Lesson 1.6 · Unit 1 · CED topic 1.4
Reference frames and relative motion
Sitting in a train that is standing at a station, you watch the train beside you begin to move, and for a moment you cannot tell whether it is the other train or your own that is rolling. Motion is always measured relative to something. That something is the observer’s reference frame, and the numbers you record for position and velocity depend on the choice. What does not depend on the choice, as this lesson shows, is acceleration, and that fact will be the foundation for Newton’s laws in Unit 2.
Describe the reference frame of a given observer. Describe the motion of objects as measured by observers in different inertial reference frames.
- The choice of reference frame will determine the direction and magnitude of quantities measured by an observer in that reference frame.
- Measurements from a given reference frame may be converted to measurements from another reference frame.
- The observed velocity of an object results from the combination of the object’s velocity and the velocity of the observer’s reference frame. Combining the motion of an object and the motion of an observer in a given reference frame involves the addition or subtraction of vectors.
- The acceleration of any object is the same as measured from all inertial reference frames.
Unless otherwise stated, the frame of reference of any problem may be assumed to be inertial. Adding or subtracting vectors to find relative velocities is restricted to motion along one dimension for AP Physics 1.
What a reference frame is
A reference frame is a point of view: an origin and a set of axes attached to an observer, together with the observer’s own motion. “The frame of the road” means measuring everything from a person standing on the road. “The frame of the train” means measuring from a passenger sitting in it. The same event gets different numbers in different frames. A coffee cup on the train table is at rest in the train’s frame and moving at 30 m/s in the road’s frame, and both statements are correct.
An inertial reference frame is one in which Newton’s first law holds: an object with no net force on it keeps a constant velocity. Any frame moving at constant velocity relative to an inertial frame is also inertial. A frame that is accelerating, such as a car braking hard or a merry-go-round, is not. Unless a question says otherwise, assume the frame is inertial; the road and a train moving smoothly at constant velocity both qualify.
Converting velocities between frames
The rule is a vector sum. If object A moves at \(v_{A/B}\) as measured by observer B, and observer B moves at \(v_{B/G}\) as measured from the ground G, then the velocity of A measured from the ground is \[ v_{A/G} = v_{A/B} + v_{B/G}. \] Read each subscript \(X/Y\) as “X as measured by Y.” The inner labels match and cancel, leaving A as measured by G, which is how to keep the chain straight. In one dimension, velocities carry signs, so the addition is a signed addition. A person walking forward at +2 m/s inside a train moving at +30 m/s relative to the ground is moving at +32 m/s relative to the ground (Figure 1.6a); a person walking toward the back of the same train at \(-2\ \text{m/s}\) relative to the train is moving at +28 m/s relative to the ground.
To change frames in the other direction, subtract the velocity of the new observer from every velocity. If you ride in a car moving at \(+15\ \text{m/s}\), a neighboring car moving at \(+25\ \text{m/s}\) relative to the road is moving at \(25 - 15 = +10\ \text{m/s}\) relative to you, and your own car, of course, is at rest in your frame. The two frames in Figure 1.6b describe the same pair of cars and the same motion.
What changes and what does not
Positions, displacements and velocities all change when you change frames, because they all depend on where the origin is and how it moves. So does the kinetic energy of an object (Unit 3): the cup that is at rest on the train has zero kinetic energy in the train’s frame and a great deal in the road’s. Acceleration, however, is the same in every inertial frame. Two frames that differ only by a constant velocity differ by a constant, and the change in velocity over any interval is the same for both. If the train is moving at constant velocity relative to the road, a coin dropped by a passenger falls with \(10\ \text{m/s}^2\) downward in both frames.
This is why Newton’s laws, which link force to acceleration, can be used in any inertial frame and give the same predictions. It is also why you cannot detect your own constant velocity without looking outside: nothing you can measure about accelerations changes.
Car A is traveling at \(+25\ \text{m/s}\) and car B at \(+15\ \text{m/s}\), both in the same direction on a straight road. Car A is 200 m behind car B. (a) What is A’s velocity relative to B? (b) How long until A catches B?
(a) \(v_{A/B} = v_{A/G} - v_{B/G} = 25 - 15 = +10\ \text{m/s}\). In B’s frame, A approaches at 10 m/s.
(b) In B’s frame, A must cover the 200 m gap at a constant 10 m/s: \(t = \dfrac{200\ \text{m}}{10\ \text{m/s}} = 20\ \text{s}\).
Check in the road frame. In 20 s, A travels \((25)(20) = 500\ \text{m}\) and B travels \((15)(20) = 300\ \text{m}\). A has gained \(500 - 300 = 200\ \text{m}\). ✓ The frame of car B made the problem a one-line division.
Car A is moving at \(+25\ \text{m/s}\) and car C is coming toward it at \(-20\ \text{m/s}\), as measured on the road. How fast does each appear to move from the other’s point of view?
Step 1. \(v_{A/C} = v_{A/G} - v_{C/G} = 25 - (-20) = +45\ \text{m/s}\). C sees A approaching at 45 m/s.
Step 2. \(v_{C/A} = -20 - 25 = -45\ \text{m/s}\). A sees C approaching at 45 m/s in the negative direction.
Interpretation. The relative speed is the sum of the two speeds when objects approach head-on. This is why a head-on crash at 25 m/s and 20 m/s feels like hitting a wall at 45 m/s from either car’s own frame.
A boat can move at 5.0 m/s relative to the water. A river flows at 2.0 m/s. How long does a trip take that goes 210 m downstream and then 210 m back upstream? What is the average speed over the round trip?
Downstream. Take the current direction as positive. \(v_{\text{boat/ground}} = 5.0 + 2.0 = +7.0\ \text{m/s}\), so \(t_1 = 210/7.0 = 30\ \text{s}\).
Upstream. \(v_{\text{boat/ground}} = -5.0 + 2.0 = -3.0\ \text{m/s}\), a speed of 3.0 m/s, so \(t_2 = 210/3.0 = 70\ \text{s}\).
Round trip. The total time is \(30 + 70 = 100\ \text{s}\) for \(420\ \text{m}\), so the average speed is \(420/100 = 4.2\ \text{m/s}\), less than the 5.0 m/s still-water speed. The current helps for a short time and hinders for a long time, so it costs more than it gives.
- State the frame. “The velocity of the ball” is incomplete without “relative to the ground” or “relative to the train.” Good answers name the frame.
- Use the subscript chain. Write \(v_{A/G} = v_{A/B} + v_{B/G}\) before substituting, and check that the inner labels cancel.
- Acceleration is the same. When a question asks what an observer in a different inertial frame would measure for acceleration, the answer is the same value. For velocity and kinetic energy it is different.
- One dimension only. Relative-velocity questions on this exam are along a line, so signs do the work.
- Pick the convenient frame. Switching to the frame of one of the objects often makes the other the only thing moving, which simplifies catch-up and passing problems.
- A bus moves at \(+20\ \text{m/s}\). A passenger walks toward the rear at 1.5 m/s relative to the bus. What is the passenger’s velocity relative to the road?
Show answer
\(v = +20 + (-1.5) = +18.5\ \text{m/s}\).
- Two trains, each 150 m long, approach on parallel tracks at \(+30\ \text{m/s}\) and \(-25\ \text{m/s}\) relative to the ground. How long do they take to pass completely, from the moment the fronts meet to the moment the rears separate?
Show answer
The relative speed is \(30 - (-25) = 55\ \text{m/s}\). The trains must move past each other by their combined length, \(300\ \text{m}\), so \(t = 300/55 = 5.5\ \text{s}\).
- A swimmer who can swim at 1.5 m/s in still water swims 50 m upstream in a river that flows at 1.0 m/s, then returns downstream. Find the time for each leg.
Show answer
Upstream the ground speed is \(1.5 - 1.0 = 0.5\ \text{m/s}\), so \(t = 50/0.5 = 100\ \text{s}\). Downstream it is \(1.5 + 1.0 = 2.5\ \text{m/s}\), so \(t = 50/2.5 = 20\ \text{s}\).
- A passenger in a train moving smoothly at constant velocity tosses a ball straight up. Describe where it lands relative to the passenger and explain using reference frames.
Show answer
It lands back in the passenger’s hand. In the train’s frame (inertial, since the train moves at constant velocity) the ball goes straight up and straight down, with acceleration \(g\) downward. The acceleration is the same in the ground frame, and the ball keeps the train’s forward velocity throughout, so it stays above the passenger.
- An airplane has an airspeed of 200 m/s (its speed relative to the air) and flies into a headwind of 30 m/s. How long does it need to cover \(1.7 \times 10^5\ \text{m}\) along the ground?
Show answer
Ground speed \(= 200 - 30 = 170\ \text{m/s}\). \(t = (1.7\times10^5)/170 = 1.0\times10^3\ \text{s}\).
- For an object that moves at constant acceleration, which quantities are the same when measured in two different inertial frames: position, velocity, acceleration, kinetic energy? Explain briefly.
Show answer
Only the acceleration. Two inertial frames differ by a constant relative velocity, which shifts every measured velocity by the same constant and so shifts position and kinetic energy as well. A constant does not change how fast the velocity changes, so the acceleration is the same.