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Dynamics · Chapter 2 of 10 · Beginner

Kinematics of Particles

Before asking why a body moves, describe how it moves. Position, velocity, and acceleration are linked by calculus, and the right coordinate system makes each problem simple.

Readiness check

This chapter is pure description of motion, built on calculus and vectors. Tick only what you can do closed-notes.

Differentiate and integrate polynomials and trig functions of time.
Resolve a vector into x and y components.
State the difference between speed and velocity.
Recall that acceleration is the rate of change of velocity.
Use radians for angles.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsRefresh derivatives and integrals in vector calculus.

3 or more weak itemsReview Chapter 1 and basic calculus before continuing.

The core idea

Velocity is the time derivative of position and acceleration is the time derivative of velocity; describing motion is choosing the coordinates that make those derivatives easy.

v = ds/dta = dv/dta ds = v dv

Kinematics is geometry in time, with no forces yet. Once you have position as a function of time, velocity and acceleration follow by differentiation; going the other way needs integration and initial conditions. For motion along a curve, splitting acceleration into a part that changes speed and a part that changes direction is what makes curved paths tractable.

The skill works when: you pick coordinates suited to the path and keep the derivative chain s → v → a consistent.

The skill breaks down when: constant-acceleration formulas are used where acceleration varies, or curved motion is treated as if it had no direction-changing acceleration.

The concept. A particle traces a path. Its position vector r locates it, the velocity v is tangent to the path (the derivative of r), and the acceleration is the derivative of v. Everything in kinematics flows from this chain.

The skills, taught in order

Kinematics is a small set of relations applied in whichever coordinates fit the path. Five skills take you from straight-line motion to curved paths in three coordinate systems.

2.1 Rectilinear motion and the calculus chain

For motion along a line, v = ds/dt and a = dv/dt. Eliminating time gives the useful third relation a ds = v dv, which links acceleration, velocity, and position directly when time is not wanted. These three cover any one-dimensional motion, constant acceleration or not.

2.2 Constant acceleration: a special case

When a is constant, integrating the chain gives the familiar formulas. Use them only after confirming the acceleration really is constant.

Relation	Formula	Eliminates
velocity-time	v = v₀ + at	position
position-time	s = s₀ + v₀t + ½at²	velocity
velocity-position	v² = v₀² + 2a(s − s₀)	time

2.3 Curvilinear motion in rectangular coordinates

On a plane, treat x and y independently: a_x and a_y each obey the one-dimensional relations. Projectile motion is the classic case, with a_x = 0 and a_y = −g, so the horizontal and vertical motions are solved separately and share only the time.

2.4 Normal and tangential coordinates

For motion along a known curve, resolve acceleration along the path and perpendicular to it: a_t = dv/dt changes the speed, and a_n = v²/ρ changes the direction, where ρ is the radius of curvature. The total acceleration is their vector sum, a = √(a_t² + a_n²). Even at constant speed a curved path has a_n ≠ 0.

2.5 Polar coordinates

When motion is described by a radius and angle (a rotating arm, an orbit), use r and θ: the velocity has components ṙ and rθ̇, and the acceleration includes the term −rθ̇² (centripetal) and the Coriolis term 2ṙθ̇. Polar coordinates shine whenever the natural description is "distance and direction from a point."

Engineering connection: trajectory design, cam and linkage motion, vehicle cornering, and the velocity and acceleration inputs that Chapter 3 turns into forces.

Worked example 1: projectile range and height

A ball is launched at 20 m/s at 35° above the horizontal from ground level. Find the time of flight, the horizontal range, and the maximum height. Neglect air resistance.

Figure 1. Projectile motion in rectangular coordinates. The horizontal motion is at constant velocity; the vertical motion is constant-acceleration under gravity. The two share only the flight time.

ProblemFind the time of flight, range, and maximum height for the launch in Figure 1.
Given / findv₀ = 20 m/s, θ = 35°, g = 9.81 m/s², launch and landing at ground level. Find t_flight, range, h_max.
AssumptionsNo air resistance, so a_x = 0 and a_y = −g; particle model.
ModelSplit into components: horizontal at constant velocity, vertical at constant acceleration, linked by time.
Equationsv₀ₓ = v₀cosθ, v₀ᵧ = v₀sinθ t_flight = 2v₀ᵧ/g h_max = v₀ᵧ²/2g
Solvev₀ₓ = 20cos35° = 16.38 m/s, v₀ᵧ = 20sin35° = 11.47 m/s. Time of flight = 2(11.47)/9.81 = 2.34 s. Range = v₀ₓ × t_flight = 16.38 × 2.34 = 38.3 m. Maximum height = 11.47²/(2 × 9.81) = 6.71 m.
CheckAt the apex the vertical velocity is zero and time is half the flight, both consistent. The 45° launch would maximise range; 35° gives a respectable but smaller distance, as expected.
ConclusionDecoupling the motion into independent x and y problems is the whole technique. The same split handles any projectile, including launches from a height by changing the vertical initial condition.

Result. Time of flight 2.34 s, range 38.3 m, maximum height 6.71 m.

Worked example 2: acceleration on a curve

A car rounds a curve of radius 120 m at 18 m/s while speeding up at 2.0 m/s². Find the normal and tangential accelerations and the magnitude and direction of the total acceleration.

Figure 2. Normal and tangential acceleration. The tangential part speeds the car up along the path; the normal part, pointing to the centre, changes its direction. Their vector sum is the total acceleration.

ProblemFind a_n, a_t, and the total acceleration for the car in Figure 2.
Given / findρ = 120 m, v = 18 m/s, a_t = 2.0 m/s² (speeding up). Find a_n, a_t, |a|, and its direction.
AssumptionsParticle model on a circular arc of known radius; given values are instantaneous.
ModelUse n-t coordinates: a_n from the speed and radius, a_t given, then combine as perpendicular components.
Equationsa_n = v²/ρ a = √(a_t² + a_n²) tanφ = a_n/a_t
Solvea_n = 18²/120 = 324/120 = 2.70 m/s². a_t = 2.0 m/s². Total a = √(2.0² + 2.70²) = √11.29 = 3.36 m/s², at φ = tan⁻¹(2.70/2.0) = 53.5° from the tangent (toward the centre).
CheckBoth components are a few m/s², a realistic cornering acceleration. The normal part exceeds the tangential, so the total leans toward the centre, as the figure shows.
ConclusionEven modest cornering speeds create significant inward acceleration, which Chapter 3 will turn into the friction force the tyres must supply. If the car held constant speed, a_t would be zero but a_n would remain.

Result. a_n = 2.70 m/s², a_t = 2.0 m/s², total a = 3.36 m/s² at 53.5° from the tangent.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Constant-a formulas always	Wrong answer when a varies	"Is the acceleration actually constant?"	If a depends on time or position, integrate the calculus chain instead.
Forgetting normal acceleration	Curved motion treated as straight	"Is the path curved?"	Any curve has a_n = v²/ρ, even at constant speed.
Coupling x and y in a projectile	Horizontal motion given an acceleration	"What is a_x?"	With no drag, a_x = 0; the components share only time.
Speed versus velocity	Direction change ignored	"Did the direction change too?"	Velocity is a vector; a turn is an acceleration even at constant speed.

Practice ladder

Level 1 · Direct skill

A particle moves with s = 2t³ − 6t (metres, seconds). Find its velocity and acceleration at t = 2 s.

Show answer

v = ds/dt = 6t² − 6 = 18 m/s; a = dv/dt = 12t = 24 m/s². The calculus chain handles variable acceleration directly, where the constant-a formulas could not.

Level 2 · Mixed concept

For the Worked Example 1 ball, what launch angle would double the maximum height while keeping the same speed?

Show answer

h_max ∝ sin²θ, so doubling height needs sin²θ' = 2sin²35°, sinθ' = √2·sin35° = 0.811, θ' = 54.2°. Height is far more sensitive to angle than range near the middle of the range.

Level 3 · Independent problem

A point on a fan blade is 0.25 m from the axis and spins at a constant 1500 rev/min. Find its speed and acceleration.

Show answer

ω = 1500 × 2π/60 = 157.1 rad/s; v = ωr = 39.3 m/s; a_t = 0 (constant speed), a_n = v²/ρ = 39.3²/0.25 = 6170 m/s² ≈ 630g. Rotating machinery generates enormous normal accelerations, which is why balance matters.

Level 4 · Transfer to real engineering

Film or observe a real curved or projectile motion (a thrown ball, a car on a roundabout). Estimate the speed and radius or launch angle, and compute a velocity and an acceleration with a stated assumption.

What good work looks like

The right coordinate system chosen, the calculus chain or n-t relations applied, and a named assumption (no drag, constant speed, constant radius) with its effect on the answer.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check whether constant-acceleration formulas are valid for this problem before I use them."

"Give me five motions; I will pick the best coordinate system for each."

"Find the range." Decoupling the components yourself is the skill.

"What is the acceleration on the curve?" Forming a_n = v²/ρ is the point.

Portfolio task

Analyse one real motion in two coordinate systems where possible (for example a projectile in rectangular, a turn in n-t), and show the velocity and acceleration agree.

Must include: the derivative chain or n-t relations, a stated coordinate choice, and a units and magnitude check.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the three rectilinear relations.

v = ds/dt, a = dv/dt, and a ds = v dv.

2. When may you use v² = v₀² + 2as?

Only when the acceleration is constant.

3. In a drag-free projectile, what are a_x and a_y?

a_x = 0 and a_y = −g; the motions share only time.

4. Give the normal and tangential accelerations.

a_n = v²/ρ (changes direction), a_t = dv/dt (changes speed).

5. When are polar coordinates natural?

When motion is described as a distance and direction from a point, such as a rotating arm or orbit.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-solve the projectile from a blank page.

+3 daysOne n-t problem with a new radius and speed.

+7 daysFeed a velocity and acceleration into kinetics, Chapter 3.

+30 daysReuse these relations for rigid bodies in Chapter 7.

Textbook mapping

Item	Mapping
Primary source	Meriam and Kraige, Engineering Mechanics: Dynamics (7th ed), Chapter 2 (Kinematics of Particles)
Cross-reference	Hibbeler, Dynamics, Ch. 12 · Beer and Johnston, Ch. 11
Core topics	2.1 Rectilinear motion · 2.2 Constant acceleration · 2.3 Rectangular coordinates · 2.4 Normal-tangential · 2.5 Polar coordinates
Engineering connection	Trajectories, cam and linkage motion, and the kinematic inputs to kinetics.
Read next	Chapter 3: Kinetics of Particles, Newton's Second Law.