AP Physics 1 · Topic 1.4

Topic 1.4: Reference Frames and Relative Motion

Unit 1: Kinematics10-15% of the multiple-choice section

Velocity depends on who measures it. To convert a velocity from one reference frame to another, add the velocity of the object in the first frame to the velocity of that frame relative to the second. AP Physics 1 keeps this in one dimension, so direction is carried entirely by a plus or minus sign.

AP Physics: Unit 1 (topics 1.4 Reference Frames and Relative Motion). AP Physics 1 Unit 1, Topic 1.4, covering learning objective 1.4.A (describe the reference frame of a given observer) and 1.4.B (describe the motion of objects as measured by observers in different inertial reference frames). A CED boundary statement restricts relative-velocity vector work to one dimension. Unit 1 carries 10 to 15 percent of the multiple-choice section.

What a reference frame is, and why it changes the answer

A reference frame is the coordinate system an observer measures from: where that observer puts the origin, and which way counts as positive. The CED's first essential knowledge for Topic 1.4 is blunt about the consequence. The choice of reference frame determines the direction and magnitude of the quantities measured by an observer in that frame.

So a question like how fast is the coffee cup moving has no answer until you name a frame. Sitting on a tray table in a cruising airliner, the cup moves at zero relative to the passenger, at roughly 250 m/s relative to the ground, and faster still relative to the Sun. All three are correct measurements, and none of them is more real than the others.

That is what learning objective 1.4.A means by describe the reference frame of a given observer. The first move in any relative-motion problem is naming the frame each given number was measured in, before any arithmetic happens.

The subscript convention that prevents sign errors

Write every velocity with two labels. vA/Bv_{A/B} means the velocity of A as measured in B's frame, read aloud as the velocity of A relative to B.

Two rules do everything AP Physics 1 asks:

  • Chain rule. vA/C=vA/B+vB/Cv_{A/C} = v_{A/B} + v_{B/C}. The inner subscripts match and drop out, leaving the outer pair. This is the CED's statement that the observed velocity of an object results from combining the object's velocity with the velocity of the observer's reference frame.
  • Reversal rule. vB/A=vA/Bv_{B/A} = -v_{A/B}. Swapping the labels flips the sign and changes nothing else. If a train moves at +18.0+18.0 m/s relative to the platform, the platform moves at 18.0-18.0 m/s relative to the train.

The chain rule is also how subtraction appears. To get the velocity of A relative to B when both were measured against the ground G, write vA/B=vA/G+vG/B=vA/GvB/Gv_{A/B} = v_{A/G} + v_{G/B} = v_{A/G} - v_{B/G}, applying the reversal rule to the second term.

Read the subscripts back as a sentence before substituting numbers. Most wrong answers in this topic are setup errors, not arithmetic errors.

AP Physics 1 keeps this in one dimension

A boundary statement in the CED fixes the scope: adding or subtracting vectors to find relative velocities is restricted to motion along one dimension for AP Physics 1.

That rules out the classic boat-crossing-a-river setup, where the boat's heading and the current run perpendicular and you build a vector triangle. Those belong to courses that allow two-dimensional relative motion, not to this exam.

What remains is simpler and still easy to get wrong. In one dimension a vector is a signed number, so the vector addition in the chain rule is ordinary addition of positives and negatives. Choose a positive direction, translate every stated speed into a signed velocity, then add. A speed of 15 m/s westward becomes 15-15 m/s once east is positive, and that minus sign is the only place direction lives.

Two-dimensional vector work has not left Unit 1. It moves to Topic 1.5, where components and projectile motion live; the Unit 1 overview maps out the order.

Inertial frames and the quantity every observer agrees on

Here is the invariance claim the CED makes for Topic 1.4: the acceleration of any object is the same as measured from all inertial reference frames.

That carries more weight than it first looks. Position, displacement, velocity, and speed are all frame-dependent. Velocity always differs between two frames in relative motion, while speed sometimes coincides: a passenger walking rearward at 9.0 m/s inside a train doing 18.0 m/s has vP/T=9.0v_{P/T} = -9.0 m/s and vP/G=+9.0v_{P/G} = +9.0 m/s, equal speeds with opposite signs. Acceleration is different in kind. Two observers in relative motion can disagree about a ball's velocity and about which way it moves, and still write down the same acceleration. Since they also agree on its mass, they agree on the net force, which is why Newton's laws come out the same in every inertial frame.

The CED defines the term in Topic 2.4: an inertial reference frame is one from which an observer would verify Newton's first law of motion. You rarely have to test for it, since a Topic 1.4 boundary statement and the exam conventions on the AP Physics 1 table of information both say the frame of reference of any problem is assumed to be inertial unless otherwise stated.

The four-step setup

  1. Name the frames. Give each one a letter: G for ground, T for train, P for passenger. Ambiguity here is what produces wrong signs three lines later.
  2. Choose a positive direction and write it down. East positive, right positive, up positive, whichever you like, and then never change it mid-problem.
  3. Label every given velocity with two subscripts and a sign. The train travels east at 18.0 m/s becomes vT/G=+18.0v_{T/G} = +18.0 m/s. A given with no frame attached is a given you have not finished reading.
  4. Chain the subscripts toward the target. Line the equation up so the inner subscripts cancel, then add. If a piece you need points the wrong way, flip it with the reversal rule instead of guessing a sign.

Finish with a check from a second frame. Both observers have to agree on where and when two objects meet, because meeting is a physical event rather than a measurement, so a disagreement means a sign went wrong.

Where relative motion shows up on the exam

Three patterns cover most of the questions.

  • Something moving on something moving. A passenger walking in a train, a person on a moving walkway, a ball rolled along the aisle of a bus. The chain rule handles all of them in one line.
  • Two objects on one line. Two cars approaching, or one overtaking another. The useful quantity is the relative velocity, because the gap between them closes at the magnitude of vA/Bv_{A/B} no matter how the total is split between the two speeds.
  • Conceptual comparison. A question describes an event and asks which observers agree on which measurement. Velocity, displacement, and speed are frame-dependent. Acceleration is not, across inertial frames.

Relative motion also matters quietly in conservation of momentum problems. Momentum values are frame-dependent, while the conservation statement holds separately in each inertial frame, so choose one frame at the start and solve the whole problem inside it.

Mistakes that cost points

  • Adding speeds instead of velocities. Two cars approaching at 25.0 m/s and 15.0 m/s have a relative speed of 40.0 m/s only because their velocities carry opposite signs. Travelling the same direction, the answer is 10.0 m/s. The arithmetic is identical; the signs are the whole difference.
  • Using a velocity with no frame attached. Every number in a relative-motion problem needs two subscripts before it enters an equation.
  • Flipping subscripts without flipping the sign. vA/Bv_{A/B} and vB/Av_{B/A} have equal magnitudes and opposite signs. They are never interchangeable.
  • Combining perpendicular velocities numerically. The CED restricts adding or subtracting vectors to find relative velocities to one dimension, so a relative-velocity calculation that needs a right triangle is out of scope. Reasoning qualitatively about a two-dimensional scenario is still fair game.
  • Assuming acceleration changes between frames. It does not, across inertial frames, and questions asking which observer measures a larger acceleration are testing exactly that.
  • Changing the positive direction partway through. A sign convention is chosen once per problem, not once per line.

A passenger walking on a moving train

A train travels east along a straight track at 18.018.0 m/s relative to the ground. A passenger walks toward the front of the train at 1.51.5 m/s relative to the train. Take east as positive. Find her velocity relative to the ground, her velocity relative to the ground if she instead walks toward the back at the same speed, and the velocity of the train relative to her in the first case.

  1. Name the frames: G for ground, T for train, P for passenger. Given, with signs: vT/G=+18.0m/sv_{T/G} = +18.0 \, \mathrm{m/s} and, walking forward, vP/T=+1.5m/sv_{P/T} = +1.5 \, \mathrm{m/s}.

  2. Chain the subscripts so the inner pair cancels: vP/G=vP/T+vT/G=(+1.5m/s)+(+18.0m/s)=+19.5m/sv_{P/G} = v_{P/T} + v_{T/G} = (+1.5 \, \mathrm{m/s}) + (+18.0 \, \mathrm{m/s}) = +19.5 \, \mathrm{m/s}, which is 19.5 m/s east.

  3. Walking toward the back changes exactly one sign: vP/T=1.5m/sv_{P/T} = -1.5 \, \mathrm{m/s}, so vP/G=(1.5m/s)+(+18.0m/s)=+16.5m/sv_{P/G} = (-1.5 \, \mathrm{m/s}) + (+18.0 \, \mathrm{m/s}) = +16.5 \, \mathrm{m/s}. She still moves east relative to the ground, just more slowly than the train does.

  4. For the train relative to the forward-walking passenger, use the reversal rule: vT/P=vP/T=1.5m/sv_{T/P} = -v_{P/T} = -1.5 \, \mathrm{m/s}. In her frame the train slides backward past her at 1.5 m/s, which is exactly what walking down an aisle feels like.

  5. Check with distances over 10.010.0 s. Relative to the ground she advances (19.5m/s)(10.0s)=195m(19.5 \, \mathrm{m/s})(10.0 \, \mathrm{s}) = 195 \, \mathrm{m} while the train advances (18.0m/s)(10.0s)=180m(18.0 \, \mathrm{m/s})(10.0 \, \mathrm{s}) = 180 \, \mathrm{m}. The 15m15 \, \mathrm{m} difference equals (1.5m/s)(10.0s)(1.5 \, \mathrm{m/s})(10.0 \, \mathrm{s}), her progress along the train itself.

Walking forward, vP/G=+19.5m/sv_{P/G} = +19.5 \, \mathrm{m/s} (19.5 m/s east). Walking backward, vP/G=+16.5m/sv_{P/G} = +16.5 \, \mathrm{m/s} (16.5 m/s east). The train's velocity relative to the forward-walking passenger is 1.5m/s-1.5 \, \mathrm{m/s}, or 1.5 m/s west.

Two cars on the same straight road

Car A drives east at 25.025.0 m/s and car B drives west at 15.015.0 m/s along the same straight road, both measured relative to the ground. They start 240240 m apart and are approaching each other. Take east as positive. Find the velocity of A relative to B, and the time until they meet.

  1. Translate the words into signed velocities: vA/G=+25.0m/sv_{A/G} = +25.0 \, \mathrm{m/s} and vB/G=15.0m/sv_{B/G} = -15.0 \, \mathrm{m/s}. Two speeds stated in words became one positive and one negative number, and that translation is most of the problem.

  2. Apply the chain and reversal rules: vA/B=vA/G+vG/B=vA/GvB/G=(+25.0m/s)(15.0m/s)=+40.0m/sv_{A/B} = v_{A/G} + v_{G/B} = v_{A/G} - v_{B/G} = (+25.0 \, \mathrm{m/s}) - (-15.0 \, \mathrm{m/s}) = +40.0 \, \mathrm{m/s}. In B's frame, A closes on B at 40.0 m/s heading east.

  3. Confirm the pair: vB/A=vA/B=40.0m/sv_{B/A} = -v_{A/B} = -40.0 \, \mathrm{m/s}, equal in magnitude and opposite in sign, as the reversal rule requires.

  4. The gap closes at the relative speed, so t=240m40.0m/s=6.0st = \frac{240 \, \mathrm{m}}{40.0 \, \mathrm{m/s}} = 6.0 \, \mathrm{s}.

  5. Check in the ground frame: in 6.0 s car A covers (25.0m/s)(6.0s)=150m(25.0 \, \mathrm{m/s})(6.0 \, \mathrm{s}) = 150 \, \mathrm{m} and car B covers (15.0m/s)(6.0s)=90m(15.0 \, \mathrm{m/s})(6.0 \, \mathrm{s}) = 90 \, \mathrm{m}, and 150m+90m=240m150 \, \mathrm{m} + 90 \, \mathrm{m} = 240 \, \mathrm{m}. Both frames place the meeting at the same instant, as they must.

  6. Change one sign to see the contrast. If A were chasing B eastward instead, vB/G=+15.0m/sv_{B/G} = +15.0 \, \mathrm{m/s} and vA/B=25.0m/s15.0m/s=+10.0m/sv_{A/B} = 25.0 \, \mathrm{m/s} - 15.0 \, \mathrm{m/s} = +10.0 \, \mathrm{m/s}, so the same two speeds would close the same gap four times more slowly, in 24 s.

vA/B=+40.0m/sv_{A/B} = +40.0 \, \mathrm{m/s}, so A approaches B at 40.0 m/s, and the cars meet after t=6.0st = 6.0 \, \mathrm{s}.

Frequently asked questions

How do I find the velocity of one object relative to another?

Subtract, using signed velocities along your chosen positive direction: the velocity of A relative to B equals the velocity of A relative to the ground minus the velocity of B relative to the ground. Equivalently, chain the subscripts so the inner labels cancel. Swapping the two labels flips the sign and changes nothing else.

Does AP Physics 1 ask for two-dimensional relative motion, like a boat crossing a river?

No. A CED boundary statement restricts adding or subtracting vectors to find relative velocities to motion along one dimension for AP Physics 1. If a relative-velocity problem has you drawing a vector triangle with perpendicular components, it is outside the scope of this exam.

Does acceleration depend on the reference frame?

Not among inertial frames. The CED states that the acceleration of any object is the same as measured from all inertial reference frames. Position, displacement, velocity, and speed are all frame-dependent, with velocity always differing under relative motion and speed sometimes coming out the same in both frames. Acceleration does not change at all, and since mass does not either, observers also agree on the net force.

What is an inertial reference frame, and do I have to check for one?

The CED defines it in Topic 2.4 as a frame from which an observer would verify Newton's first law of motion. You rarely have to check. A Topic 1.4 boundary statement and the exam conventions printed on the AP Physics 1 table of information both state that the frame of reference of any problem is assumed to be inertial unless otherwise stated.

Why do two observers get different numbers for the same motion?

Because every measurement happens inside a frame. The CED states that the choice of reference frame determines the direction and magnitude of the quantities an observer measures. A ball dropped inside a moving bus falls straight down according to a rider and follows a curved path according to someone on the sidewalk, and both descriptions are correct.