AP PHYSICS C: MECHANICS › UNIT 1, KINEMATICS › TOPIC 1.4

Reference Frames and Relative Motion

Two observers in steady relative motion write down different positions and different velocities for the same object. Differentiate their disagreement twice and it vanishes. That single fact is what lets the rest of this course exist.

What you should be able to do

1.4.A Relate the position, velocity and acceleration of an object measured in one inertial reference frame to those measured in another.

1.4.B Derive the velocity and acceleration relationships by differentiating the position relationship, and say why the acceleration term drops out.

1. A frame is a coordinate system with a clock attached

Fix an origin, fix a positive direction, and fix what the whole apparatus is bolted to. That is a reference frame. Every position, velocity and acceleration in the first three topics was measured in one, almost always the ground, and almost always without anybody saying so.

Now put a second frame, call it B, in motion at a constant velocity V with respect to the first, call it A, and set both clocks so they read zero when the origins coincide. A point in space has a coordinate in each.

x_B(t) = x_A(t) − Vt

This is the Galilean transformation. It says nothing deep on its own. Everything interesting comes from differentiating it.

Read the minus sign. If B is running in the positive direction, then in B’s own bookkeeping the world has slid backward by Vt. Coordinates in the moving frame are smaller, and they get smaller at a steady rate.

2. Differentiate once: velocities add

Take the derivative of both sides with respect to time. The derivative of a position is a velocity, and V is a constant, so its product with t differentiates to V.

dx_B/dt = dx_A/dt − Vso v_B = v_A − V

Written with subscripts that carry their own bookkeeping, where v_PA means the velocity of P as measured by A, the same statement reads as a chain in which the inner labels cancel.

v_PA = v_PB + v_BAv_AB = − v_BA

If a chain of subscripts does not cancel, the expression is wrong. That is worth more than it sounds: the notation catches the error before the arithmetic gets a chance to hide it.

3. Differentiate again: acceleration does not change at all

Differentiate the velocity relationship. V is constant, so dV/dt = 0, and the term that made the two frames disagree simply is not there any more.

a_B = dv_B/dt = dv_A/dt − dV/dt = a_A

This is the whole reason Newton’s laws are usable. They are statements about acceleration. If every observer moving at a constant velocity measures the same acceleration, then every such observer writes down the same laws. Those observers are the inertial frames, and the definition is exactly this: a frame in which dV/dt = 0 relative to another inertial frame.

Notice how narrow the escape clause is. The moment V depends on time, the term dV/dt survives, the two accelerations differ by it, and F = ma stops being true in the moving frame unless you invent a force to patch it. That patch is what a fictitious force is, and it is a bookkeeping entry, not a push.

One motion, two observers

The same scene drawn twice. On top, a frame fixed to the ground, where the posts stand still. Below, a frame fixed to the car, where the car stands still and the posts stream past. The dots are the object’s position at one-second intervals, so their spacing is the velocity that frame measures. Nothing physical changes when you switch panels.

Car relative to ground Walker relative to car

4. What the two position graphs look like

The transformation subtracts a straight line from a curve. That is worth seeing rather than being told, because it makes the two derivative results obvious at a glance.

Subtracting Vt tilts the whole graph. A tilt changes the slope everywhere by the same amount, which is the velocity result. A tilt does not bend anything, so the curvature at every point is untouched, which is the acceleration result.

Tilt the graph and watch what survives

The upper graph is the object’s position measured in the ground frame. The lower one is the same motion measured from a frame moving at V, which is the upper curve minus the dashed line. Drag the time cursor. The tangent is drawn at that instant in both frames. Watch the two slopes differ by exactly V while the two curvatures stay equal.

V, the boost

Push V until the lower curve is momentarily flat at the cursor. You have found the frame in which the object is instantaneously at rest, and it exists for every instant of every motion. What you cannot do, at any setting, is flatten the lower curve everywhere at once. The bend is not yours to remove.

5. The dropped ball, in components

A passenger on a train moving at a constant V releases a ball from rest in her own frame at height h. Work it in her frame first, because in her frame the initial velocity is zero.

x_B(t) = 0y(t) = h − ½gt²

Transform to the platform frame by adding Vt back. The vertical coordinate is untouched, because the frames are in relative motion horizontally only.

x_A(t) = Vty(t) = h − ½gt²

Eliminate t and the platform observer has a parabola, y = h − g x_A² / (2V²). The passenger has a vertical line. They disagree about the path, the speed and the distance traveled. Differentiating either pair twice gives the same acceleration, −g vertically and zero horizontally, and both get the same landing time, t = √(2h/g), at the passenger’s feet.

Use invariants as checks. Anything two inertial observers must agree on, the acceleration, the time between two events, whether a collision happened, is a quantity you can test your algebra against. If your two frames disagree about where the ball lands, one of them is wrong, and you have found that out without knowing the answer.

Check yourself

1. An object moves as x_A(t) = 2t³ − 5t in frame A, with A in meters and seconds. Frame B moves at V = 4 m/s relative to A. Write x_B(t), then find v_B and a_B, and compare them with A’s.

x_B = 2t³ − 5t − 4t = 2t³ − 9t. Then v_B = 6t² − 9 while v_A = 6t² − 5, differing by 4 at every instant, which is V. And a_B = 12t = a_A, identical. The cubic term never felt the transformation, because subtracting a linear function cannot change a second derivative.

2. For that same motion, at what time is the object at rest in frame A, and is there a frame in which it is at rest at t = 2 s? If so, give its V.

At rest in A when 6t² − 5 = 0, so t = √(5/6) ≈ 0.913 s. At t = 2 the velocity in A is 6(4) − 5 = 19 m/s, so the frame moving at V = 19 m/s sees it instantaneously at rest there. Such a frame exists for every instant, and it is generally a different frame at each one, which is exactly why no single boost can flatten a curved graph.

3. A swimmer moves at 1.2 m/s relative to the water in a river flowing at 0.8 m/s relative to the bank. She aims straight across a 36 m river. Find the crossing time and how far downstream she lands, then say what she must do to land straight across and whether the crossing takes longer.

Aiming straight across, the across-stream component is 1.2 m/s, untouched by the current, so t = 36 / 1.2 = 30 s and she is carried 0.8 × 30 = 24 m downstream. To land straight across she must aim upstream at θ = arcsin(0.8 / 1.2) = 41.8° from the perpendicular, which leaves only √(1.2² − 0.8²) = 0.894 m/s across, so t = 40.3 s. Yes, it takes longer, and the reason is that part of her effort is now spent canceling the current rather than crossing.

4. A bus decelerates at 3.0 m/s² and a bag on the floor slides forward. Analyze it in the road frame, then say precisely what goes wrong if you insist on working in the bus frame.

In the road frame, friction on the bag is small, so the bag keeps very nearly the velocity it had while the bus slows underneath it. The bag’s acceleration is near zero and nothing pushed it forward. In the bus frame, V depends on time, so dV/dt = −3.0 m/s² does not drop out, and the bag appears to accelerate forward at 3.0 m/s² with no force on it. That frame is not inertial. You can keep using it only by adding a fictitious force −m dV/dt, which is bookkeeping for the frame’s own acceleration, not an interaction with anything.

5. Show that if V is not constant, the acceleration relationship picks up a term, and identify it.

With x_B = x_A − ∫V dt, one derivative gives v_B = v_A − V(t) and the second gives a_B = a_A − dV/dt. The surviving term is the acceleration of frame B itself. Multiply it by the mass and you have the fictitious force that an observer in B must invent to keep F = ma looking true. It is exactly zero when B is inertial, which is the case handled above.

Topic 1.5, Vectors and Motion in Two Dimensions. Every relationship on this page was written for one axis. They are all vector equations already. The next topic stops pretending otherwise, and the river problem in question 3 becomes the ordinary case rather than a trick.