- Describe how particles move in space using velocity and acceleration vectors
- Explain the tangential and normal components of acceleration
- Use vector-valued functions to describe and analyze planetary motion
Motion Vectors in the Plane and in Space
The Main Idea
Vector-valued functions describe how objects move through space over time. The position function [latex]{\bf{r}}(t)[/latex] and its derivatives give you everything you need to understand motion: where the object is, how fast it’s moving, and how its motion is changing.
For position function [latex]{\bf{r}}(t)[/latex]:
- Velocity vector: [latex]{\bf{v}}(t)={\bf{r}}'(t)[/latex]
- Acceleration vector: [latex]{\bf{a}}(t)={\bf{v}}'(t)={\bf{r}}''(t)[/latex]
- Speed: [latex]v(t)=\left\Vert{\bf{v}}(t)\right\Vert=\left\Vert{\bf{r}}'(t)\right\Vert[/latex]
Two Dimensions vs. Three Dimensions
The formulas work the same way in both cases. You just add or remove the [latex]{\bf{k}[/latex] component.
In 2D: [latex]{\bf{r}}(t)=x(t){\bf{i}}+y(t){\bf{j}}[/latex]
- Velocity: [latex]{\bf{v}}(t)=x'(t){\bf{i}}+y'(t){\bf{j}}[/latex]
- Acceleration: [latex]{\bf{a}}(t)=x''(t){\bf{i}}+y''(t){\bf{j}}[/latex]
- Speed: [latex]v(t)=\sqrt{(x'(t))^{2}+(y'(t))^{2}}[/latex]
In 3D: [latex]{\bf{r}}(t)=x(t){\bf{i}}+y(t){\bf{j}}+z(t){\bf{k}}[/latex]
- Velocity: [latex]{\bf{v}}(t)=x'(t){\bf{i}}+y'(t){\bf{j}}+z'(t){\bf{k}}[/latex]
- Acceleration: [latex]{\bf{a}}(t)=x''(t){\bf{i}}+y''(t){\bf{j}}+z''(t){\bf{k}}[/latex]
- Speed: [latex]v(t)=\sqrt{(x'(t))^{2}+(y'(t))^{2}+(z'(t))^{2}}[/latex]
Understanding the Geometry
- The velocity vector is always tangent to the path. It points in the direction of motion.
- The acceleration vector points toward where the object is turning. Think of a car going around a curve—you feel pushed toward the outside, but acceleration points inward.
- Speed is the magnitude of velocity. It tells you how fast without indicating direction.
Speed can stay constant while velocity changes. Why? Because velocity includes both speed and direction. Even at constant speed, turning changes your velocity vector because your direction changes.
Problem-Solving Strategy
Start with the position function [latex]{\bf{r}}(t)[/latex]. Take the first derivative to get velocity. Take the second derivative to get acceleration. Find the magnitude of velocity to get speed.
Components of the Acceleration Vector
The Main Idea
The acceleration vector doesn’t just tell you how motion is changing—it reveals how that change happens. It splits into two perpendicular pieces: one affecting speed, the other affecting direction.
The acceleration vector lies in the plane formed by [latex]{\bf{T}}(t)[/latex] and [latex]{\bf{N}}(t)[/latex]. It can be written as:
[latex]{\bf{a}}(t)=a_{\bf{T}}{\bf{T}}(t)+a_{\bf{N}}{\bf{N}}(t)[/latex]
where:
- Tangential component [latex]a_{\bf{T}}=\frac{{\bf{v}}\cdot{\bf{a}}}{\left\Vert{\bf{v}}\right\Vert}[/latex] measures how quickly speed is changing
- Normal component [latex]a_{\bf{N}}=\frac{\left\Vert{\bf{v}}\times{\bf{a}}\right\Vert}{\left\Vert{\bf{v}}\right\Vert}[/latex] measures how quickly direction is changing
What Each Component Means
- [latex]a_{\bf{T}}[/latex] tells you if you’re speeding up or slowing down. Positive means speeding up. Negative means slowing down.
- [latex]a_{\bf{N}}[/latex] tells you how sharply you’re turning. Larger values mean tighter curves.
Alternative Formula for Normal Component
If calculating the cross product feels tedious, use:
[latex]a_{\bf{N}}=\sqrt{{\left\Vert{\bf{a}}\right\Vert}^{2}-a^{2}_{\bf{T}}}[/latex]
Understanding Centripetal Acceleration
The normal component is also called the centripetal component. Picture driving on a circular track at constant speed. Your speed isn’t changing, so [latex]a_{\bf{T}}=0[/latex]. But you’re constantly turning, so [latex]a_{\bf{N}} \neq 0[/latex]. The acceleration points toward the center of the track. You feel pushed outward—that’s your body resisting the inward acceleration.
Imagine driving into a curve while braking. Your acceleration vector points partly backward (slowing down) and partly toward the inside of the curve (turning). The tangential component captures the braking. The normal component captures the turning.
Problem-Solving Strategy
Find [latex]{\bf{v}}(t)[/latex] and [latex]{\bf{a}}(t)[/latex] first. Compute [latex]a_{\bf{T}}[/latex] using the dot product formula. Then find [latex]a_{\bf{N}}[/latex] using either the cross product or the simpler square root formula.
- Find [latex]a_{\bf{T}}[/latex] and [latex]a_{\bf{N}}[/latex] as functions of [latex]t[/latex].
- Find [latex]a_{\bf{T}}[/latex] and [latex]a_{\bf{N}}[/latex] at time [latex]t=-3[/latex].
Projectile Motion
The Main Idea
Projectile motion describes objects moving under gravity’s influence alone—think baseballs, arrows, or cannonballs. We ignore air resistance. The math becomes straightforward when you recognize that gravity only affects vertical motion, leaving horizontal motion unchanged.
The Setup
Place the origin at the launch point. The positive [latex]y[/latex]-axis points up. The positive [latex]x[/latex]-axis points forward. Gravity pulls downward with acceleration [latex]g \approx 32 \text{ ft/s}^2[/latex] (English) or [latex]g \approx 9.8 \text{ m/s}^2[/latex] (metric).
The Position Equation
Starting with acceleration [latex]{\bf{a}}(t) = -g{\bf{j}}[/latex], integrate twice to get:
[latex]{\bf{s}}(t)=-\frac{1}{2}gt^{2}{\bf{j}}+{\bf{v}}{0}t+{\bf{s}}{0}[/latex]
where [latex]{\bf{v}}{0}[/latex] is initial velocity and [latex]{\bf{s}}{0}[/latex] is initial position.
Projectile Launched at an Angle
When fired at angle [latex]\theta[/latex] with initial speed [latex]v_0[/latex] from the origin:
[latex]{\bf{s}}(t)=v_{0}t\cos\theta{\bf{i}}+(v_{0}t\sin\theta-\frac{1}{2}gt^{2}){\bf{j}}[/latex]
The [latex]{\bf{i}[/latex] component gives horizontal distance. The [latex]{\bf{j}[/latex] component gives altitude.
Key Quantities
- Range [latex]R[/latex]: maximum horizontal distance (at same altitude)
- Height [latex]H[/latex]: maximum altitude
Finding Maximum Height
The projectile reaches maximum height when vertical velocity equals zero. Set [latex]v_y(t) = v_0\sin\theta - gt = 0[/latex] and solve for [latex]t[/latex]. Substitute this [latex]t[/latex] into the position equation.
Finding Range
For launch and landing at the same altitude:
[latex]R=\frac{v^2_0\sin 2\theta}{g}[/latex]
Maximum range occurs at [latex]\theta = 45°[/latex], giving [latex]R = \frac{v^2_0}{g}[/latex].
Kepler’s Laws
The Main Idea
Johannes Kepler used astronomical observations to describe how planets move around the Sun. His three laws apply to any orbiting object—planets, comets, asteroids, even satellites around Earth.
The Three Laws
- Law of Ellipses: Every planet follows an elliptical orbit. The Sun sits at one focus of the ellipse.
- Law of Equal Areas: A line from the Sun to a planet sweeps out equal areas in equal time intervals. This means planets move faster when closer to the Sun and slower when farther away.
- Law of Harmonies: The ratio of orbital periods squared equals the ratio of orbit sizes cubed. For any two planets: [latex]\frac{T_1^2}{T_2^2} = \frac{D_1^3}{D_2^3}[/latex]
Using Kepler’s Third Law
When measuring distance in astronomical units (A.U.) and time in Earth years, the third law simplifies to:
[latex]T^2_p=D^3_p[/latex]
where [latex]T_p[/latex] is the planet’s orbital period and [latex]D_p[/latex] is its average distance from the Sun.
Beyond Planets
Newton later proved these laws using calculus and gravity. He showed the third law works for any orbital system. For a moon orbiting a planet, the formula becomes:
[latex]P^{2}=\frac{4\pi^{2}a^{3}}{G(m+M)}[/latex]
where [latex]m[/latex] and [latex]M[/latex] are the masses, [latex]a[/latex] is the orbit’s semi-major axis, and [latex]G[/latex] is the gravitational constant.
Newton proved planets follow elliptical paths by combining his laws of motion with universal gravitation. Gravity pulls toward the Sun. This force combined with the planet’s velocity creates an ellipse. Hyperbolic paths exist too—comets moving too fast to be captured orbit once and escape.