Reference frames and point mass

Frame of reference. A frame is an origin, a set of axes, and a clock — nothing more — relative to which the position of a body is recorded as it changes with time. The frame used most often in Class 11 kinematics is an inertial frame: one at rest or moving with constant velocity, so that a body free of external force neither speeds up nor turns within it. An accelerating frame (a braking bus, a spinning merry-go-round) is non-inertial, and motion described inside it can look different — even seem to need a "phantom" push — compared with the same motion viewed from an inertial frame. Every kinematics problem should begin by naming the frame (usually the ground, unless stated otherwise) and fixing an origin and a positive direction; skipping this step is the single most common source of sign errors.

Point mass. An object is treated as a point mass when its own size, shape, and internal structure are irrelevant to the question being asked — only its position $(x, y, z)$ matters. The approximation is good whenever the body's length $L$ is much smaller than the relevant travel distance $d$ in the problem, $L \ll d$; a car's few metres are negligible against a 500 km journey, but a ship's 120 m length is not negligible against a 150 m pier. There is no fixed cutoff — validity is judged case by case, against the scale of the specific question, never against the "type" of object.

Combining frames. The genuinely new tool this article adds is how to move between two frames. Suppose a body $P$ (say, a walking passenger) has some position relative to frame $B$ (the bus) and $B$ itself has a position relative to frame $A$ (the ground). Then $P$'s position relative to $A$ is just the vector sum of the two:

where: $\vec{r}_{PA}$ is the position of $P$ relative to $A$, $\vec{r}_{PB}$ is the position of $P$ relative to $B$, and $\vec{r}_{BA}$ is the position of $B$ relative to $A$ — three position vectors that simply add tip to tail, the same rule used for displacement vectors elsewhere in this unit. Because velocity is just the rate of change of position, the same addition rule carries over to velocities, and this is the form used in practice:

where: $\vec{v}_{PA}$ is the velocity of $P$ as seen from ground frame $A$, $\vec{v}_{PB}$ is the velocity of $P$ as seen from a moving frame $B$ (for example, an observer inside the bus), and $\vec{v}_{BA}$ is the velocity of frame $B$ itself relative to $A$. In one dimension the vectors reduce to signed numbers, and the "combine frames" idea is nothing more than ordinary addition — but keeping track of which velocity is measured relative to which frame is the entire skill.

The addition rule is nothing more than the way position vectors already combine, applied twice. This is the whole reasoning, without any calculus: it is built entirely on the definition of a position vector and how displacements add.

Step 1. Fix the two frames. Let $A$ be the ground frame with its own origin, and let $B$ be a second frame (say, one glued to a moving bus) whose origin is at position $\vec{r}_{BA}$ measured from $A$'s origin at some instant. Any point $P$ can be located either from $B$'s origin, giving $\vec{r}_{PB}$, or from $A$'s origin, giving $\vec{r}_{PA}$.

Step 2. Add the position vectors tip to tail. Walking from $A$'s origin to $P$ is the same as walking from $A$'s origin to $B$'s origin, and then from $B$'s origin on to $P$ — two legs of one journey, so the vectors simply add: $\vec{r}_{PA} = \vec{r}_{BA} + \vec{r}_{PB}$, which is exactly equation (1). No physics has gone in yet, only the geometric fact that two successive displacements sum to the total displacement.

Step 3. Let time pass and look at the change. Over a short time interval $\Delta t$, each position vector changes by some $\Delta \vec{r}$, and equation (1) holds at every instant, so it must also hold for the changes: $\Delta \vec{r}_{PA} = \Delta \vec{r}_{BA} + \Delta \vec{r}_{PB}$. Dividing every term by the same $\Delta t$ and using velocity $= \Delta \vec{r}/\Delta t$ converts a relation between displacements into a relation between velocities, giving equation (2), $\vec{v}_{PA} = \vec{v}_{PB} + \vec{v}_{BA}$.

Step 4. Read the result physically. The velocity of $P$ relative to the ground is what an observer in the moving frame $B$ measures for $P$, plus whatever velocity frame $B$ itself has relative to the ground. Nothing about inertial or non-inertial frames was needed for this step — it is pure vector bookkeeping, which is exactly why it works for any two frames, moving in any way relative to each other.

A passenger inside a bus moving at $20\ \text{m/s}$ relative to the ground (the same bus and speed as the frame-of-reference example) walks toward the front of the bus at $1.5\ \text{m/s}$ relative to the bus. Find the passenger's velocity relative to the ground, and relative to a second passenger sitting still in a seat near the back.

Step 1. Set up the frames and directions. Take the ground as frame $A$ and the bus as frame $B$; take the direction the bus is travelling as positive. Given: $v_{BA} = +20\ \text{m/s}$ (bus relative to ground) and $v_{PB} = +1.5\ \text{m/s}$ (walking passenger relative to bus, same direction).

Step 2. Apply the addition rule for the ground frame. $v_{PA} = v_{PB} + v_{BA} = 1.5 + 20 = 21.5\ \text{m/s}$. The walking passenger moves at $21.5\ \text{m/s}$ relative to the ground — faster than the bus itself, because their walking speed adds on top of the bus's speed.

Step 3. Find the velocity relative to the seated passenger. The seated passenger, call them $S$, is at rest in the bus, so $v_{SB} = 0$ and hence $v_{SA} = v_{BA} = 20\ \text{m/s}$. The velocity of the walking passenger relative to the seated one is $v_{PS} = v_{PA} - v_{SA} = 21.5 - 20 = 1.5\ \text{m/s}$ — exactly $v_{PB}$, as it must be, since both are stated relative to the bus. Answer: the walking passenger moves at $21.5\ \text{m/s}$ relative to the ground, and at $1.5\ \text{m/s}$ relative to the seated passenger.

• Adding or subtracting speeds without fixing a common positive direction first — relative velocity is a vector rule, not arithmetic on magnitudes alone.
• Mixing up "P relative to B" ($v_{PB}$) with "B relative to P" ($v_{BP}$); in fact $v_{BP} = -v_{PB}$.
• Believing the ground frame is the only "correct" one — any inertial frame is equally valid; the ground is just the conventional default.
• Applying the point-mass approximation as a blanket rule instead of checking $L \ll d$ for the actual distances given.

Ask the tutor: if you want more practice combining velocities across frames — boats crossing a river current, or two vehicles approaching each other — ask the AI tutor for extra worked problems.