Kinematics and kinetics
Kinematics describes motion without asking what causes it: position, velocity and acceleration, and how they change with time. Kinetics adds the cause, linking the forces on a body to the motion they produce through Newton's second law, force equals mass times acceleration. Civil engineers meet both whenever something moves: a vehicle braking on a bridge deck, a crane hoisting a load, a pile hammer falling, or a rock block rolling down a slope towards a barrier.
For straight-line motion with constant acceleration, three equations connect initial speed u, final speed v, acceleration a, distance s and time t. They are the same equations behind stopping sight distance on roads, where the braking distance is the distance needed to bring the vehicle from design speed to rest at a deceleration set by tyre-road friction. When acceleration is not constant, the same ideas still hold in calculus form: velocity is the rate of change of position, and acceleration is the rate of change of velocity.
Curved motion adds a centripetal acceleration, v squared over r, directed towards the centre of the curve. That inward acceleration needs an inward force, supplied on a road by side friction and superelevation and on a railway by cant and the flange contact. Projectile motion, used for rockfall trajectories and water jets, splits the motion into a horizontal part with constant speed and a vertical part with constant downward acceleration g.
Kinetics problems follow a fixed routine. Draw the free-body diagram, including the inertia direction you expect, write force equals mass times acceleration along each axis, and solve with the kinematic equations. D'Alembert's principle lets you treat the term m a as a reversed inertia force so that the problem looks like statics; this is exactly how earthquake forces on a building are pictured, as inertia forces at each floor.
Convert km/h to m/s (divide by 3.6) before using any kinematic equation.
v = u + a ts = u t + ½ a t²v² = u² + 2 a sNewton: ΣF = m a; centripetal acceleration a_n = v² / r
Mixing signs: if forward is positive, a braking deceleration must be entered as a negative acceleration.
Problem. A 1 200 kg car travelling at 72 km/h brakes uniformly and stops in 40 m. Find the deceleration, the stopping time and the braking force.
- Convert speed: 72 km/h = 72 / 3.6 = 20 m/s.
- From v² = u² + 2 a s with v = 0: a = −u² / (2 s) = −400 / 80 = −5.0 m/s², a deceleration of 5.0 m/s².
- Time: t = (v − u) / a = (0 − 20) / (−5) = 4.0 s.
- Braking force: F = m a = 1 200 × 5.0 = 6 000 N = 6.0 kN, acting backwards on the car.
Answer. Deceleration 5.0 m/s², time 4.0 s, braking force 6.0 kN
Quick quiz
1Which equation links speed and distance without time?
v² = u² + 2 a s.
2What force keeps a vehicle on a circular path?
An inward (centripetal) force of m v² / r, from friction, superelevation or cant.
3What is D'Alembert's principle used for?
Treating m a as a reversed inertia force so a dynamics problem can be solved like statics.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- F. P. Beer, E. R. Johnston, Vector Mechanics for Engineers: Statics and Dynamics, McGraw Hill, latest edition.
- R. C. Hibbeler, Engineering Mechanics: Statics, Pearson, latest edition.
- S. Timoshenko, D. H. Young, Engineering Mechanics, McGraw Hill, 4th edition, 1956.
Free and official sources
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