Reference Frames and Relative Motion: Study Guide (Physics C: Mechanics 1.4)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 1.4 with this title. In Physics 1, adding velocities between frames stays in one dimension. In this calculus-based course you add them as full vectors in two or three dimensions, and you prove the acceleration rule by differentiating. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

What a reference frame is

A reference frame is the viewpoint of an observer, made precise. It has:

• an origin (the observer’s zero of position),
• a set of axes with positive directions (for example +x east, +y north),
• a clock, and
• a motion of its own relative to other frames (at rest on the ground, riding in a train, flying in the air mass).

Every position, displacement, velocity and acceleration in physics is measured in some frame. Two observers who use different frames can report different sizes and different directions for the same quantity. Neither is wrong. They are describing the same motion from different places.

A quick example. You sit in a bus that moves east at 15 m/s. A cup rests on your tray. In your frame the cup has velocity 0. In the frame of a person on the pavement, the cup moves east at 15 m/s. Both statements are true, because each belongs to a different frame.

So always write down whose frame you are using, as well as the axis, before you write any number.

Notation: subscripts that name the observer

We write v_PA for “the velocity of P relative to A”, that is, as measured by an observer at rest in frame A. The first letter is the object; the second is the frame. The same pattern works for position r_PA and acceleration a_PA.

Two rules follow directly:

• Reversal: the velocity of A relative to P is the opposite vector, v_AP = −v_PA.
• Chaining: to get from P to A through a third frame B, add the vectors with matching inner letters:

v_PA = v_PB + v_BA

Read it as “P relative to B, plus B relative to A”. The inner B’s match and drop out of the result. You can extend the chain: v_PA = v_PC + v_CB + v_BA.

One dimension first

Take +x east, with all velocities measured relative to the ground G. Car A moves at v_AG = +28 m/s. Truck B moves at v_BG = +22 m/s. Car C comes the other way at v_CG = −25 m/s.

• Velocity of A relative to B: v_AB = v_AG + v_GB = v_AG − v_BG = 28 − 22 = +6.0 m/s. The driver of the truck sees the car pull ahead slowly.
• Reversed: v_BA = −6.0 m/s. From the car, the truck drifts backwards.
• Velocity of C relative to A: v_CA = v_CG − v_AG = −25 − 28 = −53 m/s. The oncoming car closes at 53 m/s.

The same quantity, the truck’s velocity, is +22 m/s for the ground and −6.0 m/s for car A. The frame decides both the size and the sign.

Converting measurements between frames

Here is where the velocity rule comes from. Let frame S be the ground and frame S′ be a cart that moves along +x at a constant velocity V relative to S. Their origins match at t = 0. After a time t, the origin O′ is a distance Vt from O. A point P is at x in S and at x′ in S′.

From the figure, x′ = x − Vt. This converts a position measured in S into a position measured in S′. Now differentiate with respect to time, remembering that V is constant:

1. Velocity: dx′/dt = dx/dt − V, so v′ = v − V. This is v_PS′ = v_PS + v_SS′ in one dimension, since v_SS′ = −V.
2. Acceleration: d²x′/dt² = d²x/dt² − dV/dt = d²x/dt², so a′ = a.

In three dimensions the same steps use vectors. The position link is r_PA = r_PB + r_BA. Differentiating once gives v_PA = v_PB + v_BA. Differentiating again gives a_PA = a_PB + a_BA. When frame B moves at constant velocity relative to A, a_BA = 0 and

a_PA = a_PB

This is why all inertial observers measure the same acceleration for an object, even though they disagree about its position and velocity.

Inertial frames and the boundary

An inertial frame is one that does not accelerate. Any frame moving at constant velocity relative to an inertial frame is also inertial. If frame B did accelerate relative to A, then a_BA ≠ 0 and the two observers would disagree about accelerations. A passenger in a braking bus sees a loose bag slide forward with no push; that is the sign of a non-inertial frame. In this course, assume every frame is inertial unless a problem tells you otherwise. The ground is close enough to inertial for the motions you will meet.

Background (not assessed in this topic): these rules hold for everyday speeds. For speeds approaching the speed of light, velocities do not simply add, and special relativity replaces them.

Worked example 1: a drone in a crosswind

Question. Take +x east and +y north, relative to the ground G. A delivery drone D flies at 12 m/s relative to the air A. A steady wind carries the air east at 5.0 m/s relative to the ground. (a) The drone points due north. Find its velocity relative to the ground and how far east it drifts while it moves 600 m north. (b) Which way must it point so that it travels due north over the ground, and how long does the 600 m trip then take?

(a) Heading north.

1. Write the chain with matching inner letters: v_DG = v_DA + v_AG.
2. Components: v_DA = (0, 12) m/s and v_AG = (5.0, 0) m/s. So v_DG = (5.0, 12) m/s.
3. Magnitude: |v_DG| = √(5.0² + 12²) = 13 m/s. Direction: tan θ = 5.0/12, so θ = 22.6° east of north.
4. Time to move 600 m north: the north component is 12 m/s, so t = 600 ÷ 12 = 50 s. Drift east = 5.0 × 50 = 250 m.

(b) Tracking due north.

1. Now v_DG must have no east component. The drone must point partly west, into the wind, at an angle θ west of north.
2. East components: −12 sin θ + 5.0 = 0, so sin θ = 5.0/12 and θ = 24.6° west of north.
3. North component: 12 cos θ = √(12² − 5.0²) = √119 ≈ 10.9 m/s. This is the ground speed.
4. Time: 600 ÷ 10.9 = 55 s.

Check. In (b) the ground speed (10.9 m/s) is less than the airspeed (12 m/s), because part of the drone’s effort cancels the wind. That is why the trip takes 55 s instead of 50 s. Adding speeds as numbers, 12 + 5.0 = 17 m/s, would be wrong in both parts: the vectors are not parallel.

Worked example 2: a bolt in a moving lift

Question. Take +y upward. A lift (elevator) E rises at a constant 2.0 m/s relative to the building B. A bolt comes loose from the ceiling, 2.5 m above the lift floor. Ignore air resistance and use g = 9.8 m/s². Describe the fall in the lift frame and in the building frame, and compare the two descriptions.

In the lift frame E (origin on the lift floor):

1. The bolt starts at rest: v_bolt,E = 0 at t = 0, at y′ = 2.5 m.
2. a_bolt,E = −9.8 m/s², so y′(t) = 2.5 − 4.9t².
3. It reaches the floor when y′ = 0: t = √(2.5/4.9) = 0.71 s.
4. Velocity relative to the lift at impact: −9.8 × 0.714 = −7.0 m/s.

In the building frame B (origin where the lift floor was at t = 0):

1. The bolt starts with the lift’s velocity: v_bolt,B = v_bolt,E + v_EB = 0 + 2.0 = +2.0 m/s.
2. Position: y(t) = y′(t) + 2.0t = 2.5 + 2.0t − 4.9t². Differentiate twice: a = −9.8 m/s², the same as in the lift frame.
3. The bolt first rises: it is at rest relative to the building at t = 2.0/9.8 = 0.20 s, 0.20 m above its release point.
4. At t = 0.714 s: v_bolt,B = −7.0 + 2.0 = −5.0 m/s. The bolt’s displacement is 2.0(0.714) − 4.9(0.714)² = −1.07 m, while the floor has risen 2.0 × 0.714 = +1.43 m.

Compare.

Quantity	Lift frame	Building frame
Initial velocity of bolt	0	+2.0 m/s
Acceleration of bolt	−9.8 m/s²	−9.8 m/s²
Time to hit floor	0.71 s	0.71 s
Velocity at impact	−7.0 m/s	−5.0 m/s
Bolt’s displacement	−2.5 m	−1.07 m

Check. The gap closed is the bolt’s displacement minus the floor’s: −1.07 − 1.43 = −2.5 m, matching the lift frame. Both frames are inertial, so acceleration and time agree; velocity and displacement do not.

Common misconceptions

• “An object has one true speed.” Speed and direction depend on the observer’s frame. The bolt hits the floor at 7.0 m/s or 5.0 m/s, depending on who measures.
• Adding speeds as numbers. Only parallel velocities add as plain numbers. In Worked example 1, 12 and 5.0 combine to 13, not 17. Use components.
• Mixing up the subscript order. v_AB is A relative to B, and v_BA = −v_AB. Check that the inner letters match in every chain.
• “Pointing direction = travel direction.” The drone points north but moves 22.6° east of north over the ground.
• “Different observers measure different accelerations.” Not if both frames are inertial. Differentiate r_PA = r_PB + r_BA to see why.
• “In the moving frame, a dropped object falls behind.” In a lift or train moving at constant velocity, a dropped object falls straight down relative to that frame (air resistance ignored). It only lands “behind” if the frame accelerates, which is non-inertial.
• Forgetting the initial velocity in a frame change. The bolt starts at rest in the lift but at +2.0 m/s in the building frame.

Where this leads

Topic 1.5, Motion in Two or Three Dimensions, uses the same vector components for projectiles. A ball thrown from a moving vehicle is a relative-motion problem first. In Unit 2, Newton’s laws are stated for inertial frames, which is why the frame matters. Earlier topics: Topic 1.3, Representing Motion. Try the practice questions now, then use the revision notes and the checklist. You can also return to the course roadmap.