Reference Frames and Relative Motion
Introduced| English | Meaning |
|---|---|
| relative/ˈrelətɪv/ | Measured or described in comparison with a specified reference. |
| reference frame/ˈrefrəns freɪm/ | A coordinate system and reference objects used to describe positions and motion. |
| resultant/rɪˈzʌltənt/ | The single vector equal to the sum of two or more vectors. |
You are sitting still — at 30 km/s
- Right now you feel motionless. Yet Earth carries you around the Sun at about $30\ \tfrac{\text{km}}{\text{s}}$.
- Both are true. "At rest" is only meaningful relative to something.
- Motion is always measured from a chosen viewpoint — a reference frame 参考系.
- Change the frame and the same object can be still, slow, or racing.
What a reference frame is
- A reference frame is a viewpoint with its own origin and axes for measuring position.
- The ground, a moving train, a flowing river — each is a valid frame.
- A velocity only means something once you say "relative to which frame".
- Physics laws work in any frame moving at constant velocity (an inertial frame).
Why can we say you are both "at rest" and "moving at $30\ \tfrac{\text{km}}{\text{s}}$" at once?
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Relative to the ground you are at rest; relative to the Sun you move with the Earth. Velocity is frame-dependent.
Select all of these that can serve as a reference frame.
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Any physical object with an origin and axes can be a frame. A bare number is not a viewpoint.
Adding velocities on a line
- A passenger walks at $+2\ \tfrac{\text{m}}{\text{s}}$ inside a train that moves at $+30\ \tfrac{\text{m}}{\text{s}}$.
- To the ground, their velocity is $30 + 2 = 32\ \tfrac{\text{m}}{\text{s}}$.
- Walk toward the back ($-2$) and the ground sees $30 - 2 = 28\ \tfrac{\text{m}}{\text{s}}$.
- In one dimension, relative 相对 velocities just add as signed numbers.
A passenger walks at $+2\ \tfrac{\text{m}}{\text{s}}$ toward the front of a train moving at $+30\ \tfrac{\text{m}}{\text{s}}$. What is the passenger's velocity relative to the ground, in $\tfrac{\text{m}}{\text{s}}$?
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Same line, same direction: $30 + 2 = 32\ \tfrac{\text{m}}{\text{s}}$.
The same passenger now walks toward the back at $2\ \tfrac{\text{m}}{\text{s}}$. Velocity relative to the ground, in $\tfrac{\text{m}}{\text{s}}$?
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Now the walk is negative: $30 + (-2) = 28\ \tfrac{\text{m}}{\text{s}}$.
Adding velocities in two dimensions
- A boat points straight across a river; the current sweeps it downstream.
- Its velocity over the ground is the vector sum of boat-velocity and water-velocity.
- The two combine tip-to-tail into a resultant 合矢量 — a diagonal path.
- The boat crosses and drifts at the same time; neither motion cancels the other.
Add the boat and current vectors
Combine a straight-across velocity with a downstream current and read off the resultant.
Across a river, velocities must be added as ____, not just as plain numbers.
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Perpendicular velocities combine tip-to-tail as vectors, giving a diagonal resultant.
The subscript rule
- Write "velocity of A relative to B" as $\vec v_{A/B}$.
- They chain: $\vec v_{A/\text{ground}} = \vec v_{A/B} + \vec v_{B/\text{ground}}$.
- Swap the order and you negate it: $\vec v_{B/A} = -\,\vec v_{A/B}$.
- Line up the inner subscripts and the outer pair is your answer.
The velocity of A relative to B is the exact opposite (negative) of the velocity of B relative to A.
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Reversing the pair reverses the direction: $\vec v_{B/A} = -\vec v_{A/B}$.
Relative velocities add as vectors, not always as plain numbers. Only when the motions are along the same line can you just add or subtract the values. Across a river the answer needs vector addition, so the boat's ground speed is larger than its speed through the water.
A person walks at $1.5\ \tfrac{\text{m}}{\text{s}}$ toward the front of a train moving at $12\ \tfrac{\text{m}}{\text{s}}$.
- Same direction, one line: add the values.
- Velocity relative to the ground $= 12 + 1.5 = 13.5\ \tfrac{\text{m}}{\text{s}}$.
Motion is measured in a reference frame. To change frames, add velocities as vectors: $\vec v_{A/\text{ground}} = \vec v_{A/B} + \vec v_{B/\text{ground}}$, and reversing a pair negates it ($\vec v_{B/A} = -\vec v_{A/B}$). On one line they add as signed numbers; in two dimensions they add tip-to-tail.
Write from ground speeds of 15 and 9
- A boat moves at 12 m/s. You walk at 3 m/s along the deck.
- Find the ground speed walking forward and walking back.
A boat moves at 12 m/s relative to the ground. You walk at 3 m/s toward the bow, then later at 3 m/s toward the stern. What is your ground speed in each case?
12 plus 3 is 15. 12 minus 3 is 9.
What is the ground speed when you walk back toward the stern?
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Opposite directions subtract, so 12 minus 3 is 9. Other options — 15 m/s (the forward result); 28 m/s (the lesson backward walk).