Physics for ME · Module 3 of 16
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 Module 2 and Math Module 4. Tick only what you can do closed-notes.
- Resolve a velocity into components (Module 2).
- Differentiate polynomials: v = ds/dt, a = dv/dt (Math Module 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.
The skills, taught in order
Kinematics is the language for describing motion before any force is named. Master these eight skills and the constant-acceleration equations stop being formulas to memorise and become a small toolkit you choose from.
3.1 Position, displacement, and path
Position is a vector from a chosen origin. Displacement is the change in position, a straight arrow from start to finish; distance is the length actually travelled. A full lap of a track covers large distance but gives zero displacement.
3.2 Average and instantaneous velocity
Average velocity is displacement over time, Δs/Δt, the trip meter. Instantaneous velocity is its limit, ds/dt, the speedometer. The same split (average versus instantaneous) carries straight over to acceleration.
3.3 Acceleration and the meaning of its sign
Acceleration is dv/dt. On its own its sign does not mean speeding up or slowing down; compare it with the velocity. Same sign as v means speeding up, opposite sign means slowing. A car braking while moving in the +x direction has positive velocity but negative acceleration.
3.4 The constant-acceleration equations
When a is constant, four equations connect the five quantities v₀, v, a, s, and t. Each one leaves a different quantity out; pick the equation missing the variable you neither know nor want.
| Equation | Leaves out | Reach for it when |
|---|---|---|
| v = v₀ + at | s | you want the final speed after a time |
| s = v₀t + ½at² | v | you want distance after a time |
| v² = v₀² + 2as | t | you want speed after a distance (braking) |
| s = ½(v₀ + v)t | a | acceleration is unknown but both speeds are known |
3.5 Free fall
Near Earth, gravity gives every object the same downward acceleration g = 9.81 m/s², regardless of mass, once air is ignored. Free fall is simply the constant-acceleration equations with a = g pointing down.
Micro-example. Drop a tool from 20 m: t = √(2h/g) = √(40/9.81) = 2.02 s, striking at v = gt = 19.8 m/s.
3.6 Projectile motion
A projectile is two motions at once: constant-velocity horizontal and free-fall vertical, sharing only the clock. Gravity edits the vertical story alone. Worked Example 1 carries this through end to end.
3.7 Motion graphs
Read downward by slope and upward by area. The slope of the position graph is velocity; the slope of velocity is acceleration. Going the other way, the area under an acceleration graph is the change in velocity, and the area under a velocity graph is the distance.
Micro-example. A velocity-time graph that holds 25 m/s for 0.8 s then ramps to zero encloses an area equal to the stopping distance, which is exactly how Worked Example 2 reads it.
3.8 Three-dimensional motion
Nothing new, just one more axis. Each of x, y, z obeys its own 1D equations and the position is (x(t), y(t), z(t)). The vector bookkeeping from Module 2 keeps the three accounts straight.
Engineering connection: Dynamics and machine-motion analysis build directly on this idea. This module is the physical home of Math Module 4 (slopes) and its integral partner (areas).
Worked example 1: 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 (Module 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.
Worked example 2: total stopping distance
A car travels at 25 m/s (90 km/h). The driver takes 0.8 s to react, then the brakes decelerate the car at 7 m/s². Find the total stopping distance, and compare it with the same stop started at 30 m/s.
- ProblemFind the total stopping distance from 25 m/s with a 0.8 s reaction time and 7 m/s² braking, then repeat from 30 m/s.
- Given / findv₀ = 25 m/s, reaction time tr = 0.8 s, braking a = −7 m/s². Find total distance d = d₁ + d₂.
- AssumptionsConstant speed during the reaction, constant deceleration while braking, dry level road, straight line.
- ModelTwo phases in sequence: reaction at constant velocity, then constant-deceleration braking to rest. Total distance is their sum.
- Equationsd₁ = v₀ tr v² = v₀² + 2a d₂ → d₂ = v₀² / (2|a|)
- Solved₁ = 25 × 0.8 = 20 m. d₂ = 25² / (2 × 7) = 625/14 = 44.6 m. Total = 64.6 m. At 30 m/s: d₁ = 24 m, d₂ = 900/14 = 64.3 m, total = 88.3 m.
- CheckBraking distance scales as v²: (30/25)² = 1.44, and 64.3/44.6 = 1.44 as it should. Reaction distance scales only with v. The faster case needs 37% more total distance for 20% more speed.
- ConclusionOnly the reaction part grows in step with speed; the braking part grows with its square, which is why following distances and speed limits are set the way they are. That v² is the same kinetic-energy factor you meet in Module 6.
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 Module 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.