Relative Motion

Every velocity has a hidden second half

You are on a train doing 30 m/s. You walk toward the front at 1.5 m/s. How fast are you going? Both answers are right. Someone sitting across the aisle sees 1.5 m/s. Someone standing on the platform sees 31.5 m/s. Nobody is mistaken: the question was incomplete, because a velocity is not a property of an object. It is a comparison between an object and something you have chosen to call "not moving".

In projectile motion you never had to think about this. The ground was the frame, always, and the only thing moving was the ball. Now two things move at once: a boat and the river under it, a plane and the air around it, you and the rain. The whole difficulty of these problems is bookkeeping: keeping straight which velocity is measured against what. Get a notation that makes that impossible to forget and most of these problems become one triangle and a bit of trigonometry.

The subscript chain

Write $\vec{v}_{A/B}$ for "the velocity of A as measured by B", where B is whatever you are treating as stationary. Read the slash as "relative to".

Suppose you know how fast you walk along the train, $\vec{v}_{\text{you/train}}$, and how fast the train moves over the ground, $\vec{v}_{\text{train/ground}}$. In one second you move 1.5 m forward along the carriage, and in that same second the carriage itself is carried 30 m forward. The platform sees both displacements happen, one after the other, so it sees you move 31.5 m. Divide by the second and you have the rule:

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

The inner labels match and drop out, leaving the outer pair: like cancelling in a fraction. That is the whole trick. If the chain of subscripts doesn't close up like that, you have made an error before doing any arithmetic.

Two consequences you will use constantly. First, reversing a pair flips the sign: $\vec{v}_{B/A} = -\vec{v}_{A/B}$. If B sees A approaching at 8 m/s east, A sees B approaching at 8 m/s west. Second, these are vector sums, not number sums. A plane whose velocity relative to the air is 200 km/h north, in air moving east at 50 km/h, has a ground velocity with components (50 east, 200 north), so its ground speed is $\sqrt{200^2 + 50^2} = \sqrt{40000+2500} = \sqrt{42500} = 206.2$ km/h, not 250.

Crossing a river

This is the situation the whole topic is built on, so get comfortable with it. A boat can do speed $u$ relative to the water. The river runs at $v$ relative to the bank, and the channel is $w$ wide. The boat's nose points at angle $\theta$ upstream from the straight-across direction.

Split into components: across the river, and along it. Across the river the current contributes nothing (the water only carries you sideways) so the across-component of the boat's ground velocity is $u\cos\theta$, and the crossing takes

$t = \frac{w}{u\cos\theta}.$

Along the river the two contributions compete: the current pushes you downstream at $v$, and your upstream aim pulls you back at $u\sin\theta$. The net downstream velocity is $v - u\sin\theta$, so the sideways drift by the time you land is

$x = (v - u\sin\theta)\,\frac{w}{u\cos\theta}.$

Those two formulas answer every river question. Three cases matter.

Quickest crossing. Minimise $t$, which means maximise $u\cos\theta$, which means $\theta = 0$. Aim straight across, and $t = w/u$ no matter how fast the river runs. That surprises people. The current is entirely sideways; it cannot slow down your progress toward the far bank.

Land directly opposite. Set the drift to zero: $v - u\sin\theta = 0$, so $\sin\theta = v/u$. The across-component left over is $u\cos\theta = u\sqrt{1-(v/u)^2} = \sqrt{u^2-v^2}$, and the trip takes $t = w/\sqrt{u^2-v^2}$. Note this only has a solution when $u > v$. A swimmer slower than the current cannot land opposite, ever.

Least drift when $u < v$. You cannot get the drift to zero, so make it as small as you can. From the drift formula, $x/w = \frac{v}{u}\sec\theta - \tan\theta$. Differentiate with respect to $\theta$ and set it to zero: $\frac{v}{u}\sec\theta\tan\theta - \sec^2\theta = 0$, which tidies to $\sin\theta = u/v$. Notice the ratio is upside down compared to the "land opposite" case, which is a good sanity check on which regime you are in.

Three river crossings, worked

Example 1. A river is 80 m wide and flows at 3.0 m/s. A boat that does 4.0 m/s in still water is aimed straight across. Find the crossing time and the drift.

Straight across means $\theta = 0$. The across-speed is the full 4.0 m/s, so

$t = \frac{80\ \text{m}}{4.0\ \text{m/s}} = 20\ \text{s}.$

During those 20 s the current carries the boat downstream at 3.0 m/s the whole time, so $x = 3.0 \times 20 = 60$ m. The boat lands 60 m downstream, 20 s after setting off. Its speed over the ground was 5.0 m/s, but that number was never needed.

Example 2. A 200 m river flows at 2.0 m/s. A boat capable of 3.0 m/s relative to the water must land at a point 50 m downstream of the point opposite its start. How long does the crossing take?

Now the drift is prescribed, so you cannot pick $\theta$ freely. Write both components in terms of the unknown time $t$. Across: $3.0\cos\theta \cdot t = 200$. Along: $(2.0 - 3.0\sin\theta)\,t = 50$.

Two equations, two unknowns. Solve each for the trig function: $\cos\theta = 200/(3t)$ and $\sin\theta = (2.0 - 50/t)/3.0 = (2t-50)/(3t)$. Now use $\sin^2+\cos^2 = 1$, which is what lets you eliminate the angle you don't care about:

$\frac{200^2 + (2t-50)^2}{9t^2} = 1.$

Multiply out: $40000 + 4t^2 - 200t + 2500 = 9t^2$, so $5t^2 + 200t - 42500 = 0$, or $t^2 + 40t - 8500 = 0$. The quadratic formula gives

(The other root is negative, so discard it.) About 74 s. The useful move here was refusing to solve for $\theta$: you squared and added to make it vanish.

Example 3. A swimmer manages 1.20 m/s relative to the water in a 60 m river running at 1.60 m/s. What is the smallest downstream drift he can achieve?

Here $u = 1.20 < v = 1.60$, so landing opposite is impossible and this is the least-drift case: $\sin\theta = u/v = 1.20/1.60 = 0.750$, giving $\theta = 48.6^\circ$ and $\cos\theta = \sqrt{1-0.5625} = 0.6614$.

Across-speed: $1.20 \times 0.6614 = 0.7937$ m/s, so the crossing takes $t = 60/0.7937 = 75.60$ s. Net downstream speed: $1.60 - 1.20 \times 0.750 = 1.60 - 0.90 = 0.700$ m/s. The drift is therefore

$x = 0.700 \times 75.60 = 52.9\ \text{m}.$

He lands 52.9 m downstream, and no other heading does better.

Rain, wind, and apparent directions

The same rule explains something you have seen. Rain falling straight down looks slanted when you run through it, because what you see is $\vec{v}_{\text{rain/you}} = \vec{v}_{\text{rain/ground}} - \vec{v}_{\text{you/ground}}$. Subtracting your own velocity tilts the apparent fall toward you.

If rain falls vertically at 8.0 m/s and you cycle east at 6.0 m/s, then in your frame the rain has a horizontal component of 6.0 m/s westward (toward your face) and 8.0 m/s down. The angle from the vertical is $\arctan(6.0/8.0) = 36.9^\circ$.

Run it backwards and it becomes a real puzzle. Example 4. A man walking east at 3.0 m/s sees the rain fall exactly vertically. When he speeds up to 6.0 m/s east, it appears to come from ahead at $30^\circ$ from the vertical. Find the rain's actual speed relative to the ground.

Let the rain's ground velocity have eastward component $R_x$ and downward component $R_y$. In his frame the horizontal part is $R_x - 3.0$, and he reports it as zero, so $R_x = 3.0$ m/s. The rain is genuinely drifting east at 3.0 m/s.

At 6.0 m/s the apparent horizontal part is $3.0 - 6.0 = -3.0$ m/s, i.e. 3.0 m/s from ahead: consistent with what he says. The vertical part is unchanged by his walking, so

The actual speed is $\sqrt{3.0^2 + 5.196^2} = \sqrt{9.0 + 27.0} = \sqrt{36.0} = 6.0$ m/s. The trick was realising that changing frames only ever changes the component along your own motion: the vertical part is the same for everyone.

Stand on one of them

The real payoff is that a well-chosen frame can delete most of a problem. If two things are moving, sit on one. It is now at rest, and only one object is left moving.

Two trains, 120 m and 180 m long, pass each other in opposite directions at 20 m/s and 25 m/s. Sit on the first train. It is stationary; the second sweeps past at $20 + 25 = 45$ m/s and must cover the sum of the lengths, $120 + 180 = 300$ m, from nose-meets-nose to tail-clears-tail. So $t = 300/45 = 6.67$ s. No absolute positions, no algebra.

Free fall is even better. Two objects both accelerate downward at $g$, so in the frame of one of them the other has zero acceleration: gravity cancels out of the relative motion. Throw a ball up at 20 m/s from the ground at the instant another is dropped from rest 50 m above it. Relative to the dropped ball, the thrown ball moves at a constant 20 m/s upward, closing a fixed 50 m gap. They meet at $t = 50/20 = 2.5$ s. That is the whole solution.

The same cancelling idea handles the four ants. Four ants sit at the corners of a 1.50 m square, each crawling at 0.020 m/s straight at the ant counterclockwise from it. By symmetry the four always form a square, so the ant you are chasing always moves perpendicular to the line between you: its motion never changes the gap. The gap therefore closes at your full 0.020 m/s, and the time is $1.50/0.020 = 75.0$ s, even though the paths are spirals.

Example 5. Ship A sails due east at 15 km/h. At that instant ship B is 20 km due north of A, sailing due south at 25 km/h. Find their closest separation and when it happens.

Sit on A. Take east as $x$, north as $y$. Then $\vec{v}_{B/A} = \vec{v}_B - \vec{v}_A = (0, -25) - (15, 0) = (-15, -25)$ km/h. In A's frame, B starts at $(0, 20)$ and travels in a straight line with that constant velocity. Its position relative to A is

$\vec{r}(t) = (-15t,\; 20 - 25t).$

The separation squared is $s^2 = 225t^2 + (20-25t)^2 = 225t^2 + 400 - 1000t + 625t^2 = 850t^2 - 1000t + 400$. Minimise by differentiating: $1700t - 1000 = 0$, so

$t = \frac{1000}{1700} = 0.588\ \text{h} \approx 0.59\ \text{h}.$

Put that back: $s^2 = 850(0.3460) - 1000(0.5882) + 400 = 294.1 - 588.2 + 400 = 105.9$, so $s = 10.29$ km.

Geometrically, the closest approach is just the perpendicular distance from A to B's straight-line relative path, which is why the frame change is worth making.

The method survives when the two paths are not at right angles. Two roads cross at $60^\circ$; car A is 400 m from the junction closing at 20 m/s, car B is 300 m out on the other road closing at 12 m/s. Let $a = 400 - 20t$ and $b = 300 - 12t$ be the two distances from the junction. The law of cosines gives the separation directly, $s^2 = a^2 + b^2 - 2ab\cos 60^\circ = a^2+b^2-ab$, and you minimise that in $t$ exactly as above. The one subtlety is that the answer comes out at $t \approx 20.4$ s, a moment after car A has gone through the junction, so $a$ is negative there. That is legitimate (a negative $a$ just means "this far past the junction") but it is worth noticing rather than assuming the cars are still approaching.

Escalators and walkways are the same rule at its gentlest. Walk up a 24 m escalator at 0.80 m/s relative to the steps while the steps rise at 0.60 m/s: your ground speed is $0.80 + 0.60 = 1.40$ m/s, so the trip takes $24/1.40 = 17.1$ s. Chain the subscripts (you relative to steps, steps relative to ground) and you cannot get the sign wrong.

How to recognise one of these

Three signals, any one of which should make you write subscripts on every velocity before doing anything else.

Two things move at once, and one carries the other. Boat and river, plane and air, walker and escalator, you and the rain. The giveaway wording is "relative to the water", "in still air", "relative to the steps". The speed you were handed is not a ground speed.

The question asks for a distance or time between two moving objects. Closest approach, time to overtake, how long to pass, when do they meet. Do not track both positions in the ground frame. Sit on one of them, get $\vec{v}_{B/A}$, and the problem collapses to one object moving in a straight line from a known starting offset.

Something is described as "apparent" or "seems to". Rain that appears to fall at an angle, a wind that seems to come from ahead, a star that appears displaced. "Appears" is a direct instruction to work in the observer's frame, and the translation is always subtracting the observer's own velocity.

One caution to carry forward: everything here assumed the frames move at constant velocity relative to one another. Swap to an accelerating frame (a braking bus, a spinning turntable) and objects start to appear to accelerate with no force pushing them. Sorting out what does and does not count as a force is the job of the next lesson, Newton's three laws and free body diagrams.