Particle Kinematics

Key Takeaways

  • Kinematics describes motion without considering the forces that cause it.
  • Position s, velocity v = ds/dt, and acceleration a = dv/dt are related by calculus.
  • For constant acceleration: v = v₀ + at, s = s₀ + v₀t + ½at², v² = v₀² + 2a(s - s₀).
  • Projectile motion: horizontal velocity is constant (ax = 0), vertical acceleration is g = 9.81 m/s² downward.
  • Curvilinear motion uses normal-tangential (n-t) components: an = v²/ρ (centripetal), at = dv/dt (tangential).
  • Relative motion: vB = vA + vB/A — velocities add as vectors.
Last updated: March 2026

Particle Kinematics

FE Exam Weight: Dynamics accounts for 9-14 questions (~10% of the exam). Kinematics (motion description) and kinetics (forces causing motion) are the two main branches.

Rectilinear Motion (Straight Line)

Fundamental Relationships

v=dsdta=dvdt=d2sdt2vdv=adsv = \frac{ds}{dt} \qquad a = \frac{dv}{dt} = \frac{d^2s}{dt^2} \qquad v \, dv = a \, ds

Constant Acceleration Equations

These are the most commonly used equations on the FE exam:

v=v0+atv = v_0 + at s=s0+v0t+12at2s = s_0 + v_0 t + \frac{1}{2}at^2 v2=v02+2a(ss0)v^2 = v_0^2 + 2a(s - s_0) s=s0+12(v0+v)ts = s_0 + \frac{1}{2}(v_0 + v)t

Gravity: For free fall, a = g = 9.81 m/s² (or 32.2 ft/s²) downward. Choose your sign convention carefully!

Variable Acceleration

When a = f(t): integrate a(t) to get v(t), then integrate v(t) to get s(t). When a = f(v): use v dv = a ds and separate variables. When a = f(s): use v dv = a(s) ds and integrate.

Projectile Motion

A projectile moves under gravity only (air resistance neglected):

ComponentAccelerationVelocityPosition
Horizontal (x)ax = 0vx = v₀ cos θx = (v₀ cos θ)t
Vertical (y)ay = -gvy = v₀ sin θ - gty = (v₀ sin θ)t - ½gt²

Maximum height: H = (v₀ sin θ)² / (2g)

Range (level ground): R = v₀² sin(2θ) / g

Maximum range occurs at θ = 45°.

Time of flight (level ground): T = 2v₀ sin θ / g

Curvilinear Motion

Normal-Tangential (n-t) Components

For a particle moving along a curved path:

at=dvdt(tangential — changes speed)a_t = \frac{dv}{dt} \quad (\text{tangential — changes speed})

an=v2ρ(normal — changes direction; always points toward center of curvature)a_n = \frac{v^2}{\rho} \quad (\text{normal — changes direction; always points toward center of curvature})

a=at2+an2a = \sqrt{a_t^2 + a_n^2}

where ρ is the radius of curvature.

For circular motion at constant speed: at = 0 and an = v²/r (centripetal acceleration).

Polar (r-θ) Components

vr=r˙vθ=rθ˙v_r = \dot{r} \qquad v_\theta = r\dot{\theta} ar=r¨rθ˙2aθ=rθ¨+2r˙θ˙a_r = \ddot{r} - r\dot{\theta}^2 \qquad a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta}

Relative Motion

The velocity of B relative to A: vB=vA+vB/A\vec{v}_B = \vec{v}_A + \vec{v}_{B/A}

Similarly for acceleration: aB=aA+aB/A\vec{a}_B = \vec{a}_A + \vec{a}_{B/A}

Example: Car A travels north at 60 km/h and Car B travels east at 80 km/h. The velocity of B relative to A: vB/A = vB - vA → magnitude = √(60² + 80²) = 100 km/h, directed southeast relative to A.

Test Your Knowledge

A ball is thrown vertically upward at 20 m/s. How high does it go? (g = 9.81 m/s²)

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Test Your Knowledge

A car travels around a circular track of radius 100 m at a constant speed of 20 m/s. What is its centripetal acceleration?

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Test Your Knowledge

A projectile is launched at 30° with v₀ = 50 m/s. What is the horizontal range on level ground? (g = 9.81 m/s²)

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