Reference Frames and Relative Motion

On a windless day, rain falls straight down. From the passenger seat of a moving car, the same drops streak across the side window at a slant, as if they were blowing toward you. Neither observer is wrong: velocity is always measured relative to something, and the two observers are measuring relative to different things.

What a reference frame is

A reference frame is the viewpoint a measurement is made from: an observer with an origin, a set of axes, and a clock, all moving together. Every position, displacement and velocity in topics 1.1 to 1.3 was measured in some frame, usually the ground's without saying so.

The choice of frame changes both the magnitude and the direction of what is measured. A passenger sitting in a train moving at $30\ \text{m/s}$ east has velocity zero in the train's frame and $30\ \text{m/s}$ east in the ground's frame. A ball rolled toward the back of the train at $2\ \text{m/s}$ moves west in the train frame and still east, at $28\ \text{m/s}$, in the ground frame. Each description is correct within its own frame, so a velocity is incomplete until you say which frame it is measured in.

An inertial reference frame is one that is not accelerating: it is at rest or moving at constant velocity relative to another inertial frame. The ground is treated as inertial (Earth's rotation is far too slow to matter for AP problems), and so is a train cruising at constant velocity on a straight track. A braking car is not. Unless a problem says otherwise, assume every frame it mentions is inertial.

Converting between frames

Write $\vec{v}_{AB}$ for "the velocity of $A$ measured in $B$'s frame". Say a passenger $P$ walks inside a train $T$ that moves over the ground $G$. The passenger's position from the ground's origin is the passenger's position from the train's origin plus the train origin's position from the ground's origin:

Differentiate with respect to time (both frames use the same clock):

In one dimension the vectors become signed numbers. A cyclist rides east at $6.0\ \text{m/s}$ and a jogger runs west at $3.0\ \text{m/s}$, both relative to the road. With east positive, the jogger's velocity relative to the cyclist is $-3.0 - 6.0 = -9.0\ \text{m/s}$: from the cyclist's seat, the jogger approaches at $9.0\ \text{m/s}$. In two dimensions the same equation is vector addition, done by components as in topic 1.1, so speeds in different directions never add as plain numbers.

Every inertial observer measures the same acceleration

Differentiate the velocity equation once more:

If the train moves at constant velocity relative to the ground, $\vec{v}_{TG}$ is constant and $\vec{a}_{TG} = 0$, so

That is why a ball dropped in a smoothly cruising train falls straight down to the floor beside your foot: in the train frame it starts at rest and accelerates at $g$ downward. A person on the platform sees the same ball start with the train's horizontal velocity and follow a curved path (topic 1.5 treats that path), but measures the same $g$ downward. Since Newton's second law links force to acceleration, this agreement is also why the laws of motion in Unit 2 work the same way in every inertial frame. In a frame that accelerates, like a braking car, the extra term $\vec{a}_{TG}$ is not zero and objects appear to accelerate with nothing pushing them: a coffee cup slides forward on the dashboard.

Worked examples

Worked exampleExample 2: crossing a river

A river $60\ \text{m}$ wide flows east at $1.2\ \text{m/s}$ relative to the ground. A kayaker paddles at $2.0\ \text{m/s}$ relative to the water, starting from the south bank. (a) If she points the kayak due north, find her velocity relative to the ground, the time to cross, and how far downstream she lands. (b) In what direction must she point the kayak to land directly opposite her starting point, and how long does that crossing take?

Show solution

Use $K$ for kayak, $W$ for water, $G$ for ground, with $+x$ east and $+y$ north. The velocity she controls is $\vec{v}_{KW}$; the one that carries her across is $\vec{v}_{KG} = \vec{v}_{KW} + \vec{v}_{WG}$.

(a) $\vec{v}_{KW} = 2.0\,\hat{j}\ \text{m/s}$ and $\vec{v}_{WG} = 1.2\,\hat{i}\ \text{m/s}$, so

$$\vec{v}_{KG} = (1.2\,\hat{i} + 2.0\,\hat{j})\ \text{m/s}, \qquad |\vec{v}_{KG}| = \sqrt{1.2^2 + 2.0^2} = 2.33\ \text{m/s},$$

directed $\tan^{-1}\left(\frac{1.2}{2.0}\right) = 31.0^\circ$ east of north. Only the north component carries her across, so $t = \frac{60\ \text{m}}{2.0\ \text{m/s}} = 30\ \text{s}$. Meanwhile the east component carries her $(1.2\ \text{m/s})(30\ \text{s}) = 36\ \text{m}$ downstream. The current changes her east-west motion but not how fast she crosses, because the two components are perpendicular.

(b) To go straight north relative to the ground, the east component of $\vec{v}_{KG}$ must be zero. She must aim partly upstream (west) so that her own westward component cancels the current:

$$2.0\sin\theta = 1.2 \quad\Rightarrow\quad \sin\theta = 0.60, \quad \theta = 36.9^\circ \text{ west of north}.$$

Her ground velocity is the remaining north component, $2.0\cos 36.9^\circ = 1.6\ \text{m/s}$ (check: $\sqrt{2.0^2 - 1.2^2} = 1.6$), so the crossing takes $\frac{60\ \text{m}}{1.6\ \text{m/s}} = 37.5\ \text{s}$. Landing opposite costs $7.5\ \text{s}$ over the straight-across heading, because part of her paddling now fights the current instead of crossing. If the current were faster than $2.0\ \text{m/s}$, $\sin\theta$ would exceed 1 and no heading could land her directly opposite.

Where students lose points

On the exam

Questions ask you to find a velocity in another frame (often by sketching the vector triangle), to compare what two observers measure, or to justify why observers agree on acceleration. Name the frame for every velocity you write. For comparisons, a full answer states which quantities differ between frames (position, velocity, and the shape of the path) and which agree (acceleration, and time elapsed), with the reason: the frames differ by a constant velocity.