3.3 Dynamics (Kinematics & Kinetics)

Kinematics — Describing Motion

Kinematics describes motion (position, velocity, acceleration) without regard to forces. For constant acceleration a, the NCEES FE Reference Handbook lists three equations every FE Civil candidate must apply fluently:

• v = v₀ + a·t
• s = s₀ + v₀·t + ½·a·t²
• v² = v₀² + 2·a·(s − s₀)

Here v is velocity (ft/s or m/s), a is acceleration (ft/s² or m/s²), s is position, and t is time. Pick the equation that omits the variable you don't have. Velocity is the time derivative of position, v = ds/dt, and acceleration is a = dv/dt; when acceleration is not constant you must integrate or differentiate rather than use the three formulas above.

Projectile motion

A projectile combines constant horizontal velocity (no horizontal force, neglecting air resistance) with vertical free fall at the gravitational acceleration g = 32.2 ft/s² (9.81 m/s²):

• Horizontal: x = v₀·cos θ · t
• Vertical: y = v₀·sin θ · t − ½·g·t²

The two motions share only time t. At the peak, vertical velocity is zero. A classic trap is forgetting that horizontal velocity stays constant throughout the flight.

The time of flight, range, and maximum height all follow from these two equations. For a projectile launched and landing at the same elevation, the maximum height is H = (v₀ sin θ)²/(2g) and the horizontal range is R = v₀² sin(2θ)/g, which is maximized at a launch angle of 45°. On the FE Civil exam these surface in transportation (vehicle dynamics) and water-jet contexts, so recognizing the underlying projectile model is as important as the algebra.

Kinetics — Newton's Second Law

Kinetics relates forces to motion through Newton's second law:

• ΣF = m·a

where m is mass (slugs in US Customary units, kg in SI) and a is acceleration. A pervasive units trap: weight W = m·g, so mass in slugs = W / 32.2 when weight is in pounds. Never plug a weight in pounds directly in for mass.

Friction during motion

Once a body slides, kinetic friction F = μk·N acts opposite the velocity. On an incline of angle θ, a block sliding down has net force m·a = W·sin θ − μk·W·cos θ, giving a = g(sin θ − μk·cos θ).

Circular & rotational motion

For a body moving in a circle of radius r, the inward (centripetal) acceleration is a_n = v²/r, requiring a net inward force F = m·v²/r. Rotational kinematics mirror the linear forms with angular quantities: ω = ω₀ + α·t and θ = ω₀·t + ½·α·t², where ω is angular velocity (rad/s) and α is angular acceleration (rad/s²). Tangential speed relates to angular speed by v = r·ω, and tangential acceleration by a_t = r·α. The total acceleration of a point on a rotating body combines the tangential component (a_t, changing speed) and the normal/centripetal component (a_n = v²/r = r·ω², changing direction) at right angles, so its magnitude is √(a_t² + a_n²).

Energy & Momentum Methods

Two powerful shortcuts avoid solving for acceleration and time.

Work-energy principle

The net work done on a body equals its change in kinetic energy:

• W_net = ΔKE = ½·m·v₂² − ½·m·v₁²

Work by a constant force is W = F·d·cos θ. Gravitational potential energy is PE = m·g·h; a spring stores ½·k·x². When only conservative forces act, mechanical energy is conserved: KE₁ + PE₁ = KE₂ + PE₂.

Impulse-momentum

Linear momentum is p = m·v. The impulse-momentum theorem:

• F·Δt = m·v₂ − m·v₁

If no net external impulse acts, momentum is conserved — the basis for collision problems: m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'. The coefficient of restitution e (0 ≤ e ≤ 1) characterizes the collision: e = 1 is perfectly elastic (kinetic energy conserved) and e = 0 is perfectly plastic (the bodies stick together and move with a common velocity). It is defined as the ratio of relative separation velocity to relative approach velocity, e = (v₂' − v₁')/(v₁ − v₂). Momentum is always conserved in a collision, but kinetic energy is conserved only when e = 1.

Worked Example — Stopping Distance

A car travels at v₀ = 60 ft/s and decelerates at a = −10 ft/s². How far does it travel before stopping (v = 0)?

Use v² = v₀² + 2a·s with v = 0:

• 0 = (60)² + 2(−10)·s → 0 = 3600 − 20·s → s = 3600/20 = 180 ft

Alternatively by work-energy: the friction work ½·m·v₀² is dissipated; the mass cancels, confirming the result depends only on speed and deceleration — a useful check.

A second worked check by impulse-momentum: if the deceleration came from a constant braking force F over time Δt, then F·Δt = m·Δv, and the time to stop is Δt = v₀/|a| = 60/10 = 6 s. Choosing the right method — kinematics for distance, work-energy when forces and distances are known, impulse-momentum when forces and times are known — is the real skill the exam tests, not any single formula.