Relative Velocity showed that velocities add as vectors when you switch viewpoints: a boat crossing a river, a plane in the wind. With Position, Velocity and Acceleration Vectors we can see exactly what changes between frames and, just as important, what doesn't.
Why it works
Let frame S′ move at constant velocity $\vec{u}$ relative to frame S, with the origins together at $t = 0$. A point at $\vec{r}$ in S is at $$\vec{r}\,' = \vec{r} - \vec{u}t$$ in S′ (the Galilean transformation; time is the same in both). Differentiate once: $\vec{v}\,' = \vec{v} - \vec{u}$. Differentiate again: $\vec{a}\,' = \vec{a} - \dfrac{d\vec{u}}{dt} = \vec{a}$, because $\vec{u}$ is constant.
Key idea
Frames moving at constant velocity relative to each other disagree about positions and velocities ($\vec{v}\,' = \vec{v} - \vec{u}$) but agree about accelerations. Since $\vec{F} = m\vec{a}$, they also agree about forces: Newton's laws hold in every such frame. These are the inertial frames.
Worked example: A walk on a train
A train moves east at $20\ \text{m/s}$. A passenger walks toward the back at $1.5\ \text{m/s}$ relative to the train. What is her velocity relative to the ground?
- Take east positive. In the train frame her velocity is $v' = -1.5\ \text{m/s}$; the frame moves at $u = +20\ \text{m/s}$.
- $v = v' + u = -1.5 + 20 = 18.5\ \text{m/s}$.
Answer: $18.5\ \text{m/s}$ east.
Worked example: Rain on a car window
Rain falls straight down at $6\ \text{m/s}$. A car drives at $8\ \text{m/s}$. How fast, and at what angle from vertical, do drops move past the side window?
- Car frame: $\vec{v}\,' = \vec{v} - \vec{u} = \langle 0, -6 \rangle - \langle 8, 0 \rangle = \langle -8, -6 \rangle\ \text{m/s}$.
- Speed: $\sqrt{64 + 36} = 10\ \text{m/s}$.
- Angle from vertical: $\arctan(8/6) = 53.1^\circ$, slanting toward the back of the car.
Answer: $10\ \text{m/s}$ at $53.1^\circ$ from vertical.
Common mistake
Thinking acceleration also changes between frames. A ball dropped in a smoothly moving train falls straight down at your feet with $a = 9.8\ \text{m/s}^2$, exactly as on the platform; a person on the platform sees a parabola, but with the same downward acceleration. Check: differentiate $\vec{v}\,' = \vec{v} - \vec{u}$. A constant $\vec{u}$ has zero derivative. (Only a frame that speeds up, slows down or turns, such as a braking car, changes the acceleration, and then Newton's laws need correction terms.)
Galilean frames work at everyday speeds. Near the speed of light, time and length change between frames too, and Einstein's relativity replaces $\vec{r}\,' = \vec{r} - \vec{u}t$.