule, this equals $\vec{v}_{A/B} - \vec{v}_{A/B} - \vec{v}_{B/C} = -\vec{v}_{B/C} = \vec{v}_{C/B}$. The point: mixing up the subscript order does not give you what you want unless you carefully track the signs. **(b)** Let D = bird, A = air, T = train, G = ground. $$\vec{v}_{D/G} = \vec{v}_{D/A} + \vec{v}_{A/T} + \vec{v}_{T/G}$$ Each inner subscript (A, T) appears as both a "denominator" and a "numerator" in adjacent terms. **(c)** Each link in the chain is a position addition: $\vec{r}_{A/C} = \vec{r}_{A/B} + \vec{r}_{B/C}$. You can insert as many intermediate points as you like: $\vec{r}_{A/D} = \vec{r}_{A/B} + \vec{r}_{B/C} + \vec{r}_{C/D}$. Differentiating gives the velocity chain. It works for any number of frames because vector addition is associative --- you can chain displacements through as many intermediate points as needed.

Layer 4: Debug

A student solving a river-crossing problem writes: "The boat speed is 5 m/s and the river speed is 3 m/s, so the boat's ground speed is $5 + 3 = 8$ m/s."

(a) Describe a situation where this calculation is correct.

(b) Describe a situation where this calculation is wrong, and find the actual ground speed.

(c) What general principle distinguishes cases where speed addition works from cases where it does not?

Check your answer **(a)** The calculation is correct when the boat moves in the same direction as the current --- for example, heading downstream. The velocity vectors are parallel, so the magnitudes add: $|\vec{v}_{B/G}| = |\vec{v}_{B/W}| + |\vec{v}_{W/G}| = 5 + 3 = 8$ m/s. **(b)** The calculation is wrong when the boat heads across the river (perpendicular to the current). The velocity vectors are perpendicular, and the ground speed is $\sqrt{5^2 + 3^2} = \sqrt{34} \approx 5.8$ m/s --- not 8 m/s. **(c)** Speed addition (adding magnitudes) works only when the velocities are parallel (same or opposite directions). When the velocities point in different directions, you must add them as vectors. The magnitude of a vector sum is generally not the sum of the magnitudes: $|\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}|$, with equality only when the vectors are parallel.

Reflection

Think back over this section.

What does "velocity" mean without specifying "relative to what"?

You might also consider: in everyday life, you rarely state a reference frame. When you say "the car is going 60 mph," relative to what? When is the unstated frame obvious enough to omit, and when does leaving it out cause real confusion?

Looking Ahead

You have now added reference frames to your computational toolkit. Velocity addition is vector addition --- simple in principle, but demanding in bookkeeping. The subscript chain $\vec{v}{A/B} + \vec{v}$ will reappear whenever two frames of reference are in play, which in real physics is almost always.} = \vec{v}_{A/C

In the next section, we turn to a different question. So far in this chapter, we have decomposed vectors into $x$ and $y$ components --- directions fixed in space. But when an object moves along a curve, it is often more natural to decompose acceleration into a part along the path (which changes speed) and a part perpendicular to the path (which changes direction). These tangential and normal components will reveal something surprising: turning is a form of acceleration, even when speed stays constant.