Visighta

Vector Kinematics in Two and Three Dimensions

3.1 Position, Velocity, and Acceleration Vectors

A Drone in the Sky

[Video: A drone flies a looping path against a clear sky. The camera tracks it from below. At one moment it heads northeast, engines whining as it accelerates. Then it banks left, sweeping northward while visibly slowing. The path it traces is a smooth curve --- not a straight line, not a circle, but something organic and complex. The video freezes mid-flight. A label appears: "At this instant, the drone is heading northeast at 8 m/s and speeding up. A moment later, it is heading north at 5 m/s and slowing down. How do we describe this?"]

In Chapter 2, you described motion with a single number that changed over time: $x(t)$. Velocity was positive or negative --- rightward or leftward, upward or downward. A sign was enough to capture direction because there were only two directions to choose from.

That drone is not on a line. It moves northeast, then north, then wherever it pleases. Its speed changes and its direction changes, sometimes at the same time. A positive or negative number cannot encode "northeast." You need something richer.

You need a vector.

Before you read on: A ball rolls along a curved path on a table. At the instant shown, which direction does the velocity point?

  • (a) Along the curve, tangent to the path in the direction of motion
  • (b) Toward the center of curvature of the path
  • (c) In the direction the ball came from (backward along the path)

Commit to your answer before continuing.

[Interactive: Predict-Then-Reveal. An animation shows a ball tracing a smooth curve on a 2D surface. The ball freezes at a marked instant on the curve. Three arrows are drawn at that point: one tangent to the path (forward), one pointing toward the center of curvature, and one pointing backward. The student selects which arrow represents the velocity. After submitting, the correct answer is revealed: (a). A brief explanation appears: "The velocity always points tangent to the path, in the direction the object is moving at that instant. It does not point toward where the object came from or where the path is curving. This is a geometric fact that holds for any smooth path, and we will see exactly why in this section."]

If you chose (a), your intuition is good. If you chose (b) or (c), you are not alone --- and by the end of this section, you will understand geometrically why (a) is the only possibility.

From Numbers to Arrows

In one dimension, position was a single number: $x(t)$. In two or three dimensions, position is a vector-valued function of time:

$$\vec{r}(t) = x(t)\,\hat{\imath} + y(t)\,\hat{\jmath}$$

or in three dimensions, $\vec{r}(t) = x(t)\,\hat{\imath} + y(t)\,\hat{\jmath} + z(t)\,\hat{k}$.

This is the position vector --- an arrow from the origin to the object's location at time $t$. As $t$ changes, the tip of the arrow traces out the path.

Newton was among the first to treat velocity and acceleration as directed quantities with both magnitude and direction, an insight that made his laws of motion possible. Before him, speed and direction were discussed separately, which made curved motion nearly impossible to analyze.

The derivative relationships you learned in Chapter 2 carry over unchanged --- the only difference is that now each quantity is a vector:

$$\vec{v}(t) = \frac{d\vec{r}}{dt}, \qquad \vec{a}(t) = \frac{d\vec{v}}{dt}$$

In components, this means:

$$\vec{v}(t) = \frac{dx}{dt}\,\hat{\imath} + \frac{dy}{dt}\,\hat{\jmath} = v_x(t)\,\hat{\imath} + v_y(t)\,\hat{\jmath}$$

$$\vec{a}(t) = \frac{dv_x}{dt}\,\hat{\imath} + \frac{dv_y}{dt}\,\hat{\jmath} = a_x(t)\,\hat{\imath} + a_y(t)\,\hat{\jmath}$$

Each component obeys the same derivative chain you already know: $x(t) \xrightarrow{d/dt} v_x(t) \xrightarrow{d/dt} a_x(t)$, and likewise for $y$. The vector versions are simply the component versions stacked together.

Pause and connect: In Chapter 2, the chain was $x(t) \leftrightarrow v(t) \leftrightarrow a(t)$, with each link being a derivative or an integral. That chain has not changed. We have just replaced scalar functions with vector functions. The calculus is the same; the objects are richer.

Seeing the Vectors: Arrows on a Path

Definitions are necessary, but they are not enough. You need to see what these vectors look like as an object moves along a curved path. The following interactive is the heart of this section.

[Interactive: Vector Motion Explorer. A dot traces a smooth 2D curve (a figure-eight or a looping path) at varying speed. At every moment, three arrows are drawn at the dot's location:

  • Position vector $\vec{r}$ (blue): from the origin to the dot.
  • Velocity vector $\vec{v}$ (green): tangent to the path, length proportional to speed.
  • Acceleration vector $\vec{a}$ (red): showing how the velocity is changing.

Controls: a play/pause button, a step-forward button (advance by small $\Delta t$ increments), and a slider to scrub through the entire path. Below the path, a readout shows the numerical values of $\vec{r}$, $\vec{v}$, and $\vec{a}$ in component form $(v_x, v_y)$ and as magnitude-angle pairs $(|\vec{v}|, \theta)$.

Guided prompts appear as the student explores:]

Prompt 1: Play the animation and watch the green arrow (velocity). Does it ever point away from the direction of motion? Pause at several points and check.

Prompt 2: Find a moment where the path curves to the left. Which direction does the red arrow (acceleration) point --- left, right, forward, or backward?

Prompt 3: Find a moment where the dot is speeding up. What is the relationship between the green arrow and the red arrow? Now find a moment where the dot is slowing down. What changed?

Prompt 4: Step forward through a section of the curve, one frame at a time. Watch the green arrow rotate. The red arrow points in the direction the green arrow is turning toward. Verify this at several points.

If you spent time with that interactive, you discovered three things:

  1. The velocity vector always points tangent to the path, in the direction of motion.
  2. The acceleration vector points toward the concave side of the curve --- the inside of the bend.
  3. When the object speeds up, the acceleration has a component along the velocity. When it slows down, the acceleration has a component opposite to the velocity.

These are not formulas to memorize. They are geometric facts --- consequences of what "derivative of a vector" means.

Why Velocity Is Tangent to the Path

This is worth understanding, not just accepting. Consider a small time interval from $t$ to $t + \Delta t$. During that interval, the object moves from $\vec{r}(t)$ to $\vec{r}(t + \Delta t)$. The displacement is

$$\Delta \vec{r} = \vec{r}(t + \Delta t) - \vec{r}(t)$$

This displacement vector points from the old position to the new position --- it is a short chord along the path. As $\Delta t \to 0$, this chord gets shorter and shorter, and its direction approaches the tangent to the path at the point $\vec{r}(t)$.

The velocity is:

$$\vec{v}(t) = \lim_{\Delta t \to 0} \frac{\vec{r}(t + \Delta t) - \vec{r}(t)}{\Delta t}$$

Dividing by $\Delta t$ scales the length but does not change the direction. So $\vec{v}(t)$ points in the direction of $\Delta \vec{r}$ in the limit --- which is the tangent direction.

This is the same limit argument you used in Chapter 1 when the derivative of $x(t)$ gave the instantaneous rate of change. The only new ingredient is that the "rate of change" is now a vector, so it has a direction --- and that direction is tangent to the curve.

Before you read on: If the velocity is always tangent to the path, what does the magnitude of the velocity vector represent?

Check your answer The magnitude of the velocity vector is the **speed**: $|\vec{v}| = \sqrt{v_x^2 + v_y^2}$. Speed tells you how fast the object is moving; the direction of $\vec{v}$ tells you which way. Together, they give the full velocity. This is exactly the distinction the opening question hinted at: speed alone cannot describe motion in two dimensions. You need both magnitude and direction.

Where Does the Acceleration Point?

The acceleration vector is the derivative of the velocity vector:

$$\vec{a}(t) = \frac{d\vec{v}}{dt} = \lim_{\Delta t \to 0} \frac{\vec{v}(t + \Delta t) - \vec{v}(t)}{\Delta t}$$

Think about what $\vec{v}(t + \Delta t) - \vec{v}(t)$ looks like. If the object is moving along a curve, then between time $t$ and $t + \Delta t$, the velocity vector can change in two ways:

  • Its magnitude can change (the object speeds up or slows down).
  • Its direction can change (the object turns).

The acceleration vector $\vec{a}$ captures both of these changes simultaneously. If the speed is changing, $\vec{a}$ has a component along the path. If the direction is changing, $\vec{a}$ has a component perpendicular to the path, pointing toward the concave side of the curve.

This is why, in the interactive, you saw the red arrow pointing "into the bend." On a curved path, the velocity vector is continually rotating toward the inside of the curve. The acceleration must point inward to produce that rotation.

This is a preview of a deep idea that Section 3.6 will develop fully: acceleration can be split into a tangential part (changing speed) and a normal part (changing direction). For now, the key insight is that acceleration on a curve does not generally point in the direction of motion.

Three Ways to See the Same Vector

A vector is a single physical object --- an arrow with a magnitude and a direction. But we can describe that arrow in multiple ways, and switching between descriptions is one of the most useful skills in physics.

Here are three representations of the same velocity vector, shown side by side:

Representation Example What it emphasizes
Arrow on the path A green arrow tangent to the curve at the object's location Geometric meaning: direction of motion, visual sense of speed from arrow length
Component pair $(v_x, v_y) = (3, 4)$ m/s Algebraic accessibility: each component is a plain number you can differentiate or integrate separately
Magnitude and angle $ \vec{v}

[Interactive: Representation Toggle. A single velocity vector is displayed on a 2D plane. Three buttons let students toggle between "Arrow view," "Component view" (showing $v_x$ and $v_y$ as horizontal and vertical arrows that form a right triangle with $\vec{v}$ as the hypotenuse), and "Magnitude-angle view" (showing $|\vec{v}|$ and $\theta$). A slider lets students rotate the vector and watch all three representations update in real time. The student can also toggle on all three views simultaneously to see how they relate.]

These are not three different vectors. They are three descriptions of one vector. Being fluent means being able to start from any one and quickly produce the others:

  • From components to magnitude and angle: $|\vec{v}| = \sqrt{v_x^2 + v_y^2}$, $\theta = \arctan(v_y / v_x)$
  • From magnitude and angle to components: $v_x = |\vec{v}|\cos\theta$, $v_y = |\vec{v}|\sin\theta$
  • From either to the arrow: draw it

Before you read on: A velocity vector has components $v_x = -4$ m/s and $v_y = 3$ m/s. Without a calculator, estimate: What is the speed? What quadrant is the velocity pointing into?

Check your answer Speed: $|\vec{v}| = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ m/s. Direction: $v_x < 0$ and $v_y > 0$, so the velocity points into the second quadrant --- to the left and upward. The angle above the negative $x$-axis is $\arctan(3/4) \approx 37°$, which means the velocity points $180° - 37° = 143°$ from the positive $x$-axis (or equivalently, $37°$ above the leftward direction).

A Worked Example: Circular Path

Let's compute all three kinematic vectors for a concrete case and see how the geometry plays out.

An object moves along a circular path of radius $R = 2$ m, completing one full circle every $T = 4$ seconds. Its position as a function of time is:

$$\vec{r}(t) = 2\cos!\left(\frac{\pi t}{2}\right)\hat{\imath} + 2\sin!\left(\frac{\pi t}{2}\right)\hat{\jmath}$$

Step 1: Velocity. Differentiate component by component:

$$\vec{v}(t) = \frac{d\vec{r}}{dt} = -\pi\sin!\left(\frac{\pi t}{2}\right)\hat{\imath} + \pi\cos!\left(\frac{\pi t}{2}\right)\hat{\jmath}$$

Step 2: Acceleration. Differentiate once more:

$$\vec{a}(t) = \frac{d\vec{v}}{dt} = -\frac{\pi^2}{2}\cos!\left(\frac{\pi t}{2}\right)\hat{\imath} - \frac{\pi^2}{2}\sin!\left(\frac{\pi t}{2}\right)\hat{\jmath}$$

Step 3: Interpret. Notice that $\vec{a}(t) = -\frac{\pi^2}{4}\,\vec{r}(t)$. The acceleration points in the direction opposite to the position vector --- that is, directly toward the center of the circle. This is centripetal acceleration, and you have just derived it from calculus, not from a formula sheet.

Also check: $|\vec{v}| = \pi$ m/s at every instant (the speed is constant), and $|\vec{a}| = \pi^2/2$ m/s$^2$ at every instant. The object moves at constant speed, yet it has a nonzero acceleration that continually redirects the velocity vector. This is a concrete demonstration of the idea from the previous section: acceleration can change direction without changing speed.

What changed? What stayed the same? Compare this with Chapter 2. On a straight line, constant speed meant zero acceleration. On a circle, constant speed coexists with nonzero acceleration. The new ingredient is curvature. Whenever the path curves, the velocity vector must rotate, and that rotation requires acceleration --- even if the speed never changes.

A Faded Example: Parabolic Path

Now try one where some steps are left for you.

An object moves along the path $\vec{r}(t) = (3t)\,\hat{\imath} + (4t - t^2)\,\hat{\jmath}$, where $t$ is in seconds and distances are in meters.

Step 1: Find $\vec{v}(t)$.

Differentiate each component of $\vec{r}(t)$.

Check your answer $$\vec{v}(t) = \frac{d}{dt}(3t)\,\hat{\imath} + \frac{d}{dt}(4t - t^2)\,\hat{\jmath} = 3\,\hat{\imath} + (4 - 2t)\,\hat{\jmath}$$ The $x$-component of velocity is constant at $3$ m/s. The $y$-component starts at $4$ m/s and decreases at a rate of $2$ m/s per second.

Step 2: Find $\vec{a}(t)$.

Differentiate $\vec{v}(t)$.

Check your answer $$\vec{a}(t) = 0\,\hat{\imath} + (-2)\,\hat{\jmath} = -2\,\hat{\jmath}$$ The acceleration is constant, pointing purely in the negative $y$-direction. There is no $x$-component of acceleration. This is the signature of projectile-like motion --- constant acceleration in one direction, zero acceleration in the other. You will study this systematically in Section 3.4.

Step 3: Find the speed at $t = 1$ s.

Compute $|\vec{v}(1)|$.

Check your answer At $t = 1$ s: $\vec{v}(1) = 3\,\hat{\imath} + 2\,\hat{\jmath}$. $$|\vec{v}(1)| = \sqrt{3^2 + 2^2} = \sqrt{13} \approx 3.6 \text{ m/s}$$ The velocity points into the first quadrant, at an angle $\theta = \arctan(2/3) \approx 34°$ above the horizontal.

Step 4: At what time is the object at its highest point?

The highest point occurs when the $y$-component of velocity is zero. (Why? Because at the peak, the object momentarily stops rising --- the vertical velocity is zero, just as in Chapter 2.)

Check your answer Set $v_y = 4 - 2t = 0$, which gives $t = 2$ s. At $t = 2$ s, $\vec{v}(2) = 3\,\hat{\imath} + 0\,\hat{\jmath}$. The velocity is purely horizontal --- the object is still moving to the right at 3 m/s, but it has stopped rising. This is the 2D version of the "ball at the top" from Section 2.1. The velocity is not zero at the peak (the object is still moving horizontally), but the *vertical component* of velocity is zero.

The Geometry That Matters

Let's step back and collect the geometric insights from this section. These are the facts you should carry forward:

1. The velocity vector is always tangent to the path. This follows directly from the definition of the derivative. The direction of $\vec{v}$ tells you which way the object is heading at that instant. The magnitude $|\vec{v}|$ tells you how fast.

2. The acceleration vector points toward the concave side of the curve. Whenever the path bends, the velocity vector must rotate, and the acceleration provides that rotation by pointing inward. If the object is also speeding up or slowing down, the acceleration tilts forward or backward along the path.

3. Speed and velocity are different things. Speed is a scalar: $|\vec{v}| = \sqrt{v_x^2 + v_y^2}$. Velocity is a vector: it has both magnitude and direction. Two objects can have the same speed but entirely different velocities if they are heading in different directions.

4. Acceleration does not have to point in the direction of motion. This is perhaps the biggest conceptual shift from 1D to 2D. On a line, acceleration either aligns with or opposes the velocity. On a curve, acceleration can point in any direction --- and the component perpendicular to the velocity is what causes turning.

Practice

Layer 1: Concrete

Problem 1. An object has position $\vec{r}(t) = (t^2 + 1)\,\hat{\imath} + (3t - 2t^2)\,\hat{\jmath}$, with distances in meters and time in seconds.

(a) Find $\vec{v}(t)$ and $\vec{a}(t)$.

(b) Find the speed at $t = 0$ and at $t = 1$ s.

(c) At what time does the $y$-component of velocity equal zero?

Check your answer **(a)** $$\vec{v}(t) = 2t\,\hat{\imath} + (3 - 4t)\,\hat{\jmath}$$ $$\vec{a}(t) = 2\,\hat{\imath} - 4\,\hat{\jmath}$$ The acceleration is constant --- it does not depend on time. It points into the fourth quadrant (to the right and downward). **(b)** At $t = 0$: $\vec{v}(0) = 0\,\hat{\imath} + 3\,\hat{\jmath}$, so $|\vec{v}(0)| = 3$ m/s. The object starts by moving purely upward. At $t = 1$: $\vec{v}(1) = 2\,\hat{\imath} + (-1)\,\hat{\jmath}$, so $|\vec{v}(1)| = \sqrt{4 + 1} = \sqrt{5} \approx 2.24$ m/s. The object is now moving to the right and downward. **(c)** Set $v_y = 3 - 4t = 0$, giving $t = 3/4 = 0.75$ s. At this instant the object is moving purely horizontally (to the right, since $v_x(0.75) = 1.5$ m/s $> 0$). This is the highest point in the $y$-direction.

Problem 2. A particle moves in a circle of radius 5 m at a constant speed of 10 m/s.

(a) What is the magnitude of the acceleration?

(b) What is the direction of the acceleration?

Check your answer **(a)** For uniform circular motion, the acceleration has magnitude $|\vec{a}| = v^2/R = (10)^2/5 = 20$ m/s$^2$. (You saw this relationship emerge from the circular-path example earlier. Section 3.7 will derive it systematically.) **(b)** The acceleration points toward the center of the circle at every instant. It is perpendicular to the velocity, which is tangent to the circle.

Layer 2: Pattern

[Interactive: Sketch the Vectors. An animated dot traces a smooth 2D curve. At five marked instants (labeled A through E), the animation pauses. For each pause, the student draws a velocity arrow and an acceleration arrow on the screen. The system then reveals the correct arrows and compares. Marks on the curve include: a straight section (A), a gently curving section with increasing speed (B), a sharp turn at constant speed (C), the top of an arc where the dot is slowing (D), and a point where the dot moves along a straight line but is decelerating (E).]

After completing the interactive, answer: At which of the five points is the acceleration perpendicular to the velocity? At which is it parallel? At which is it neither?

Check your answer - **Point A (straight, constant speed):** Acceleration is zero (no change in speed or direction). Neither parallel nor perpendicular --- there is simply no acceleration vector to compare. - **Point B (gentle curve, speeding up):** Acceleration has both a tangential component (speeding up) and a normal component (turning). It is neither purely parallel nor purely perpendicular to the velocity. - **Point C (sharp turn, constant speed):** Speed is not changing, so the tangential component of acceleration is zero. The entire acceleration is normal (perpendicular to the velocity), pointing into the turn. - **Point D (top of arc, slowing):** The dot is both turning and decelerating. Acceleration has a normal component (into the curve) and a tangential component (opposing the velocity). Neither purely parallel nor perpendicular. - **Point E (straight line, decelerating):** No turning, so no normal component. The acceleration is purely tangential, pointing opposite to the velocity --- it is antiparallel.

Layer 3: Structure

Can the velocity vector ever be perpendicular to the acceleration vector? If so, what does the motion look like?

Check your answer Yes. This happens whenever the object is changing direction but *not* changing speed. The most familiar example is **uniform circular motion**: the velocity is always tangent to the circle, and the acceleration is always directed toward the center, perpendicular to the velocity. More generally, $\vec{v} \perp \vec{a}$ means the acceleration has no component along the direction of motion. Since the tangential component of acceleration is what changes speed ($a_T = d|\vec{v}|/dt$), having $\vec{v} \perp \vec{a}$ implies $a_T = 0$, which means the speed is constant. The acceleration is purely centripetal --- all it does is turn the velocity vector without lengthening or shortening it. So the answer is: $\vec{v} \perp \vec{a}$ corresponds to motion at constant speed along a curved path.

Layer 4: Debug

A student analyzing circular motion draws the velocity vector pointing radially outward --- away from the center of the circle. "The object is moving outward from the center," the student explains, "so the velocity must point outward."

What is wrong with this reasoning?

Check your answer The student is confusing the *position* of the object relative to the center with the *direction of motion*. The object is indeed located away from the center (the position vector points outward), but the object is not moving outward --- it is moving *around* the circle. At every instant, the object's motion is along the circle, which means the velocity is tangent to the circle, not radial. The confusion comes from conflating "where the object is" (described by $\vec{r}$) with "which way the object is going" (described by $\vec{v}$). These are different vectors pointing in different directions. For circular motion, $\vec{r}$ points outward from the center, and $\vec{v}$ is perpendicular to $\vec{r}$, tangent to the circle. A quick test: if the velocity pointed radially outward, the object would be moving away from the center in a straight line. It would not stay on the circle. The fact that the object *does* stay on the circle tells you the velocity must be tangent, not radial.

Reflection

Think about what you have read and explored in this section.

What does the velocity vector tell you that the speed alone does not?

Consider this too: in Chapter 2, velocity was a signed number, and you could read direction from the sign. Now velocity is an arrow. What did you gain by making this upgrade? What, if anything, did you lose?

Finally, a confidence check: how comfortable are you with the idea that an object can accelerate without changing speed? If this still feels strange, revisit the circular-motion example above and trace the logic again. This idea will be central to everything that follows in this chapter.

Looking Ahead

You have just extended the language of kinematics from one dimension to two and three. Position, velocity, and acceleration are now vectors --- arrows with both magnitude and direction. The derivative relationships connecting them are unchanged from Chapter 2; only the objects they act on are richer.

But you may have noticed something in the worked examples: when we computed $\vec{v}(t)$ and $\vec{a}(t)$, we differentiated each component independently. The $x$-derivative knew nothing about the $y$-derivative. Each component told its own story, and the full vector was just those stories placed side by side.

That independence is not an accident. It is the hero idea of this entire chapter: motion in perpendicular directions can be analyzed separately and then recombined. In the next section, you will see how the choice of coordinate axes affects the component description of a vector --- and why the physics does not depend on that choice. The vector is real; the components are a matter of perspective.