Notes on Motion in Two or Three Dimensions — Comprehensive Outline

3.1 Position and Velocity Vectors

Motion in two or three dimensions requires describing position, velocity, and acceleration as vectors, not just scalars on a single line.
Position vector from the origin to a point P:
$\mathbf{r} = x\,\hat{\mathbf{x}} + y\,\hat{\mathbf{y}} + z\,\hat{\mathbf{z}}.$
Coordinates of P are the components of r: (x, y, z).
Displacement during a time interval Δt: (\Delta \mathbf{r} = \mathbf{r}2 - \mathbf{r}1).
Average velocity over an interval:
$\mathbf{v}_{\text{av}} = \frac{\Delta \mathbf{r}}{\Delta t}.$
Instantaneous velocity is the limit of the average velocity as Δt → 0, and is a vector:
$\mathbf{v} = \frac{d\mathbf{r}}{dt}.$
Magnitude of velocity (speed):
$\mathbf{v}| = \sqrt{vx^2 + vy^2 + vz^2}.$ (vx = \frac{dx}{dt}, \; vy = \frac{dy}{dt}, \; vz = \frac{dz}{dt}). $</li><li>In general, the instantaneous velocity is tangent to the particle’s path at every point (the derivative of position traces the path).</li><li>In planar motion (xy-plane), with (z = 0), the velocity components reduce to (vx = \dot{x}, vy = \dot{y}) and speed to $ |\mathbf{v}| = \sqrt{\dot{x}^2 + \dot{y}^2}. $</li><li>The path is a curve; the instantaneous velocity is the tangent to that curve at each point.</li><li>Example context (multi-dimensional rover): coordinates x(t), y(t) give a path in the plane; instantaneous velocity has components (\dot{x}, \dot{y}) and time derivatives give velocity components.</li><li>{{Key equations}} $ \mathbf{r} = x\hat{\mathbf{x}} + y\hat{\mathbf{y}} + z\hat{\mathbf{z}},\quad \mathbf{v}{\text{av}} = \frac{\Delta \mathbf{r}}{\Delta t},\quad \mathbf{v} = \frac{d\mathbf{r}}{dt},vx = \frac{dx}{dt},\quad vy = \frac{dy}{dt},\quad vz = \frac{dz}{dt}, $ $ |\mathbf{v}| = \sqrt{vx^2 + vy^2 + v_z^2}. $</li></ul><div data-type="horizontalRule"><hr></div><h4 id="98e94806-292b-419a-a386-152d9cfe78ac" data-toc-id="98e94806-292b-419a-a386-152d9cfe78ac" collapsed="false" seolevelmigrated="true">3.2 The Acceleration Vector</h4><ul><li>Acceleration describes how velocity changes with time; it has both magnitude and direction.</li><li>Average acceleration over a time interval: $ \mathbf{a}_{\text{av}} = \frac{\Delta \mathbf{v}}{\Delta t}. $</li><li>Instantaneous acceleration is the limit as Δt → 0 of the average, i.e., $ \mathbf{a} = \frac{d\mathbf{v}}{dt}. $</li><li>Components of acceleration along x, y, z: $ ax = \frac{d^2 x}{dt^2},\quad ay = \frac{d^2 y}{dt^2},\quad a_z = \frac{d^2 z}{dt^2}. $</li><li>The acceleration vector is the derivative of the velocity vector and is therefore independent of the path being straight or curved.</li><li>Tangent vs. curvature interpretation:<ul><li>The velocity vector is tangent to the path.</li><li>The acceleration vector may point toward the concave side of a curved path (normal component) or along the velocity (tangential component).</li></ul></li><li>Decomposition of acceleration into components parallel and perpendicular to the velocity:<ul><li>Parallel component (affects speed): $ a_{\parallel} = \frac{dv}{dt}. $</li><li>Perpendicular component (affects direction): for circular motion, the perpendicular (normal) component magnitude is $ a_{\perp} = \frac{v^2}{R}, $ where (R) is the radius of curvature of the path. Alternatively, for general motion the total acceleration is not necessarily aligned with velocity.</li></ul></li><li>In two dimensions (xy-plane), with (ax, ay): $ \mathbf{a} = ax\hat{\mathbf{x}} + ay\hat{\mathbf{y}}. $</li><li>Uniform circular motion (special case):<ul><li>Speed is constant; acceleration is purely centripetal (toward the center).</li><li>Magnitude: $ a = \frac{v^2}{R} = \omega^2 R = \frac{4\pi^2 R}{T^2}, $ where (v = \omega R) and (T) is the period; also (\omega = 2\pi/T).</li><li>Direction is toward the center of the circle (radial inward).</li></ul></li><li>Nonuniform circular motion: tangential component exists, with $ a{\parallel} = \frac{dv}{dt},\quad a{\perp} = \frac{v^2}{R}, $ and the total acceleration magnitude is $ |\mathbf{a}| = \sqrt{a{\parallel}^2 + a{\perp}^2}. $</li><li>Example highlights:<ul><li>If a particle moves along a curved path with constant speed, the acceleration is purely perpendicular to velocity (toward the concave side).</li><li>If speed changes, the acceleration has a nonzero component along the velocity (tangential) as well as a normal component.</li></ul></li><li>Centered concepts: acceleration points toward the concave side of a turning path; velocity is tangent to the path at every instant.</li><li>{{Key equations}} $ \mathbf{a} = \left( ax, ay, az \right),\quad ax = \frac{d^2 x}{dt^2},\ ay = \frac{d^2 y}{dt^2},\ az = \frac{d^2 z}{dt^2}, $ $ a{\parallel} = \frac{dv}{dt},\quad a{\perp} = \frac{v^2}{R},\quad |\mathbf{a}| = \sqrt{a{\parallel}^2 + a{\perp}^2}, $ $ a = \frac{v^2}{R} = \omega^2 R = \frac{4\pi^2 R}{T^2}. $</li></ul><div data-type="horizontalRule"><hr></div><h4 id="58e14aea-296f-4433-ade1-ffeb054c9cf7" data-toc-id="58e14aea-296f-4433-ade1-ffeb054c9cf7" collapsed="false" seolevelmigrated="true">3.3 Projectile Motion</h4><ul><li>A projectile is a body given an initial velocity and then acted upon by gravity (neglecting air resistance here).</li><li>In projectile motion (no air resistance): motion is confined to a vertical plane defined by the initial velocity; the horizontal and vertical motions can be treated separately.</li><li>Equations for horizontal (x) and vertical (y) components (no air resistance):<ul><li>Horizontal motion (constant velocity): $ x(t) = x0 + v{0x} t, \ vx(t) = v{0x}, $</li><li>Vertical motion (constant downward acceleration): $ y(t) = y0 + v{0y} t - \frac{1}{2}gt^2,\quad vy(t) = v{0y} - gt, $ where (g) is the acceleration due to gravity (positive in magnitude, often taken as (g\approx 9.80\,\mathrm{m/s^2}) downward).</li></ul></li><li>Initial velocity components from initial speed (v0) and launch angle (\alpha0): $ v{0x} = v0 \cos\alpha0,\quad v{0y} = v0 \sin\alpha0. $</li><li>Position and velocity of a projectile (general form): $ x(t) = v{0x} t,\quad y(t) = v{0y} t - \tfrac{1}{2}gt^2, $ $ vx(t) = v{0x},\quad vy(t) = v{0y} - gt. $</li><li>Speed at time t: $ v(t) = \sqrt{vx(t)^2 + vy(t)^2}. $</li><li>Trajectory (path) in the x–y plane (eliminating t) has the general parabolic form (for equal launch/landing height): $ y = x \tan\alpha0 - \frac{g x^2}{2 v0^2 \cos^2\alpha_0}. $</li><li>Key results for a projectile with equal initial and final heights (no air resistance):<ul><li>Maximum height: $ h = \frac{v0^2 \sin^2\alpha0}{2g}, $</li><li>Range (horizontal distance when it returns to the same height): $ R = \frac{v0^2 \sin(2\alpha0)}{g}. $</li></ul></li><li>If initial and final heights differ, solve for the time when y(t) equals the final height and use x(t) to find horizontal range; this often involves solving a quadratic in time.</li><li>Examples illustrate:<ul><li>A horizontal launch off a cliff: the horizontal range equals the horizontal speed times the flight time, while vertical motion determines flight time and height drop.</li><li>The dart-monkey problem demonstrates the independence of vertical and horizontal motions and that the dart will hit the falling monkey as long as both bodies fall the same vertical distance during the same time interval.</li></ul></li><li>Parabolic trajectory is the hallmark of ideal projectile motion; air resistance or Earth curvature would modify these results.</li><li>{{Key equations}} $ x(t) = x0 + v{0x} t,\quad vx(t) = v{0x}, $ $ y(t) = y0 + v{0y} t - \frac{1}{2}gt^2,\quad vy(t) = v{0y} - gt, $ $ v(t) = \sqrt{vx^2 + vy^2}, $ $ v{0x} = v0 \cos\alpha0,\quad v{0y} = v0 \sin\alpha0, $ $ x = v{0x} t,\quad y = v{0y} t - \frac{1}{2}gt^2, $ $ y = x \tan\alpha0 - \frac{g x^2}{2 v0^2 \cos^2\alpha0}, h = \frac{v0^2 \sin^2\alpha0}{2g},\quad R = \frac{v0^2 \sin(2\alpha_0)}{g}. $</li></ul><div data-type="horizontalRule"><hr></div><h4 id="e8559162-6b40-42ad-a282-258cfaf34022" data-toc-id="e8559162-6b40-42ad-a282-258cfaf34022" collapsed="false" seolevelmigrated="true">3.4 Motion in a Circle</h4><ul><li>Circular motion basics:<ul><li>Uniform circular motion: speed is constant, velocity changes direction, thus there is acceleration toward the circle’s center (centripetal/normal acceleration).</li><li>Magnitude of centripetal acceleration: $ a = \frac{v^2}{R} = \omega^2 R. $ Since (v = \omega R), also (a = \frac{4\pi^2 R}{T^2}) where (T) is the period and (\omega = 2\pi/T).</li><li>Direction of centripetal acceleration is toward the center of the circle.</li></ul></li><li>Nonuniform circular motion:<ul><li>There is still a radial (normal) component a_rad = v^2/R toward the center.</li><li>There is a tangential component a_tan = dv/dt equal to the rate of change of speed; its direction is the same as velocity if speeding up, opposite if slowing down.</li><li>The total acceleration is the vector sum of the radial and tangential components.</li></ul></li><li>Applications and examples discuss how a vehicle on a curved path experiences centripetal acceleration and how the speed affects the required radius for a given centripetal acceleration.</li><li>Formulas for circular motion: $ a_{ ext{rad}} = \frac{v^2}{R} = \omega^2 R,\quad v = \omega R, $ $ a = \frac{v^2}{R} = \frac{4\pi^2 R}{T^2}. $</li><li>Uniform circular motion vs. projectile motion: both can have similar magnitudes for certain quantities, but circular motion has a continuously changing direction of acceleration toward the center, while projectile motion has a constant downward vertical acceleration with a horizontal component that remains constant (no circular path).</li><li>{{Key equations}} $ a_{ ext{rad}} = \frac{v^2}{R} = \omega^2 R,\quad v = \omega R,\quad a = \frac{v^2}{R} = \frac{4\pi^2 R}{T^2}. $</li></ul><div data-type="horizontalRule"><hr></div><h4 id="d378b3aa-29eb-4f95-9c54-acca42a24a6b" data-toc-id="d378b3aa-29eb-4f95-9c54-acca42a24a6b" collapsed="false" seolevelmigrated="true">3.5 Relative Velocity</h4><ul><li>Relative velocity describes how velocity depends on the observer’s frame of reference.</li><li>Straight-line (one dimension) relative velocity:<ul><li>If frame A is at rest with respect to the ground and frame B moves with velocity (v{B/A}), then the velocity of a point P relative to A is$ v{P/A} = v{P/B} + v{B/A}. $</li><li>Example: a passenger walking at 3.0 m/s inside a train moving at 60.0 km/h; different observers (on the ground, in the train, in another train) measure different velocities for the same passenger.</li></ul></li><li>General (two or three dimensions) relative velocity (Galilean transformation):<ul><li>If frame B moves with velocity (\mathbf{v}{B/A}) with respect to frame A, then for any particle P:$ \mathbf{v}{P/A} = \mathbf{v}{P/B} + \mathbf{v}{B/A}. $</li><li>In vector form, the velocity of P relative to A equals the velocity relative to B plus the velocity of B relative to A: $ \mathbf{v}{P/A} = \mathbf{v}{P/B} + \mathbf{v}_{B/A}. $</li></ul></li><li>The velocity addition rule reduces to the one-dimensional form when all velocities lie along the same line.</li><li>Einstein’s remark: at high speeds (near c) the simple Galilean addition breaks down; special relativity modifies velocity addition (not needed for these problems but noted).</li><li>Examples emphasize relative velocity in circular or linear tracks, and how to construct velocity vectors from different frames.</li><li>{{Key equations}} $ \mathbf{v}{P/A} = \mathbf{v}{P/B} + \mathbf{v}_{B/A},\quad \text{(Galilean velocity transform)} $</li><li>[[Note on multiple frames]] When three frames A, B, and C are involved: $ \mathbf{v}{P/A} = \mathbf{v}{P/C} + \mathbf{v}{C/B} + \mathbf{v}{B/A}. $</li></ul><div data-type="horizontalRule"><hr></div><h4 id="9e8cc1b8-8ae6-432f-8775-1ea99d631c8a" data-toc-id="9e8cc1b8-8ae6-432f-8775-1ea99d631c8a" collapsed="false" seolevelmigrated="true">3.5 Relative Velocity (continued) – 2D Space</h4><ul><li>Relative velocity in two or three dimensions uses vector addition of velocities from different frames.</li><li>If a passenger walks across a moving car (frames A: ground, B: train) the velocity of the passenger relative to A is $ \mathbf{v}{P/A} = \mathbf{v}{P/B} + \mathbf{v}_{B/A}. $</li><li>The magnitude and direction of the relative velocity can be found from the vector sum (e.g., Pythagorean relation in the right-triangle diagram when the three velocity vectors are perpendicular).</li><li>Example problems show how to compute the velocity of a object relative to ground when given its velocity relative to another frame and the frame’s velocity relative to ground.</li><li>The “relative velocity seen from above” approach yields the same results through vector addition in the plane.</li><li>{{Key equations}} $ \mathbf{v}{P/A} = \mathbf{v}{P/B} + \mathbf{v}{B/A},\quad \mathbf{v}{P/A} \text{ magnitude/angle from components as needed.} $</li></ul><div data-type="horizontalRule"><hr></div><h4 id="29feef77-ac4a-4809-82dd-255eb0ae39c7" data-toc-id="29feef77-ac4a-4809-82dd-255eb0ae39c7" collapsed="false" seolevelmigrated="true">Summary: Core Concepts and Formulas</h4><ul><li>Position and motion: The position vector (\mathbf{r}) has components (x, y, z); displacement is the change in position; velocity is the instantaneous rate of change of position; speed is the magnitude of velocity.</li><li>Velocity is always tangent to the path; instantaneous velocity is the derivative of position with respect to time; its components are time derivatives of coordinates: $ vx = \frac{dx}{dt},\quad vy = \frac{dy}{dt},\quad v_z = \frac{dz}{dt}. $</li><li>Acceleration is the time derivative of velocity; its components are $ ax = \frac{d^2 x}{dt^2},\; ay = \frac{d^2 y}{dt^2},\; a_z = \frac{d^2 z}{dt^2}. $</li><li>The acceleration vector has parallel and perpendicular components relative to velocity:<ul><li>Parallel:$ a_{\parallel} = \frac{dv}{dt}. $</li><li>Perpendicular: for circular motion,$ a_{\perp} = \frac{v^2}{R}. $</li><li>Magnitude:$ |\mathbf{a}| = \sqrt{a{\parallel}^2 + a{\perp}^2}. $</li></ul></li><li>Uniform circular motion:<ul><li>Magnitude of centripetal acceleration:$ a = \frac{v^2}{R} = \omega^2 R = \frac{4\pi^2 R}{T^2}. $</li><li>Direction is toward the center (radial inward).</li></ul></li><li>Projectile motion (no air resistance): horizontal motion with constant velocity and vertical motion with constant acceleration (-g).<ul><li>Equations of motion: $ x(t) = x0 + v{0x} t, \quad vx(t) = v{0x}, $ $ y(t) = y0 + v{0y} t - \frac{1}{2} g t^2, \quad vy(t) = v{0y} - g t. $</li><li>Initial components from speed and launch angle:$ v{0x} = v0 \cos\alpha0,\quad v{0y} = v0 \sin\alpha0. $</li><li>Trajectory (for equal launch/landing heights):$ y = x \tan\alpha0 - \frac{g x^2}{2 v0^2 \cos^2\alpha_0}. $</li><li>Maximum height and range (equal elevations): $ h = \frac{v0^2 \sin^2\alpha0}{2g}, \quad R = \frac{v0^2 \sin(2\alpha0)}{g}. $</li></ul></li><li>The relative velocity concept (Galilean transformation) shows that velocity depends on the frame of reference. For frames A and B moving relative to each other with velocity (\mathbf{v}{B/A}):$ \mathbf{v}{P/A} = \mathbf{v}{P/B} + \mathbf{v}{B/A}.$$
The special-relativity note: the simple vector addition of velocities holds at everyday speeds, but relativity modifies velocity addition at relativistic speeds (not needed for standard mechanics problems but mentioned for completeness).

Practical notes from the chapter:

A curved path does not imply the speed is changing; the acceleration can be entirely perpendicular to velocity (e.g., uniform circular motion).
A curved path with changing speed has both a parallel (speed change) and a perpendicular (direction change) acceleration component.
In projectile motion, the independence of horizontal and vertical motions allows simple solutions and reveals the parabolic trajectory in vacuum.
Relative velocity is a powerful tool for analyzing motion in different frames (e.g., passengers in trains, aircraft with crosswinds, etc.).