Physics for ME · Chapter 3 of 16 · Beginner
Kinematics: Motion in 1D, 2D, and 3D
Describe motion before explaining it: position, velocity, and acceleration as functions of time, with projectiles as the classic 2D case.
Readiness check
From Chapter 2 and Math Chapter 4. Tick only what you can do closed-notes.
- Resolve a velocity into components (Chapter 2).
- Differentiate polynomials: v = ds/dt, a = dv/dt (Math Chapter 4).
- Solve quadratic equations reliably.
- Sketch parabolas and straight lines from equations.
- Keep track of signs with a declared positive direction.
The core idea
Velocity is the slope of position; acceleration is the slope of velocity. Everything else is bookkeeping.
v = v₀ + ats = v₀t + ½at²v² = v₀² + 2asThe three constant-acceleration equations cover an enormous share of practice. In 2D, the components live separate lives: gravity touches only the vertical one, which is the whole secret of projectile motion.
What this chapter covers
- 3.1 Position, displacement, and path: where, versus how far traveled.
- 3.2 Average and instantaneous velocity: trip meter versus speedometer.
- 3.3 Acceleration: the slope of velocity, with sign meaning.
- 3.4 Constant-acceleration equations: the big three and when they apply.
- 3.5 Free fall: g as a constant acceleration downward.
- 3.6 Projectile motion: independent horizontal and vertical stories.
- 3.7 Motion graphs: reading s-t, v-t, a-t like an engineer.
- 3.8 3D motion as components: nothing new, one more axis.
Engineering connection: MIT 8.01 begins exactly here; Dynamics and machine motion analysis build on it.
Worked example: a projectile, fully described
A test launcher fires a ball at v₀ = 25 m/s, 40° above horizontal, from ground level. Find the flight time, the range, and the maximum height.
- ProblemFind flight time, range, and peak height for the launch in Figure 1.
- Given / findv₀ = 25 m/s at 40°; level ground; g = 9.81 m/s².
- AssumptionsNo air drag (Chapter 1 showed drag matters at car scale; for a dense ball over 60 m it is a modest correction); launch and landing at the same height.
- ModelSplit the launch: vx = 25 cos 40° = 19.15 m/s stays constant; vy = 25 sin 40° = 16.07 m/s fights gravity.
- Equationst = 2vy/g R = vxt h = vy²/2g
- Solvet = 2(16.07)/9.81 = 3.28 s. R = 19.15 × 3.28 = 62.7 m. h = 16.07²/(2 × 9.81) = 13.2 m.
- CheckRange formula cross-check: R = v₀² sin 80°/g = 625 × 0.985/9.81 = 62.7 m. Symmetry: peak at t/2 = 1.64 s, when vy = 16.07 − 9.81(1.64) ≈ 0.
- ConclusionTwo independent 1D problems solved one 2D flight. Sprinkler design, material chutes, and sports machines all run this exact decomposition.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Velocity confused with acceleration | "It moves, so it accelerates"; v = 0 read as a = 0 | "Is the speedometer reading changing?" | At a projectile's peak vy = 0 while a = −g. The slope of v, not v itself. |
| Constant-acceleration formulas everywhere | The big three applied to drag or spring motion | "Is a actually constant here?" | Check the force story first. Varying a needs the ODE tools. |
| Gravity given to the horizontal axis | Range computed with decelerating vx | "Which component does gravity touch?" | Only the vertical. Horizontal velocity is constant without drag. |
| Signs improvised mid-solution | Negative times, upward g | "Which direction did I declare positive?" | Declare once (up positive, g = −9.81), keep it, and let the algebra carry the signs. |
Practice ladder
A cart accelerates from rest at 2.5 m/s² for 6 s. Find its final speed and distance covered.
Show answer
v = 15 m/s; s = ½ × 2.5 × 36 = 45 m.
A machine part is dropped from a 20 m gantry. How long until impact, and how fast does it hit?
Show answer
t = √(2h/g) = √(40/9.81) = 2.02 s; v = gt = 19.8 m/s (about 71 km/h). The v² = 2gh route gives the same speed: dropped tools are projectiles.
A conveyor launches packages horizontally at 3 m/s from a height of 1.2 m onto a lower belt. Where must the lower belt's catch point be placed?
Show answer
Fall time: t = √(2 × 1.2/9.81) = 0.495 s. Horizontal travel: 3 × 0.495 = 1.48 m from the discharge point. Horizontal launch means vy0 = 0; the two axes never mix.
Film a thrown or launched object (ball, water jet, chalk) against a known background scale. Extract three or four positions frame by frame, estimate v₀ and the launch angle, and predict the range; compare with the landing you filmed.
What good work looks like
A position table from frames, fitted v₀ and θ, predicted versus observed range with percent error, and one sentence on the largest error source (frame rate, scale, drag).
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Build a small projectile calculator (spreadsheet or Python): inputs v₀, θ, launch height; outputs flight time, range, peak height, and the trajectory plot. Validate against the worked example (3.28 s, 62.7 m, 13.2 m) and the Level 3 conveyor case.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Define velocity and acceleration as derivatives, with units.
v = ds/dt (m/s); a = dv/dt (m/s²): slopes of the position and velocity curves.
2. Write the three constant-acceleration equations and their precondition.
v = v₀ + at; s = v₀t + ½at²; v² = v₀² + 2as: valid only while a is constant.
3. What is true at the top of a projectile's flight?
vy = 0 but a = −g still; vx is unchanged throughout.
4. How do you read distance from a v-t graph, and velocity change from an a-t graph?
As areas under the curves: the integral link from Math Chapter 5.
5. Why do the two projectile axes decouple?
Gravity has no horizontal component, so each axis obeys its own 1D equations, sharing only the time.
Textbook mapping
| Item | Mapping |
|---|---|
| Main source | OpenStax University Physics Vol. 1, Motion Along a Straight Line and Motion in Two and Three Dimensions |
| Benchmark | MIT 8.01 Classical Mechanics (kinematics opens the course) |
| Core topics | 3.1 Position and path · 3.2 Velocity · 3.3 Acceleration · 3.4 Constant-a equations · 3.5 Free fall · 3.6 Projectiles · 3.7 Motion graphs · 3.8 3D as components |
| Engineering connection | Dynamics, cam and mechanism motion, machine cycle analysis. |
| Read next | Chapter 4: Newton's Laws and Force Models. |