CELE Engineering Mechanics — Dynamics: KinematicsSummary

Position (s, x, y) is the location of a body measured from a reference point. Velocity (v) is the rate of change of position: v = ds/dt. Acceleration (a) is the rate of change of velocity: a = dv/dt = v(dv/ds). These are vector quantities; direction matters. In rectilinear motion, we use sign convention: typically rightward or downward is positive. The instantaneous relationships are: v = ds/dt and a = dv/dt. The combination a = v(dv/ds) is useful when acceleration is given as a function of position.

When acceleration is constant, three fundamental equations relate velocity (v), initial velocity (v₀), acceleration (a), time (t), and displacement (s): (1) v = v₀ + at, (2) s = v₀t + ½at², (3) v² = v₀² + 2as. These equations are derived by integrating a = constant. Each equation omits one of the five variables; choose the equation that omits the variable you don't know or need. For example, if time is unknown and not required, use v² = v₀² + 2as. Sign convention is critical: if a body decelerates, a is negative in the chosen positive direction.

Free fall is rectilinear motion under gravity alone (no air resistance). The acceleration is a = g = 9.81 m/s² downward (often approximated as 10 m/s² for quick calculations). If we take downward as positive, then a = +9.81 m/s². If we take upward as positive, then a = −9.81 m/s². A common scenario: an object is dropped (v₀ = 0) from height h; find the time to hit the ground and the impact velocity. Using s = h = ½gt², we solve t = √(2h/g), and v = gt = √(2gh). Another scenario: an object is thrown upward with initial speed v₀; it rises to height H = v₀²/(2g) and returns in time t = 2v₀/g.

When acceleration is not constant but is given as a function of time or position—for example, a = f(t) or a = f(s)—the kinematic equations must be replaced by integration. From a = dv/dt, integrate to get v = ∫a(t)dt + v₀. From v = ds/dt, integrate again to get s = ∫v(t)dt + s₀. If acceleration depends on position, a = f(s), use the relation a = v(dv/ds), which separates as v dv = a(s) ds, then integrate both sides: ½(v² − v₀²) = ∫a(s)ds. These problems require calculus skill and appear in advanced scenarios on the PRC examination.

A projectile is an object launched with an initial velocity v₀ at angle θ above the horizontal. The key insight is that horizontal and vertical motions are independent and superposed. Horizontal motion (no acceleration): x = v₀ cos(θ) · t. Vertical motion (constant acceleration g downward): y = v₀ sin(θ) · t − ½gt². These are the parametric equations; eliminate t to get the trajectory equation. Key results (for launch and landing at the same height): time of flight T = (2v₀ sin θ)/g, range R = (v₀² sin 2θ)/g, and maximum height H = (v₀² sin² θ)/(2g). Maximum range occurs at θ = 45°. At that angle, R_max = v₀²/g.

Just as rectilinear motion uses s, v, a, rotational motion uses θ (angle in radians), ω (angular velocity in rad/s), and α (angular acceleration in rad/s²). The equations mirror the rectilinear kinematic equations: (1) ω = ω₀ + αt, (2) θ = ω₀t + ½αt², (3) ω² = ω₀² + 2αθ. Angles must be in radians. A full revolution is 2π radians. Conversion from rpm (revolutions per minute) to rad/s: ω (rad/s) = (2π N)/60, where N is in rpm. For example, 300 rpm = (2π × 300)/60 = 31.42 rad/s. These equations apply when angular acceleration is constant; variable angular acceleration requires integration.

For a point at distance r from the axis of rotation, the linear quantities are related to angular quantities as follows: (1) arc length s = rθ (where θ is in radians), (2) linear velocity v = rω, (3) tangential acceleration a_t = rα, and (4) normal (centripetal) acceleration a_n = v²/r = rω². The tangential acceleration is due to the change in speed; the normal acceleration is due to the change in direction. The total acceleration of the point is a = √(a_t² + a_n²), directed toward the center plus the tangential direction. These relationships allow us to convert between linear and rotational descriptions of motion.

In kinematics, the choice of positive direction is arbitrary but must be consistent throughout a problem. For rectilinear motion, typically rightward or upward is chosen as positive. Once chosen, all quantities (s, v, a) must be signed accordingly. For example, if rightward is positive and a car is moving leftward, then v is negative. If the car is slowing down while moving leftward, the acceleration is rightward (positive). The key is: acceleration has the same sign as the change in velocity, not necessarily the direction of motion. For rotational motion, the right-hand rule is standard: curl your right hand's fingers in the direction of rotation, and your thumb points in the positive angular direction. Sign errors are among the most common mistakes on PRC examinations.

For a projectile launched horizontally from height h, the time to impact is found from y = −½gt²: t = √(2h/g). The horizontal distance traveled is x = v₀ cos(θ) · t. The vertical velocity at impact is v_y = −gt (negative because downward). The horizontal velocity remains constant: v_x = v₀ cos(θ). The magnitude of velocity at impact is v = √(v_x² + v_y²), and the angle below horizontal is β = arctan(|v_y|/v_x). For projectiles launched at angle θ and landing at the same height, the time of flight is T = (2v₀ sin θ)/g, and the velocity at impact has the same magnitude as the launch velocity but is directed downward at angle θ below the horizontal.

For a projectile launched at angle θ with initial speed v₀, eliminating time from the parametric equations x = v₀ cos(θ) · t and y = v₀ sin(θ) · t − ½gt² gives the trajectory equation: y = x tan(θ) − (gx²)/(2v₀² cos²(θ)). This is a parabola. The maximum height occurs when v_y = 0, which happens at time t = (v₀ sin θ)/g. Substituting back: H = (v₀² sin² θ)/(2g). The projectile reaches this height at horizontal distance x = (v₀² sin θ cos θ)/g = (v₀² sin 2θ)/(2g), which is half the range. For a projectile to clear an obstacle of height h at horizontal distance d, the trajectory must satisfy y(d) ≥ h. This often requires solving a quadratic equation in the initial angle or speed.

Position s, velocity v, and acceleration a are related by successive differentiation: a = dv/dt = d²s/dt². Conversely, integrating acceleration gives velocity, and integrating velocity gives position. For constant acceleration, applying this chain yields the three kinematic equations. For variable acceleration, explicit integration is required. This relationship is foundational to all kinematics and connects to calculus fundamentals.

Rotational kinematics mirrors rectilinear kinematics: θ ↔ s, ω ↔ v, α ↔ a, with the same functional forms for equations. The linear motion of a point at radius r is linked to the angular motion by v = rω, a_t = rα, a_n = rω². Understanding this analogy allows students to apply the same problem-solving strategy to both types of motion.

Projectile motion is a superposition of two independent rectilinear motions: uniform horizontal motion and uniformly accelerated vertical motion. The two are linked by a common time parameter t. Analyzing projectile problems separately in x and y directions simplifies the solution.

The kinematic equations (v = v₀ + at, etc.) are valid only for constant acceleration. When acceleration depends on time, position, or velocity, the differential form a = dv/dt or a = v(dv/ds) must be integrated. Recognizing whether a problem involves constant or variable acceleration is the first critical step.

All kinematic quantities—position, velocity, and acceleration—are signed relative to the chosen positive direction. A positive acceleration does not always mean the body is speeding up; it means the velocity is increasing in the positive direction. Deceleration (slowing down) occurs when acceleration opposes velocity, which means a and v have opposite signs.

In problems where time is unknown or irrelevant, the equation v² = v₀² + 2as provides a direct relationship between velocity and position without time. This is derived from combining the three kinematic equations and is useful for 'distance to stop' problems or maximum range calculations.

A vehicle traveling at speed v₀ brakes with constant deceleration a. The stopping distance is found from v² = v₀² + 2as with v = 0: s = v₀²/(2a). For example, a car traveling at 30 m/s (108 km/h) with deceleration a = 6 m/s² requires stopping distance s = 900/12 = 75 m. This calculation is essential for road design, highway markings, and traffic safety analysis. The NSCP 2015 and local traffic codes reference such calculations for intersection design and sight distance requirements.

When explosives are used in construction (e.g., rock blasting), fragments and debris are ejected as projectiles. Engineers must calculate the range and impact location to establish safe exclusion zones. If a blast ejects debris at v₀ = 40 m/s at angle θ = 35°, the range is R = (1600 × sin 70°)/(9.81) ≈ 153 m. Proper analysis prevents injuries and property damage.

Construction equipment often undergoes rotational and linear motion. A crawler excavator bucket rotates at ω = 20 rpm while the boom extends. Converting ω = (2π × 20)/60 = 2.09 rad/s. If the bucket is at radius r = 3 m from the pivot, the linear speed is v = 2.09 × 3 = 6.27 m/s, and the centripetal acceleration is a_n = 6.27²/3 = 13.1 m/s². These calculations inform power requirements, structural stress analysis, and operator safety training.

During construction, tools or materials may accidentally fall from height h. The time to impact is t = √(2h/g), and the impact velocity is v = √(2gh). For h = 20 m, t = √(40/9.81) ≈ 2.02 s and v = √(2 × 9.81 × 20) ≈ 19.8 m/s. This analysis informs safety protocols, protective device design, and incident investigation.

A ferry crossing a river with current velocity v_c must maintain course. If the ferry's speed relative to water is v_f at angle θ to the current, the resultant motion is projectile-like in 2D. Analysis ensures safe crossing, minimizes drift, and optimizes travel time. This is a vector addition problem solved using kinematic principles.

Elevator design involves constant acceleration phases for comfort and energy efficiency. An elevator accelerates at a = 1.5 m/s² for time t₁, then moves at constant velocity, then decelerates at a = −1.5 m/s² for time t₂. Total distance and time calculations use kinematic equations. These parameters affect passenger comfort (humans typically tolerate 0.1g ≈ 0.98 m/s² without discomfort) and energy consumption.

A mass sliding down a slope with constant acceleration a can be analyzed using kinematic equations to predict how long it takes to reach the base and at what speed. For slope angle θ and coefficient of friction μ, a = g(sin θ − μ cos θ). Analysis of velocity and displacement helps assess hazard zones and design mitigation measures (e.g., retaining walls, drainage systems).

Pollutant or smoke propagation in a tunnel involves motion at constant velocity (in steady-state) or variable velocity (during transient phases). Kinematics predicts concentration distribution and evacuation times. Design of tunnel ventilation systems relies on these calculations.

A car's acceleration profile (a as a function of time or speed) determines fuel consumption and performance. If a car accelerates from 0 to 100 km/h (27.8 m/s) in 8 seconds with roughly constant acceleration, a ≈ 3.48 m/s², and the distance covered is s = ½at² = ½(3.48)(64) ≈ 111 m. This analysis is used in traffic flow models, emissions calculations, and road design.

Industrial machinery (pumps, generators, compressors) in civil engineering plants (water treatment, power plants) starts and stops with angular acceleration. A motor accelerates from rest to 1500 rpm in 30 seconds: α = (2π × 1500)/(60 × 30) = 5.24 rad/s². During startup, transient stresses develop in shafts and bearings; kinematic analysis predicts them.