Classical-Mechanics · Unit 13 · Video 1 · Interactive Practice

No Absolute Rest: Galilean Transformations and the Physics of Moving Observers

IKey Formulas

Formula	Name	What you need
$\vec{r}\,' = \vec{r} - \vec{R}$	Galilean coordinate transformation	Position in $S$ and the origin separation $\vec{R}$
$\vec{V} = \dfrac{d\vec{R}}{dt}$	Relative velocity of the frames	How $\vec{R}$ changes with time
$\vec{v}\,' = \vec{v} - \vec{V}$	Law of addition of velocities	Velocity in $S$ and the frame velocity $\vec{V}$
$\vec{a}\,' = \vec{a}$ when $\vec{V}$ is constant	Relatively inertial frames	$d\vec{V}/dt = \vec{0}$

Key Insight: $\vec{V}$ is the velocity of $S'$ relative to $S$ , and it is subtracted — every disagreement between the two observers about velocity traces back to that single vector.

IITwo Origins, One Object

Two observers measure different position vectors for the same object — differing by exactly $\vec{R}$ .

💡 Subtract two position vectors and $\vec{R}$ cancels — displacement, and therefore velocity and acceleration, never depends on where an origin sits.

IIIRelatively Inertial Frames

With $\vec{V}$ constant, uniform motion stays uniform — so neither observer can claim to be the one truly at rest.

t 0.0 s Box

💡 Differentiating $\vec{v}\,' = \vec{v} - \vec{V}$ once more with $\vec{V}$ constant gives $\vec{a}\,' = \vec{a}$ , so the two frames measure identical accelerations for any motion, not just uniform ones.

IVWhat Airplane B Sees

Airplane B's pilot clocks airplane A at $256$ m/s — faster than either plane's own ground speed.

Step 1

Step 1 — Choose axes and read the angles

\hat{i}\ \text{east},\qquad \hat{j}\ \text{north}

\theta_A = \tfrac{\pi}{4} = 45\degree, \qquad \theta_B = -\tfrac{\pi}{4} = -45\degree

Step 2 — Components of $\vec{v}_A$

\vec{v} = v\cos\theta\,\hat{i} + v\sin\theta\,\hat{j}

\vec{v}_A = 160\cos 45\degree\,\hat{i} + 160\sin 45\degree\,\hat{j} = 80\sqrt{2}\,\hat{i} + 80\sqrt{2}\,\hat{j}

Step 3 — Components of $\vec{v}_B$ , and the frame velocity

\vec{v}_B = 100\sqrt{2}\,\hat{i} - 100\sqrt{2}\,\hat{j}

The observer rides in B, so frame

S'

moves with B:

\vec{V} = \vec{v}_B

Step 4 — Subtract, component by component

\vec{v}_A\,' = \vec{v}_A - \vec{V} = \big(80\sqrt{2} - 100\sqrt{2}\big)\hat{i} + \big(80\sqrt{2} + 100\sqrt{2}\big)\hat{j}

\vec{v}_A\,' = -20\sqrt{2}\,\hat{i} + 180\sqrt{2}\,\hat{j}\ \mathrm{m/s}

Subtracting B's negative

\hat{j}

-component adds to A's.

Step 5 — Magnitude

\big(20\sqrt{2}\big)^2 = 800, \qquad \big(180\sqrt{2}\big)^2 = 64800

\left|\vec{v}_A\,'\right| = \sqrt{65600} \approx 256\ \mathrm{m/s}

Step 6 — Direction, after the quadrant check

\tan\theta_A' = \frac{180\sqrt{2}}{-20\sqrt{2}} = -9 \quad\Longrightarrow\quad \arctan(-9) = -83.7\degree

x' \lt 0

and

y' \gt 0

place the vector in quadrant II:

\theta_A' = 180\degree - 83.7\degree = 96.3\degree

VQuiz Questions

Problem 1 · Transform a Position

Given: In frame $S$ an object sits at $\vec{r} = 9\hat{i} + 5\hat{j}$ m, and the origin of frame $S'$ sits at $\vec{R} = 6\hat{i} + \hat{j}$ m — find the position vector $\vec{r}\,'$ measured by the $S'$ observer.

✅ Correct!

\vec{r}\,' = \vec{r} - \vec{R} = (9-6)\hat{i} + (5-1)\hat{j} = 3\hat{i} + 4\hat{j}

m, so the

S'

observer finds the object

5

m away.

❌ Check the sign. That is

\vec{r} + \vec{R}

. The triangle reads

\vec{r} = \vec{R} + \vec{r}\,'

, so

\vec{r}\,'

is what is left after

\vec{R}

is removed.

❌ Reversed. That is

\vec{R} - \vec{r} = -\vec{r}\,'

, the arrow pointing from the object back to

O'

❌ Not quite. Apply

\vec{r}\,' = \vec{r} - \vec{R}

one component at a time.

Show solution

The three vectors close a triangle: going from $O$ straight to the object gives $\vec{r}$ , while going $O \to O' \to$ object gives $\vec{R} + \vec{r}\,'$ . Both paths end at the same point, so

\vec{r} = \vec{R} + \vec{r}\,' \quad\Longrightarrow\quad \vec{r}\,' = \vec{r} - \vec{R}

Component by component:

\vec{r}\,' = (9 - 6)\hat{i} + (5 - 1)\hat{j} = 3\hat{i} + 4\hat{j}\ \mathrm{m}

Its magnitude is $\left|\vec{r}\,'\right| = \sqrt{3^2 + 4^2} = 5$ m, while the $S$ observer reports $\left|\vec{r}\right| = \sqrt{81 + 25} \approx 10.3$ m. The two observers disagree about position because position is measured from an origin.

Problem 2 · Which Vector Gets Subtracted

Given: A river flows east at $3$ m/s. Measured from the bank (frame $S$ , with $\hat{i}$ east) a boat's velocity is $\vec{v} = 5\hat{i} + 2\hat{j}$ m/s. A raft drifting with the current carries frame $S'$ — find the boat's velocity as measured from the raft.

✅ Correct! The raft carries

\vec{V} = 3\hat{i}

m/s, so

\vec{v}\,' = \vec{v} - \vec{V} = 2\hat{i} + 2\hat{j}

m/s.

❌ Sign convention.

\vec{V}

is the velocity of

S'

relative to

S

, and the law subtracts it:

\vec{v}\,' = \vec{v} - \vec{V}

❌ That is $\vec{v}$ itself. The raft is moving, so it is not the same frame as the bank — the transformation cannot leave the velocity untouched.

❌ Reversed. You computed

\vec{V} - \vec{v}

, which is the velocity of the raft as seen from the boat.

❌ Not quite. Identify

\vec{V}

first: the raft drifts with the water, so

\vec{V} = 3\hat{i}

m/s.

Show solution

Frame $S'$ rides on the raft, and the raft drifts with the current, so the velocity of $S'$ relative to $S$ is

\vec{V} = 3\hat{i}\ \mathrm{m/s}

The law of addition of velocities subtracts that frame velocity:

\vec{v}\,' = \vec{v} - \vec{V} = (5 - 3)\hat{i} + (2 - 0)\hat{j} = 2\hat{i} + 2\hat{j}\ \mathrm{m/s}

The east component shrinks because the raft is already carrying the observer east, while the north component is untouched — $\vec{V}$ has no $\hat{j}$ part.

Problem 3 · The Two Airplanes

Given: Airplane A flies northeast at $160$ m/s and airplane B flies southeast at $200$ m/s, with $\hat{i}$ east, $\hat{j}$ north, and angles measured counterclockwise from $+x$ (the video's example). An observer flies in B — find the velocity of A in that observer's frame, then its direction.

What is the velocity vector?

What direction does it point?

✅ Correct! From B's cockpit, A streaks past at

256

m/s pointing

96.3\degree

from east — just

6.3\degree

west of due north.

❌ You added.

\vec{v}_A + \vec{v}_B

gives that vector; the law of addition of velocities subtracts the frame velocity.

❌ Reversed. That is

\vec{v}_B - \vec{v}_A

— the velocity of B as seen from A, not of A as seen from B.

❌ Watch the double negative. B's

\hat{j}

-component is

-100\sqrt{2}

, and

80\sqrt{2} - (-100\sqrt{2}) = 180\sqrt{2}

, not

-20\sqrt{2}

❌ Not quite. Write both velocities in the ground frame first, then subtract

\vec{V} = \vec{v}_B

component by component.

❌ That is the raw calculator value.

\arctan

always returns an angle in quadrant I or IV; here

x' \lt 0

and

y' \gt 0

, so the vector lies in quadrant II.

❌ Not quite. Take

\arctan

y'/x'

, then correct the quadrant using the signs of the components.

Show solution

Step 1: Components in the ground frame. Northeast is $\theta_A = 45\degree$ , southeast is $\theta_B = -45\degree$ , and $\cos 45\degree = \sin 45\degree = \tfrac{\sqrt{2}}{2}$ :

\vec{v}_A = 80\sqrt{2}\,\hat{i} + 80\sqrt{2}\,\hat{j}, \qquad \vec{v}_B = 100\sqrt{2}\,\hat{i} - 100\sqrt{2}\,\hat{j}

Step 2: Identify the frame velocity. The observer rides in B, so $\vec{V} = \vec{v}_B$ .

Step 3: Subtract.

\vec{v}_A\,' = \big(80\sqrt{2} - 100\sqrt{2}\big)\hat{i} + \big(80\sqrt{2} + 100\sqrt{2}\big)\hat{j} = -20\sqrt{2}\,\hat{i} + 180\sqrt{2}\,\hat{j}\ \mathrm{m/s}

Step 4: Magnitude. $\big(20\sqrt{2}\big)^2 = 800$ and $\big(180\sqrt{2}\big)^2 = 64800$ :

\left|\vec{v}_A\,'\right| = \sqrt{800 + 64800} = \sqrt{65600} \approx 256\ \mathrm{m/s}

This beats both ground speeds because the planes' north-south motions oppose each other.

Step 5: Direction, with the quadrant check.

\tan\theta_A' = \frac{180\sqrt{2}}{-20\sqrt{2}} = -9 \quad\Longrightarrow\quad \arctan(-9) = -83.7\degree

Since $x' = -20\sqrt{2} \lt 0$ and $y' = 180\sqrt{2} \gt 0$ , the vector is in quadrant II, so

\theta_A' = 180\degree - 83.7\degree = 96.3\degree

Problem 4 · What Survives the Transformation

Given: A train moves along a straight track at constant $\vec{V} = 25\hat{i}$ m/s relative to the ground. A passenger throws a ball, and in the train frame $S'$ the ball's acceleration is $\vec{a}\,' = -2\hat{i} + 3\hat{j}$ m/s $^2$ — find the acceleration the ground observer measures.

✅ Correct! A constant

\vec{V}

has

d\vec{V}/dt = \vec{0}

, so acceleration is identical in both frames — and both observers infer the same net force.

❌ Wrong quantity transformed.

\vec{V}

shifts velocities, not accelerations; adding it here also mixes m/s with m/s

^2

❌ Wrong quantity transformed. Subtracting

\vec{V}

is the velocity law; acceleration transforms with

d\vec{V}/dt

, which vanishes here.

❌ That is the frame's relative acceleration.

\vec{A} = d\vec{V}/dt = \vec{0}

describes the two frames, not the ball.

❌ Not quite. Differentiate

\vec{v} = \vec{v}\,' + \vec{V}

once with

\vec{V}

constant.

Show solution

Start from the law of addition of velocities and differentiate with respect to the single universal time $t$ :

\vec{v}\,' = \vec{v} - \vec{V} \quad\Longrightarrow\quad \frac{d\vec{v}\,'}{dt} = \frac{d\vec{v}}{dt} - \frac{d\vec{V}}{dt}

The train's velocity is constant, so $d\vec{V}/dt = \vec{A} = \vec{0}$ and

\vec{a}\,' = \vec{a} = -2\hat{i} + 3\hat{j}\ \mathrm{m/s^2}

The two frames are relatively inertial. They disagree about the ball's position and velocity, but agree exactly on its acceleration — so they agree on the net force $m\vec{a}$ acting on it. That agreement is the Principle of Relativity.

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