1.4 Reference frames and relative motion

How fast you are moving depends on who is watching. Learn to switch between observers with one simple rule: add the velocities as vectors.

Little picture stories

Press Next (or Play) to walk through each story one small step at a time. Keep asking: who is watching?

1. Standing still on a moving walkway

2. Walking on the moving walkway

You walk forward at 2 m/s on a walkway that moves forward at 1 m/s. How fast are you going relative to the floor?

3. Flip who is watching, flip the sign

A car drives east at 20 m/s. Seen from inside the car, a lamp post by the road moves:

4. Check yourself

Think of your answer first, then tap to see it.

a) A walkway moves at 1 m/s. Mia stands still on it. What is her velocity relative to the walkway? Relative to the floor?

Show answer

Relative to the walkway: 0. Relative to the floor: 1 m/s, the walkway's speed.

b) Mia walks forward at 1.5 m/s along that walkway (1 m/s). How fast does Leo, standing on the floor, see her go?

Show answer

1.5 m/s + 1 m/s = 2.5 m/s forward. Her walk and the walkway add up.

c) You ride in a car going +15 m/s. What is the velocity of a parked tree, as you see it from the car?

Show answer

−15 m/s: the tree seems to rush backward. Flip who is watching, flip the sign.

Already know this?

Three quick questions. Get all three right on the first try and you can skip ahead.

A train moves at +20 m/s. A passenger walks at +1.5 m/s relative to the train. Her velocity relative to the ground:

Car A goes east at 30 m/s, car B east at 25 m/s. B's velocity relative to A:

A boat heads straight across a river at 4 m/s (relative to the water). The current is 3 m/s. Its speed relative to the bank:

Walking on a train

Read the story as text

You are on a train that rolls past a station at 20 m/s. You get up and walk toward the front of the train at 1.5 m/s to find a seat.

Your friend sitting in the train sees you walk slowly past at 1.5 m/s. A person standing on the platform sees you zoom by at 21.5 m/s. A bird flying beside the train at 20 m/s sees you creep forward at only 1.5 m/s too.

Nobody is wrong. Each person measures from their own place. So the question is: how do we find a velocity as seen by one observer when we know it as seen by another?

The idea: velocity is always "relative to something"

A reference frame is the point of view you measure from: an observer with a ruler and a clock. The ground is the frame we use most. The train, a boat, or the moving water can be frames too.

In pictures

  • Every speed is measured by someone, from somewhere: Leo on the floor, Mia on the walkway, you in the car.
  • Riding on something moving: its motion and your own motion add up (walking forward on the walkway: 1.5 + 1 = 2.5).
  • Walking against it: they partly cancel (Tom walking backward at the walkway's own speed stays in one spot).
  • Swap who is watching and the sign flips: from the car, the tree rushes backward.
  • Sideways motions (boat and river) add as arrows, tip to tail.

Below is the same idea in words, and then with numbers.

Position and velocity depend on the frame. The rule for velocities is short:

vA rel C = vA rel B + vB rel C

Read it as a chain: "you relative to the ground = you relative to the train + the train relative to the ground." The middle letters (B) match up and "cancel", like a chain of links.

Two more facts you will use all the time:

In 2D, add the velocities as arrows (vectors), not as plain numbers. When the two velocities are at right angles, use the Pythagorean theorem for the size and a tangent for the direction.

far bank near bank (start) boat rel water 2.0 m/s current 1.5 m/s boat rel ground 2.5 m/s downstream = +x (east)

Same idea with numbers: crossing a river

A river is 60 m wide and flows east at 1.5 m/s. A boat moves at 2.0 m/s relative to the water. The driver points the boat straight across (north). Take east = +x (downstream) and north = +y (across).

Show all steps as text
  1. Write the two velocities as components. Boat relative to water: (0, 2.0) m/s. Water relative to ground: (1.5, 0) m/s. The rule vboat,ground = vboat,water + vwater,ground works one component at a time.
  2. Add them: vboat,ground = (0 + 1.5, 2.0 + 0) = (1.5, 2.0) m/s. x parts add to x parts, y parts add to y parts.
  3. Time to cross: only the y part carries the boat across. t = 60 m ÷ 2.0 m/s = 30 s. The current is sideways, so it does not help or hurt the trip across.
  4. Drift downstream: x = 1.5 m/s × 30 s = 45 m. The boat rides the water east for the whole 30 s.
  5. Speed seen from the bank: √(1.5² + 2.0²) = √6.25 = 2.5 m/s, at tan⁻¹(1.5/2.0) = 36.9° east of straight across. The two parts are at right angles, so Pythagoras gives the size.
  6. Check: distance travelled on the ground = 2.5 m/s × 30 s = 75 m, and √(60² + 45²) = 75 m. ✓ Two independent methods agree, so the numbers are right.

What if the driver wants to land straight across? Point the boat upstream at angle θ (from straight across) so its upstream part cancels the current: 2.0 sin θ = 1.5, so sin θ = 0.75 and θ = 48.6°. Now the across part is only 2.0 cos 48.6° = 1.32 m/s, so the trip takes 60 ÷ 1.32 = 45.4 s. Landing straight across costs time.

Play: change who is watching

Use Watch from to jump between observers. Watch how positions and velocity arrows change, while the physics stays the same. Positive is to the right (east).

positionvelocityacceleration

Sim 1: walking on a train

Three observers: the Train, the Walker walking along it, and a person standing on the Platform at x = 40 m. Ground means the station itself. Try: walker velocity −2 m/s (walking to the back). Then watch from the train, and from the walker.

Sim 2: a boat crossing a river

East (+x) is downstream, north (+y) is across. Heading 0° means "pointed straight across"; negative angles point upstream. The Leaf floats with the water (watch from it to see the boat as the water sees it), and the Dock is the spot directly across from the start. Find the heading that lands the boat straight across (x = 0 at the far bank).

Try to beat the current: set current 1.5 m/s and boat speed 2.0 m/s, then change only the current. Does the crossing time change when heading is 0°? (It should not.)

Worked examples

Walking to the back of the train basic

A train moves east at 20 m/s. A passenger walks toward the back of the train at 1.5 m/s relative to the train. What is the passenger's velocity relative to the ground?

Show solution
Show all steps as text

Take east as +. Passenger relative to train: −1.5 m/s. Train relative to ground: +20 m/s.

vP,G = vP,T + vT,G = −1.5 + 20 = +18.5 m/s (east).

The passenger still moves east relative to the ground, just a bit slower than the train.

Two cars on a highway basic

Car A drives east at 30 m/s. Car B drives east at 25 m/s. (a) What is B's velocity relative to A? (b) What if B drives west at 25 m/s instead?

Show solution
Show all steps as text

East is +. The rule: vB,A = vB,G + vG,A = vB,G − vA,G.

(a) vB,A = 25 − 30 = −5 m/s. From car A, car B seems to drift backward (west) at 5 m/s.

(b) vB,A = −25 − 30 = −55 m/s. Head-on traffic seems to rush at you at 55 m/s. That is why head-on crashes are so dangerous.

An airplane in a crosswind medium

A plane flies at 200 m/s relative to the air, pointed due north. A wind blows toward the east at 50 m/s. Find the plane's velocity relative to the ground (size and direction).

Show solution
Show all steps as text

Components (east = +x, north = +y): plane rel air (0, 200), air rel ground (50, 0). Sum: (50, 200) m/s.

Speed: √(50² + 200²) = √42 500 = 206 m/s.

Direction: tan⁻¹(50/200) = 14.0° east of north.

Landing straight across AP

The river is 60 m wide and flows east at 1.5 m/s. The boat's speed relative to the water is 2.0 m/s. (a) Which way must the boat point to land directly across from its start? (b) How long does the trip take? (c) Compare with pointing straight across.

Show solution
Show all steps as text

(a) The boat's east-west part relative to the water must cancel the current: 2.0 sin θ = 1.5, so sin θ = 0.75 and θ = 48.6° upstream (west of north).

(b) Across part: 2.0 cos 48.6° = 2.0 × 0.661 = 1.32 m/s. Time: 60 ÷ 1.32 = 45.4 s.

(c) Pointing straight across takes 60 ÷ 2.0 = 30 s but lands 45 m downstream. Pointing straight across is always the fastest way over; aiming upstream trades time for landing spot.

Note: if the current were faster than the boat (say 2.5 m/s), sin θ would have to be 1.25, which is impossible. The boat could not land directly across.

Rain on a car window AP

Rain falls straight down at 8.0 m/s relative to the ground. A car drives east at 12 m/s. At what angle from vertical do raindrops streak across the side window, and how fast do they move relative to the car?

Show solution
Show all steps as text

vrain,car = vrain,ground + vground,car = (0, −8.0) + (−12, 0) = (−12, −8.0) m/s.

Speed: √(144 + 64) = √208 = 14.4 m/s.

Angle from vertical: tan⁻¹(12/8.0) = 56.3°, slanting toward the back of the car. The faster you drive, the more slanted the streaks.

Practice (AP style)

  1. A passenger walks toward the front of a train at 2.0 m/s relative to the train. The train moves at 15 m/s relative to the ground in the same direction. What is the passenger's speed relative to the ground?

    Show answer

    (C). Same direction, so v = 2.0 + 15 = 17 m/s. (A) subtracts by mistake; (D) multiplies.

  2. Car A moves east at 25 m/s. Truck B moves west at 20 m/s. What is the velocity of the truck relative to car A?

    Show answer

    (D). East +: vB,A = vB,G − vA,G = −20 − 25 = −45 m/s, so 45 m/s west.

  3. A boat moves at 3.0 m/s relative to the water and points straight across a river 90 m wide. The current is 4.0 m/s. How long does the crossing take?

    Show answer

    (C). Only the across part matters: t = 90 ÷ 3.0 = 30 s. (A) wrongly uses the ground speed √(3² + 4²) = 5 m/s.

  4. For the boat in question 3, how far downstream does it land?

    Show answer

    (C). Drift = 4.0 m/s × 30 s = 120 m. (D) is the total distance travelled, √(90² + 120²) = 150 m.

  5. The boat keeps pointing straight across, but the river's current gets faster. What happens to the time to cross?

    Show answer

    (C). Perpendicular motions are independent. The longer path is covered at a faster speed, and the two effects cancel exactly.

  6. A passenger on a train moving at constant velocity drops a ball. Which statement is correct?

    Show answer

    (B). In the train frame the ball starts at rest and falls straight. On the ground it already has the train's horizontal velocity, so it moves like a projectile: a curve. It lands right below the release point, at the passenger's feet.

  7. In question 6, how does the ball's acceleration measured by the passenger compare with the acceleration measured by the person on the ground?

    Show answer

    (A). Frames that move at constant velocity relative to each other agree on acceleration. Adding a constant velocity does not change how velocity changes.

  8. Short free response. A plane has an airspeed of 150 m/s. The pilot wants to travel due north to a city 600 km away. A steady wind blows toward the east at 30 m/s.

    (a) In which direction must the pilot point the plane? (b) What is the plane's speed relative to the ground? (c) How long does the trip take, in minutes?

    Show answer

    (a) The plane's west part must cancel the wind: 150 sin θ = 30, sin θ = 0.20, θ = 11.5° west of north.

    (b) North part: √(150² − 30²) = √21 600 = 147 m/s (this is the ground speed, since the east-west parts cancel).

    (c) t = 600 000 m ÷ 147 m/s = 4 080 s ≈ 68 min.

    Scoring idea: 1 point for a vector diagram or components that cancel the wind, 1 point for the angle, 1 point for using Pythagoras with subtraction (not 150 + 30 or √(150² + 30²)), 1 point for the time with units.

AP question types for this topic

Four short free response questions, one of each AP Physics 1 free response type, all about this topic. Write your answer first, then open the worked answer and score yourself with the points shown.

1. Mathematical Routines

A small plane flies from one town to another a distance D away, then straight back. Its airspeed (speed relative to the air) is v. A steady wind of speed w blows along the route, from the first town toward the second. Pilot Ines times the round trip.

(a) Derive an expression for the total round-trip time in terms of D, v and w.

(b) Calculate it for D = 36 km, v = 50 m/s and w = 10 m/s, and compare with no wind.

(c) Show from your expression that any wind along the route makes the round trip longer.

Show worked answer and scoring
  1. (a) Out: ground speed v + w; back: v − w. T = D/(v + w) + D/(v − w) = 2Dv/(v2 − w2). 1 point for the ground speeds, 1 point for the expression.
  2. (b) Out: 36 000 ÷ 60 = 600 s. Back: 36 000 ÷ 40 = 900 s. T = 1500 s (25 min). No wind: 72 000 ÷ 50 = 1440 s (24 min). 1 point.
  3. (c) No wind: T = 2D/v = 2Dv/v2. With wind the denominator v2 − w2 is smaller, so T is bigger: the plane spends longer at the slow speed than at the fast one. 1 point.

Total: 4 points.

2. Translation Between Representations

A moving walkway carries riders forward at 1.0 m/s relative to the ground. Noor walks on it backwards (against its motion) at 1.5 m/s relative to the walkway. Take forward as positive.

xt
A
xt
B
xt
C
xt
D

Position–time graphs. A: steep line rising. B: gentle line rising. C: gentle line falling. D: steep line falling.

(a) Which graph shows Noor's position as seen by Jonas, who stands still on the floor beside the walkway? Justify with a velocity equation.

(b) Which graph shows her position as seen by Priya, who stands still on the walkway? Explain.

(c) Describe Jonas's graph if Noor slows to 1.0 m/s relative to the walkway, and say what Jonas sees.

Show worked answer and scoring
  1. (a) C. vNG = vNW + vWG = −1.5 + 1.0 = −0.5 m/s: a gentle falling line. 1 point for C, 1 point for the equation.
  2. (b) D. Priya moves with the walkway, so she sees only vNW = −1.5 m/s, a steeper falling line. 1 point.
  3. (c) −1.0 + 1.0 = 0, so Jonas's graph is a flat horizontal line: he sees Noor walking but staying in the same place beside him. 1 point for flat, 1 point for the explanation.

Total: 5 points.

3. Experimental Design and Analysis

Ines wants to measure the speed of a long paper belt pulled steadily across a table by a motor. She has a battery toy car with several speed settings, a motion sensor fixed to the table and a metre stick.

(a) Describe how she can use the car to find the belt speed.

(b) Write the velocity equation she is testing, and say what to plot to get a straight line and what the slope and intercept mean.

(c) Her data (belt moving the same way as the car) are below. Find the belt speed, and the setting at which a car driving backwards on the belt would stay still.

vcar, still paper (m/s)0.200.400.600.80
vsensor (m/s)0.490.710.891.11

Show worked answer and scoring
  1. (a) With the belt stopped, measure each setting with the sensor: the car's speed relative to the paper. Then run the belt and measure the car's speed relative to the table at each setting, three times each. 1 point for measuring both speeds, 1 point for several settings with repeats.
  2. (b) vcar,table = vcar,belt + vbelt. Plot sensor speed against still-paper speed: slope should be 1, intercept = belt speed. 1 point for the equation, 1 point for slope and intercept.
  3. (c) Best fit: slope 1.02 (close to 1), intercept 0.29, so belt speed ≈ 0.29 m/s. A car driving backwards at about 0.29 m/s relative to the belt has zero velocity relative to the table. 1 point for the belt speed, 1 point for the backwards setting.

Total: 6 points.

4. Qualitative/Quantitative Translation

Ayo swims across a river 40 m wide, always pointing straight across, at 1.6 m/s relative to the water. The current flows at 1.2 m/s. Ayo claims: “The current does not change how long it takes me to cross.”

(a) Explain in words whether Ayo is right.

(b) Derive the crossing time and calculate it, plus how far downstream Ayo lands and the length of the path.

(c) Ayo's path is longer with the current. Connect your numbers to explain why the time is still the same.

Show worked answer and scoring
  1. (a) Yes. The current only adds velocity along the river, at right angles to the crossing. Perpendicular components are independent, so the across velocity, and so the time, is unchanged. 1 point.
  2. (b) t = d/u = 40 ÷ 1.6 = 25 s. Drift = 1.2 × 25 = 30 m. Path = √(402 + 302) = 50 m. 1 point for 25 s, 1 point for 30 m and 50 m.
  3. (c) Ground speed = √(1.62 + 1.22) = 2.0 m/s; 50 m ÷ 2.0 m/s = 25 s. The path grew by the same factor as the speed. In t = d/u the current speed does not appear at all, which matches (a). 1 point for the check, 1 point for the link.

Total: 5 points.

Common mistakes

The mistake: Adding speeds as plain numbers in 2D: "2 m/s across plus 1.5 m/s current = 3.5 m/s."

Why it is wrong: Velocities are vectors. At right angles they combine with Pythagoras: √(2² + 1.5²) = 2.5 m/s.

How to spot it: Your answer is bigger than any arrow-tip-to-tail drawing could make. Draw the arrows first.

The mistake: Forgetting signs in 1D, so "train 20 east, walk 1.5 to the back" becomes 21.5 m/s.

Why it is wrong: Walking to the back is the negative direction: 20 + (−1.5) = 18.5 m/s.

How to spot it: Ask "should this make me faster or slower than the train?" before you add.

The mistake: Thinking a faster current makes the crossing take longer (or shorter).

Why it is wrong: The current is perpendicular to the crossing direction. It only changes where you land, not when.

How to spot it: For time across, divide the width by the across part of velocity only.

The mistake: Mixing up vA rel B and vB rel A.

Why it is wrong: They are opposite: vB rel A = −vA rel B. From the car, the road moves backward.

How to spot it: Check that the inside letters chain: vA,C = vA,B + vB,C. If the letters do not link, flip one term.

The mistake: Thinking a dropped object on a moving train lands behind you.

Why it is wrong: The object already moves forward with the train when you let go, and nothing slows it horizontally (ignoring air). It lands at your feet.

How to spot it: Work in the train's frame first: there, everything starts at rest.