Particle Kinematics

FE Fundamentals of Engineering Exam

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Particle Kinematics

Key Takeaways Explain This Topic AI

Kinematics describes motion geometrically; position s, velocity v = ds/dt, and acceleration a = dv/dt are linked by calculus, with v dv = a ds when time is absent.

The four constant-acceleration equations (v = v₀ + at, s = s₀ + v₀t + ½at², v² = v₀² + 2aΔs, s = s₀ + ½(v₀+v)t) apply ONLY when a is constant.

Projectile motion decouples into constant-velocity horizontal motion (ax = 0) and free-fall vertical motion (ay = −g = −9.81 m/s²).

Range on level ground is R = v₀²sin(2θ)/g, maximized at θ = 45°; maximum height is H = (v₀ sin θ)²/(2g).

Curvilinear motion splits acceleration into tangential at = dv/dt (changes speed) and normal an = v²/ρ (changes direction, points to the center).

Relative velocity adds as vectors: v_B = v_A + v_{B/A}; this handles two bodies moving simultaneously.

Last updated: June 2026

FE Exam Weight: Dynamics is 9–14 questions (about 10%) of the 110-question FE Other Disciplines exam. It has two branches: kinematics (describing motion) and kinetics (relating forces to motion). Every formula here is in the searchable NCEES FE Reference Handbook, so success depends on choosing the right relation and substituting cleanly under roughly 3 minutes per problem.

Rectilinear Motion

Motion along a straight line is governed by three calculus relationships: v = d s d t , a = d v d t = d 2 s d t 2 , v d v = a d s v = \frac{ds}{dt}, \qquad a = \frac{dv}{dt} = \frac{d^2s}{dt^2}, \qquad v\,dv = a\,ds v = d t d s ​ , a = d t d v ​ = d t 2 d 2 s ​ , v d v = a d s

The last form, v dv = a ds , eliminates time and is the right tool whenever a problem relates speed to position without mentioning time.

The Constant-Acceleration Equations

When acceleration is constant, these four equations cover everything: v = v 0 + a t v = v_0 + at v = v 0 ​ + a t s = s 0 + v 0 t + 1 2 a t 2 s = s_0 + v_0 t + \tfrac{1}{2}at^2 s = s 0 ​ + v 0 ​ t + 2 1 ​ a t 2 v 2 = v 0 2 + 2 a ( s − s 0 ) v^2 = v_0^2 + 2a(s - s_0) v 2 = v 0 2 ​ + 2 a ( s − s 0 ​ ) s = s 0 + 1 2 ( v 0 + v ) t s = s_0 + \tfrac{1}{2}(v_0 + v)\,t s = s 0 ​ + 2 1 ​ ( v 0 ​ + v ) t

Free fall is the most common special case: a = g = 9.81 m/s² (32.2 ft/s²) directed downward. Pick a sign convention (e.g., up positive) and apply it consistently to every term.

Trap: These four equations are valid only for constant acceleration . If a depends on time, velocity, or position, you must integrate: a = f(t) → integrate twice; a = f(v) → separate v dv = a dv path; a = f(s) → integrate v dv = a(s) ds.

Projectile Motion

Neglecting air resistance, a projectile experiences gravity alone. The motion decouples into independent horizontal and vertical parts that share only the time t:

• Component Acceleration Velocity Position
• Horizontal x 0 vx = v₀ cos θ (constant) x = (v₀ cos θ)t
• Vertical y −g vy = v₀ sin θ − gt y = (v₀ sin θ)t − ½gt²

Key level-ground results derived from these:

Maximum height: H = (v₀ sin θ)²/(2g) (occurs when vy = 0).

Range: R = v₀² sin(2θ)/g, maximized at θ = 45° .

Time of flight: T = 2v₀ sin θ / g.

Worked Example — Projectile Range

A ball launches at v₀ = 50 m/s, θ = 30°, on level ground (g = 9.81 m/s²). R = v 0 2 sin ⁡ 2 θ g = 50 2 sin ⁡ 60 ° 9.81 = 2500 ( 0.866 ) 9.81 = 2165 9.81 = 220.7 m R = \frac{v_0^2 \sin 2\theta}{g} = \frac{50^2 \sin 60°}{9.81} = \frac{2500(0.866)}{9.81} = \frac{2165}{9.81} = 220.7\text{ m} R = g v 0 2 ​ sin 2 θ ​ = 9.81 5 0 2 sin 60° ​ = 9.81 2500 ( 0.866 ) ​ = 9.81 2165 ​ = 220.7 m

Maximum height: H = (50 sin 30°)²/(2·9.81) = (25)²/19.62 = 625/19.62 = 31.9 m. Note that θ = 30° and θ = 60° give the same range (sin 2θ is identical) but different heights — a classic distractor.

• Curvilinear Motion
• Normal–Tangential (n–t) Components

When a particle follows a curved path, split acceleration into the direction of motion and perpendicular to it: a t = d v d t ( changes the speed ) , a n = v 2 ρ ( changes the direction ) a_t = \frac{dv}{dt}\;(\text{changes the speed}), \qquad a_n = \frac{v^2}{\rho}\;(\text{changes the direction}) a t ​ = d t d v ​ ( changes the speed ) , a n ​ = ρ v 2 ​ ( changes the direction )

The normal component an always points toward the center of curvature (centripetal), and ρ is the radius of curvature. The total acceleration magnitude is a = √(at² + an²).

For circular motion at constant speed , at = 0 and only an = v²/r remains — pure centripetal acceleration. This is the form behind banked-curve and orbit questions.

Polar (r–θ) Components

For motion described by radius and angle: v r = r ˙ , v θ = r θ ˙ , a r = r ¨ − r θ ˙ 2 , a θ = r θ ¨ + 2 r ˙ θ ˙ v_r = \dot{r}, \quad v_\theta = r\dot\theta, \qquad a_r = \ddot{r} - r\dot\theta^2, \quad a_\theta = r\ddot\theta + 2\dot{r}\dot\theta v r ​ = r ˙ , v θ ​ = r θ ˙ , a r ​ = r ¨ − r θ ˙ 2 , a θ ​ = r θ ¨ + 2 r ˙ θ ˙

The 2ṙθ̇ term is the Coriolis acceleration — easy to forget and a frequent trap on cam/slotted-arm problems.

Relative Motion

When two bodies move at once, velocities and accelerations combine as vectors : v ⃗ B = v ⃗ A + v ⃗ B / A , a ⃗ B = a ⃗ A + a ⃗ B / A \vec{v}_B = \vec{v}_A + \vec{v}_{B/A}, \qquad \vec{a}_B = \vec{a}_A + \vec{a}_{B/A} v B ​ = v A ​ + v B / A ​ , a B ​ = a A ​ + a B / A ​

Worked Example — Relative Velocity

Car A drives north at 60 km/h; car B drives east at 80 km/h. The velocity of B relative to A is v_{B/A} = v_B − v_A. The components are 80 east and 60 south (subtracting A's north), so: ∣ v B / A ∣ = 80 2 + 60 2 = 6400 + 3600 = 10,000 = 100 km/h |v_{B/A}| = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10{,}000} = 100\text{ km/h} ∣ v B / A ​ ∣ = 8 0 2 + 6 0 2 ​ = 6400 + 3600 ​ = 10 , 000 ​ = 100 km/h

directed southeast relative to A. Always subtract as vectors — never simply subtract magnitudes.

Choosing the Right Kinematics Tool

The FE exam rewards quickly recognizing which relationship to use:

• Given / asked Use
• Constant a, need v–t or s–t The four constant-acceleration equations
• a as a function of time Integrate a(t) → v(t) → s(t)
• a or v related to position, no time v dv = a ds
• Projectile, level or sloped Decoupled x (constant v) and y (free fall)
• Circular / curved path n–t components: an = v²/ρ, at = dv/dt
• Two bodies moving at once Relative motion: v_B = v_A + v_{B/A}
• Worked Example — Variable Acceleration

A particle starts from rest with acceleration a = 6t (m/s²). Find its velocity and position at t = 4 s. Integrating a: v = ∫6t dt = 3t² = 3(16) = 48 m/s . Integrating again: s = ∫3t² dt = t³ = 4³ = 64 m . Because acceleration is not constant, the constant-a equations would give wrong answers here — this is the single most common kinematics trap. Always confirm a is constant before reaching for v = v₀ + at.