CLASSICAL-MECHANICS · Interactive Practice | Unit 3 · Video 1

Where, When, How Far: The Mathematical Language of Motion

IKey Formulas

Formula	Name	Description
$\vec{r}(t) = x(t)\,\hat{i}$	Position vector	Locates the object relative to the chosen origin
$x_0 \equiv x(t = 0)$	Initial position	Position coordinate at time $t = 0$
$\Delta t = t_2 - t_1$	Time interval	Elapsed time between two instants (scalar, in seconds)
$\Delta \vec{r} = \vec{r}(t_2) - \vec{r}(t_1) = \Delta x\,\hat{i}$	Displacement	Change in position vector (vector, can be negative)

IIVisualization 1 — The Position Vector

The sign of $x(t)$ sets the arrow's direction; its magnitude sets the arrow's length.

Position x(t) [m] 3

💡 The origin has no physical meaning of its own — it is only the reference point we chose. Sliding it would renumber every $x(t)$ without changing any actual motion.

IIIVisualization 2 — Displacement

Displacement is end minus start — so it turns negative the moment the object finishes left of where it began.

Start x(t₁) [m] 2 End x(t₂) [m] -3

💡 Displacement depends only on the start and end points, not the path between them — unlike distance, the total path length, which is never negative.

IVVisualization 3 — The Time Interval

Elapsed time is a scalar: a duration with magnitude but no direction in space.

First instant t₁ [s] 2 Second instant t₂ [s] 7

💡 With magnitude but no spatial direction, $\Delta t$ is a scalar — the odd one out among the vector quantities $\vec{r}$ and $\Delta\vec{r}$ above. Its SI unit is the second (s).

VQuiz Questions

Question 1

An object is located at position $x(t) = -2$ m on the axis. Its position vector is $\vec{r}(t) = x(t)\,\hat{i}$ .

Which best describes the position vector $\vec{r}(t)$ ?

✅ Correct! The length is the magnitude (2 m) and the negative sign flips the direction to the left.

❌ Not quite. An arrow length is a magnitude, so it is never negative. The negative sign of x(t) changes the direction, not the length.

Show solution

Solution:

The position vector is $\vec{r}(t) = x(t)\,\hat{i} = (-2)\,\hat{i}$ .

The magnitude is $|x(t)| = 2$ m — length is always non-negative.
The sign of $x(t)$ sets the direction: since $x(t) = -2 < 0$ , the arrow points opposite to $\hat{i}$ , i.e. to the left.

So $\vec{r}(t)$ is an arrow of length 2 m pointing left (opposite to $\hat{i}$ ).

Question 2

An object moves from $x(t_1) = 2$ m to $x(t_2) = -3$ m.

What is the displacement $\Delta \vec{r}$ ?

✅ Correct! Δx = x(t₂) − x(t₁) = −3 − 2 = −5 m, so Δr = −5 î.

❌ Not quite. You have the magnitude right, but the sign is reversed — subtract final minus initial: −3 − 2 = −5.

❌ Not quite. Use Δx = x(t₂) − x(t₁) = −3 − 2. Be careful with the two negative quantities.

Show solution

Solution:

Displacement is the change in position, end minus start: $\Delta x = x(t_2) - x(t_1) = (-3) - (2) = -5 \text{ m}$ $\Delta \vec{r} = \Delta x\,\hat{i} = -5\,\hat{i}$

The negative sign means the object moved 5 m in the direction opposite to $\hat{i}$ (to the left). A common mistake is to reverse the subtraction ( $2 - (-3) = +5$ ) — always compute final minus initial.

Question 3

True or False: For the trip from $x(t_1) = 2$ m to $x(t_2) = -3$ m, the distance traveled and the magnitude of the displacement are both 5 m, so distance and displacement are always equal.

✅ Correct! They match here only because the motion is straight with no reversal. In general distance ≥ |displacement|.

❌ Not quite. They are equal for this specific path, but not in general — any back-and-forth motion makes the distance larger than the magnitude of the displacement.

Show solution

Solution: False.

For this particular straight-line trip with no reversals, the distance (5 m) does happen to equal the magnitude of the displacement (5 m). But that is a coincidence of this path, not a general rule.

Displacement is a vector: the net change in position, $\Delta x\,\hat{i}$ . It depends only on the start and end points and can be negative.
Distance is a scalar: the total path length, always $\geq 0$ .

If the object first went from 2 m out to 5 m and then back to −3 m, the displacement would still be −5 î (magnitude 5 m), but the distance traveled would be $|5-2| + |{-3}-5| = 3 + 8 = 11$ m. They are not always equal.

Question 4

Consider these three quantities from one-dimensional kinematics: position $x(t)$ , the time interval $\Delta t = t_2 - t_1$ , and the displacement $\Delta \vec{r}$ .

Which statement correctly classifies them?

✅ Correct! Only the time interval is a scalar; position and displacement both carry a direction along the axis.

❌ Not quite. A quantity is a vector when it has a direction. Position and displacement point along the axis (±î), but elapsed time has no spatial direction.

Show solution

Solution:

Position $\vec{r}(t) = x(t)\,\hat{i}$ has both magnitude and direction (the arrow can point in the $+\hat{i}$ or $-\hat{i}$ direction) — it is a vector.
Time interval $\Delta t = t_2 - t_1$ has magnitude (a duration) but no spatial direction — it is a scalar, measured in seconds.
Displacement $\Delta \vec{r} = \Delta x\,\hat{i}$ is the change in the position vector and points in a direction along the axis — it is a vector.

So: position — vector, time interval — scalar, displacement — vector.

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