CELE Engineering Mechanics — Dynamics: KinematicsDetailed Explanation
The Dynamics: Kinematics chapter rewards slow, careful thinking over quick pattern matching, especially on Professional Regulation Commission (PRC) — Board of Civil Engineering's scenario-based CELE items. This detailed explanation walks through the full derivation of every core idea, then links each one to a worked example pulled from recent CELE Engineering Mechanics papers.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Engineering Mechanics section sits under a "Core" weighting, and Dynamics: Kinematics is the 7th chapter in the 8-chapter CELE Engineering Mechanics rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Engineering Mechanics.
Dynamics: Kinematics - Detailed Explanation
Kinematics is the branch of dynamics that describes the geometry of motion — position, displacement, velocity, and acceleration — without reference to the forces that produce it. For the PRC Civil Engineer Licensure Examination, kinematics problems appear consistently in the Engineering Mechanics section and typically account for a significant share of board exam questions in that subject area. Mastery of kinematics is also a prerequisite for the kinetics topics (Newton's second law, work-energy, impulse-momentum) that follow. This chapter covers the three major domains tested on the board exam: (1) rectilinear (straight-line) motion with both constant and variable acceleration, (2) projectile motion, and (3) rotational (angular) kinematics, including the linkage between angular and linear quantities. All problems are worked in SI units consistent with Philippine engineering practice.
Concepts
Rectilinear Motion: Fundamental Definitions and Constant-Acceleration Equations
Rectilinear motion is translation along a straight line. The three primary kinematic quantities are defined by calculus: • Position s (m) — location along the reference line • Velocity v = ds/dt (m/s) — rate of change of position • Acceleration a = dv/dt = v(dv/ds) (m/s²) — rate of change of velocity Displacement Δs = s₂ − s₁ is a vector quantity (signed); distance is the total path length (always positive). When acceleration is constant (the most frequently tested case), the three kinematic equations are derived by integrating a = const: v = v₀ + at … (1) s = v₀t + ½at² … (2) v² = v₀² + 2as … (3) Each equation involves four of the five quantities {v₀, v, a, t, s}. Select the equation that contains the three knowns and the one unknown. For free fall (body released from rest or thrown vertically), set a = g = 9.81 m/s² (positive downward, or assign your own sign convention and be consistent). A body thrown upward decelerates at 9.81 m/s²; at maximum height v = 0. For variable acceleration (a is a function of t, v, or s), integrate directly: v = v₀ + ∫a dt, s = s₀ + ∫v dt or use the chain-rule form a = v(dv/ds) and separate variables.
Examples
Deceleration is substituted as a negative acceleration because it opposes motion (positive direction = forward). Both Eqs. (2) and (3) confirm s = 50 m — cross-checking with a second equation is good exam practice.
Scenario
BOARD-TYPE PROBLEM (Constant Deceleration): A car travels at 20 m/s and brakes uniformly to a complete stop in 5 s. Determine (a) the deceleration and (b) the braking distance.
Solution
Given: v₀ = 20 m/s, v = 0, t = 5 s (a) Using Eq. (1): v = v₀ + at 0 = 20 + a(5) a = −4 m/s² → deceleration = 4 m/s² (b) Using Eq. (2): s = v₀t + ½at² s = 20(5) + ½(−4)(5²) s = 100 − 50 = 50 m Verification using Eq. (3): v² = v₀² + 2as 0 = (20)² + 2(−4)(s) s = 400/8 = 50 m ✔
Taking downward as positive simplifies free-fall problems. Both methods for impact speed agree, confirming correctness. The cliff height 60.1 m ≈ 60 m is a clean answer typical of board problems.
Scenario
BOARD-TYPE PROBLEM (Free Fall): A stone is dropped from a cliff and strikes the ground 3.5 s later. Find (a) the height of the cliff and (b) the impact speed. Use g = 9.81 m/s².
Solution
Define downward as positive. v₀ = 0 (dropped from rest). (a) s = v₀t + ½gt² s = 0 + ½(9.81)(3.5²) s = ½(9.81)(12.25) = 60.1 m (b) v = v₀ + gt = 0 + 9.81(3.5) = 34.3 m/s (downward) Or: v² = 2gs = 2(9.81)(60.1) = 1179.6 → v = 34.3 m/s ✔
For variable acceleration given as a function of time, integrate sequentially: first for velocity, then for position. The integration limits correspond to the time interval of interest.
Scenario
BOARD-TYPE PROBLEM (Variable Acceleration): The acceleration of a particle is a = (6t − 4) m/s², with v₀ = 2 m/s and s₀ = 0 at t = 0. Find v and s at t = 3 s.
Solution
v = v₀ + ∫₀ᵗ a dt = 2 + ∫₀³ (6t − 4) dt = 2 + [3t² − 4t]₀³ = 2 + (27 − 12) − 0 = 2 + 15 = 17 m/s s = s₀ + ∫₀ᵗ v dt First find v(t): v(t) = 2 + 3t² − 4t s = ∫₀³ (2 + 3t² − 4t) dt = [2t + t³ − 2t²]₀³ = (6 + 27 − 18) − 0 = 15 m
Applications
- Braking distance calculations for highway design (DPWH geometric design standards)
- Elevator acceleration/deceleration analysis in building structural systems
- Pile-driving analysis: ram velocity at impact (free fall of drop hammer)
- Train deceleration into stations for transit infrastructure planning
- Determining stopping sight distance in road design per DPWH standards
Misconceptions
- Confusing displacement with distance — displacement is a signed (vector) quantity; distance is scalar and always positive.
- Using a = g = 9.81 m/s² as positive when upward is taken as positive — the correct substitution is a = −9.81 m/s².
- Assuming that when velocity is zero, acceleration is also zero — at the peak of a vertical throw, v = 0 but a = −g ≠ 0.
- Applying the constant-acceleration equations when acceleration is variable — always check whether a is truly constant before using the trio.
- Forgetting the ½ factor in s = v₀t + ½at² — a very common arithmetic error under exam pressure.
Related Concepts
- Kinetics: Newton's second law (ΣF = ma links kinematics to forces)
- Work-Energy theorem (requires knowing velocity as a kinematic output)
- Impulse-Momentum (change in momentum requires velocity from kinematics)
- Projectile motion (kinematics in two dimensions)
- Rotational kinematics (angular analog of rectilinear equations)
Common Exam Questions
Example
Car A starts from rest with a = 2 m/s². Car B, traveling at constant 15 m/s, passes the same point 5 s earlier. When and where does A overtake B?
Approach
Write separate position equations s₁(t) and s₂(t) for each body. Set s₁ = s₂ (meeting) or s₂ − s₁ = constant (given head start) and solve for t. Then substitute back for position or velocity.
Question Type
Two-body meeting or overtaking problems
Example
A ball is thrown upward at 15 m/s from a 20 m high platform. Find time of flight to ground and impact speed.
Approach
Take upward positive. Set up s = v₀t − ½gt². Substitute the final height (negative if below launch point) and solve the resulting quadratic for t. Pick the physically meaningful (positive) root.
Question Type
Vertical throw with different launch and landing heights
Example
From a given v-t graph with linear segments, compute total displacement and identify phases of constant acceleration.
Approach
Acceleration = slope of v-t graph. Displacement = area under v-t graph. Velocity = slope of s-t graph. Use geometry (triangles, trapezoids) to compute areas.
Question Type
Finding acceleration from v-t or s-t graphs
Key Points To Remember
- Choose a positive direction at the start and maintain it throughout the problem.
- Deceleration means the acceleration vector opposes the velocity vector — substitute a negative value into the equations.
- Equation (3), v² = v₀² + 2as, is especially useful when time t is unknown and not required.
- At maximum height for a vertically thrown body, v = 0 instantaneously but a = −g ≠ 0.
- Displacement can be zero even when distance is not (e.g., a ball thrown up and caught at the same height).
- For variable acceleration: if a = f(t), integrate; if a = f(v) or a = f(s), use separation of variables or the chain-rule relation.
- Average velocity = total displacement / total time; average speed = total distance / total time.
Projectile Motion
A projectile is any body given an initial velocity and then left to move under gravity alone (air resistance neglected). The key insight is the independence principle: horizontal and vertical motions are completely independent and are analyzed separately; they share only the elapsed time t. With launch speed v₀ at angle θ above the horizontal and origin at launch point (positive x to the right, positive y upward): Horizontal: aₓ = 0, vₓ = v₀cosθ (constant), x = v₀cosθ · t Vertical: aᵧ = −g, vᵧ = v₀sinθ − gt, y = v₀sinθ · t − ½gt² For launch and landing at the same elevation (y = 0 at landing): Time of flight: T = 2v₀sinθ / g Range: R = v₀²sin2θ / g Maximum height: H = v₀²sin²θ / (2g) Maximum range is at θ = 45° (since sin2θ is maximized at 2θ = 90°). Complementary angles (e.g., 30° and 60°) give the same range. For an elevated launch or target (y ≠ 0 at landing), use the parametric equations directly and solve the resulting quadratic in t. The resultant velocity at any instant: v = √(vₓ² + vᵧ²), direction = arctan(vᵧ/vₓ) below horizontal (during descent).
Examples
Note: sin2θ = sin(80°) ≠ sin²θ. This is the most common error in projectile range problems. Always double-check which formula to apply: sin2θ for range, sin²θ for maximum height.
Scenario
BOARD-TYPE PROBLEM: A ball is launched at v₀ = 30 m/s at θ = 40° from horizontal. Find (a) range, (b) maximum height, and (c) time of flight. Use g = 9.81 m/s².
Solution
(a) R = v₀²sin2θ / g = (30²)(sin80°) / 9.81 = (900)(0.9848) / 9.81 = 886.3 / 9.81 = 90.3 m (b) H = v₀²sin²θ / (2g) = (900)(sin40°)² / (2 × 9.81) = (900)(0.6428)² / 19.62 = (900)(0.4132) / 19.62 = 371.9 / 19.62 = 18.96 m ≈ 18.9 m (c) T = 2v₀sinθ / g = 2(30)(sin40°) / 9.81 = 2(30)(0.6428) / 9.81 = 38.57 / 9.81 = 3.93 s
With a horizontal launch, v₀ᵧ = 0 and the vertical motion is pure free fall. The range formula R = v₀²sin2θ/g cannot be used here because launch and landing heights differ. Use the parametric equations directly.
Scenario
BOARD-TYPE PROBLEM (Elevated target): A projectile is fired horizontally at 25 m/s from the edge of a cliff 45 m high. Find (a) time of flight, (b) horizontal range, and (c) speed at impact.
Solution
θ = 0° (horizontal launch), v₀ₓ = 25 m/s, v₀ᵧ = 0, y₀ = 45 m above ground. Take downward as positive for vertical motion. (a) y = ½gt² → 45 = ½(9.81)t² t² = 90/9.81 = 9.174 → t = 3.03 s (b) x = v₀ₓ · t = 25(3.03) = 75.7 m (c) vᵧ = gt = 9.81(3.03) = 29.7 m/s (downward) v = √(vₓ² + vᵧ²) = √(25² + 29.7²) = √(625 + 882.1) = √1507 = 38.8 m/s
When the target coordinates (x, y) are given and the launch speed is fixed, the substitution tanθ via the identity sec²θ = 1 + tan²θ converts the equation to a quadratic in tanθ, yielding two solutions (low and high trajectory). The smaller angle is the minimum required angle.
Scenario
BOARD-TYPE PROBLEM (Required angle): A football is kicked at 20 m/s and must clear a 3 m crossbar 18 m away. What is the minimum launch angle?
Solution
At x = 18 m, t = x / (v₀cosθ) = 18 / (20cosθ) Height at that time: y = v₀sinθ · t − ½gt² 3 = 20sinθ · [18/(20cosθ)] − ½(9.81)[18/(20cosθ)]² 3 = 18tanθ − ½(9.81)(324/400cos²θ) 3 = 18tanθ − (9.81 × 324)/(800cos²θ) 3 = 18tanθ − 3.97/cos²θ Using the identity 1/cos²θ = 1 + tan²θ: 3 = 18tanθ − 3.97(1 + tan²θ) 3.97tan²θ − 18tanθ + 6.97 = 0 Let u = tanθ: u = [18 ± √(324 − 4 × 3.97 × 6.97)] / (2 × 3.97) u = [18 ± √(324 − 110.7)] / 7.94 u = [18 ± √213.3] / 7.94 = [18 ± 14.61] / 7.94 u₁ = 32.61/7.94 = 4.11 → θ₁ = 76.3° u₂ = 3.39/7.94 = 0.427 → θ₂ = 23.1° The minimum angle is θ = 23.1°.
Applications
- Ballistic trajectory calculations in military and geotechnical blasting engineering
- Design of spillway trajectories (flip-bucket energy dissipators) for dams
- Sports facility design: clearance heights for long-jump pits, javelin runways
- Fluid jets in hydraulic engineering (jet trajectory from an orifice)
- Analysis of falling debris hazard zones in construction site safety (DOLE regulations)
Misconceptions
- Using sin2θ and sin²θ interchangeably — sin2θ = 2sinθcosθ is used for range; sin²θ appears in the height formula.
- Assuming maximum range always occurs at 45° — this is true ONLY for level ground. For elevated or depressed targets, the optimum angle changes.
- Forgetting that at maximum height, the projectile still has horizontal velocity vₓ = v₀cosθ ≠ 0.
- Thinking complementary angles give the same maximum height — they give the same range but different heights.
- Applying the range formula when launch and landing heights differ — always check this condition first.
Related Concepts
- Rectilinear motion (projectile motion is two simultaneous rectilinear motions)
- Relative motion (projectile from a moving vehicle requires adding velocities)
- Newton's second law in kinetics (gravity force mg causes vertical acceleration g)
- Vector decomposition (initial velocity must be resolved into components)
Common Exam Questions
Example
A shell is fired at 600 m/s at 30°. Find the maximum range. Answer: R = (600²)(sin60°)/9.81 = 31,780 m ≈ 31.8 km.
Approach
Apply the three standard formulas directly. Confirm that launch and landing heights are equal before using the closed-form formulas. Convert angle to degrees/radians as required.
Question Type
Find range, height, or time of flight given v₀ and θ
Example
What angle gives maximum range equal to 80% of v₀²/g? Set sin2θ = 0.8, so 2θ = 53.1°, θ = 26.6°.
Approach
Substitute in the range formula and solve for sin2θ or sinθ. If both coordinates (x, y) of the target are given, use the parametric equations and the tanθ substitution to get a quadratic.
Question Type
Find launch angle for a required range or height
Example
A ball thrown at 15 m/s and 30° from a 10 m high roof — find horizontal distance from building base.
Approach
Use x = v₀cosθ · t and y = v₀sinθ · t − ½gt². Set y = Δh (the height difference), solve the quadratic in t, substitute into x equation for range.
Question Type
Projectile launched from or to a different elevation
Key Points To Remember
- Horizontal velocity vₓ = v₀cosθ is constant throughout the flight (no horizontal force).
- Vertical velocity vᵧ = v₀sinθ − gt decreases at rate g; equals zero at maximum height.
- Range formula R = v₀²sin2θ / g applies ONLY when launch and landing heights are equal.
- Complementary launch angles (θ and 90°−θ) produce the same horizontal range.
- Maximum range = v₀²/g, achieved at θ = 45°.
- At maximum height, vᵧ = 0 but vₓ ≠ 0 (the projectile is still moving horizontally).
- For a projectile launched horizontally (θ = 0): vₓ = v₀, vᵧ = −gt, x = v₀t, y = −½gt².
Rotational (Angular) Kinematics and Linkage to Linear Motion
Rotation about a fixed axis is described by three angular quantities that are the exact analogs of the rectilinear set: Angular position: θ (radians, rad) Angular velocity: ω = dθ/dt (rad/s) Angular acceleration: α = dω/dt (rad/s²) For constant angular acceleration α, the equations mirror the rectilinear trio: ω = ω₀ + αt … (R1) θ = ω₀t + ½αt² … (R2) ω² = ω₀² + 2αθ … (R3) Unit conversions (critical for board exam): 1 revolution = 2π radians ω [rad/s] = (2π/60) × N [rpm] = πN/30 Linkage between angular and linear quantities for a point at radius r from the axis: Arc length: s = rθ Tangential (linear) velocity: v = rω Tangential acceleration: aₜ = rα (along the circular path) Normal (centripetal) acceleration: aₙ = v²/r = rω² (directed toward the center) Total linear acceleration: a = √(aₜ² + aₙ²) Note: aₙ is always directed toward the rotation axis (centripetal); it exists even at constant angular velocity (α = 0, ω ≠ 0). aₜ is zero only when α = 0.
Examples
The centripetal acceleration at full speed dominates the tangential acceleration by a factor of ~314. This is characteristic of rotating machinery at operating speed — the centripetal component governs the design of connecting elements (bolts, welds) on rotating parts.
Scenario
BOARD-TYPE PROBLEM: A flywheel accelerates from rest to 300 rpm in 10 s. Find (a) angular acceleration, (b) number of revolutions made, and (c) tangential and centripetal accelerations of a point 0.5 m from the axis at t = 10 s.
Solution
Convert: ω₀ = 0, ω = 300 rpm × (2π/60) = 31.42 rad/s (a) α = (ω − ω₀)/t = (31.42 − 0)/10 = 3.14 rad/s² (b) θ = ω₀t + ½αt² = 0 + ½(3.14)(10²) = 157.1 rad Rev = θ/(2π) = 157.1/6.283 = 25 revolutions (c) At r = 0.5 m, ω = 31.42 rad/s: aₜ = rα = 0.5(3.14) = 1.57 m/s² aₙ = rω² = 0.5(31.42²) = 0.5(987.2) = 493.6 m/s² a_total = √(1.57² + 493.6²) ≈ 493.6 m/s² (aₙ dominates at full speed)
The equation ω² = ω₀² + 2αθ is the direct analog of v² = v₀² + 2as, and is ideal when time is not given. Converting revolutions to radians before substitution is the critical first step.
Scenario
BOARD-TYPE PROBLEM (Deceleration): A motor wheel running at 600 rpm coasts to rest over 40 complete revolutions. Find (a) angular deceleration and (b) time to stop.
Solution
ω₀ = 600 × (2π/60) = 62.83 rad/s, ω = 0 θ = 40 rev × 2π = 251.3 rad (a) Using ω² = ω₀² + 2αθ: 0 = (62.83)² + 2α(251.3) α = −(3947.8)/(502.6) = −7.85 rad/s² Deceleration = 7.85 rad/s² (b) Using ω = ω₀ + αt: 0 = 62.83 + (−7.85)t t = 62.83/7.85 = 8.0 s
In gear trains, the contact (pitch) point has the same linear (tangential) velocity for both gears. The gear ratio inversely relates angular velocities to radii: the larger gear always rotates more slowly.
Scenario
BOARD-TYPE PROBLEM (Gear system): Gear A (radius 80 mm) meshes with Gear B (radius 200 mm). If Gear A rotates at 1200 rpm, find (a) angular velocity of Gear B and (b) linear velocity of the pitch point.
Solution
For meshing gears, the pitch-point velocity is the same for both gears: v = rₐ · ωₐ = r_B · ω_B ωₐ = 1200 × (2π/60) = 125.66 rad/s (a) ω_B = (rₐ/r_B) × ωₐ = (0.08/0.20) × 125.66 = 50.27 rad/s N_B = 50.27 × (60/2π) = 480 rpm [Gear ratio: N_B/Nₐ = rₐ/r_B = 80/200 = 0.4 ✔] (b) v = rₐ · ωₐ = 0.08 × 125.66 = 10.05 m/s
Applications
- Analysis of rotating machinery: turbines, motors, pumps in mechanical and industrial systems
- Gear train design for power transmission in construction equipment (cranes, excavators)
- Belt-and-pulley systems (belt velocity = rω of both pulleys for a non-slipping belt)
- Centrifugal force calculations in rotating structural components (flywheels, shafts)
- Tire/wheel analysis: relating vehicle speed to wheel rpm for road design
- Foundation design for rotating machinery (vibration analysis requires knowledge of ω and α)
Misconceptions
- Using degrees instead of radians in rotational kinematic equations — the equations require θ in radians.
- Thinking centripetal acceleration is zero when the body rotates at constant speed — aₙ = rω² ≠ 0 as long as ω ≠ 0.
- Confusing centripetal (aₙ) and centrifugal — centripetal is the actual inward acceleration; centrifugal is a fictitious outward force concept in a rotating frame.
- Forgetting that aₙ and aₜ are perpendicular — total acceleration a = √(aₙ² + aₜ²), not their sum.
- Not converting rpm to rad/s before calculation — the most common numerical error in rotational problems.
Related Concepts
- Rectilinear kinematics (same equation structure with different variable names)
- Kinetics of rotation (requires rotational inertia I and torque T = Iα)
- Circular motion in projectile and relative motion contexts
- Gear trains and mechanical advantage
- Vibration fundamentals (angular frequency ω = 2πf)
Common Exam Questions
Example
A shaft speeds up from 200 rpm to 800 rpm in 30 s. Find α in rad/s² and number of revolutions made.
Approach
List knowns: ω₀, ω (or N in rpm → convert!), α or t or θ (in rad — convert from rev). Identify which rotational equation contains exactly the three knowns and the desired unknown. Substitute and solve.
Question Type
Find revolutions, angular velocity, or angular acceleration from given data
Example
A 0.3 m radius disk accelerates from rest at α = 5 rad/s². Find aₙ and aₜ of a rim point after 4 s.
Approach
First solve the rotational kinematics problem to get ω and α at the required instant. Then apply v = rω, aₜ = rα, aₙ = rω² at the given radius r.
Question Type
Find linear acceleration components of a point on a rotating body
Example
A 150 mm pulley drives a 450 mm pulley. If the small pulley rotates at 900 rpm, what is the speed of the large pulley?
Approach
For a common belt or meshing gears, set the linear velocity at the contact point equal: r₁ω₁ = r₂ω₂. Solve for the unknown angular velocity, then proceed with kinematics if required.
Question Type
Belt-and-pulley or gear problems
Key Points To Remember
- Rotational equations are structurally identical to rectilinear equations — substitute θ↔s, ω↔v, α↔a.
- Always convert rpm to rad/s before substituting into rotational equations.
- Centripetal acceleration aₙ = rω² = v²/r — present whenever the body moves in a curved path at any speed.
- Tangential acceleration aₜ = rα — present only when angular velocity is changing.
- The angle θ in all rotational kinematic equations must be in radians.
- Number of revolutions N_rev = θ/(2π) where θ is in radians.
- For a decelerating rotor, α is negative; energy methods can also determine stopping revolutions.
Practice Problems
When distance and initial/final velocities are known but time is not, Eq. (3) — v² = v₀² + 2as — is the most direct starting equation. Time is then found from Eq. (1).
Problem
PROBLEM 1 (Rectilinear — Two Bodies): A train accelerates uniformly from 10 m/s to 30 m/s over a distance of 400 m. Find (a) the acceleration and (b) the time for this phase.
Solution
(a) v² = v₀² + 2as (30)² = (10)² + 2a(400) 900 = 100 + 800a a = 800/800 = 1.0 m/s² (b) v = v₀ + at 30 = 10 + (1.0)t t = 20 s Check with s = v₀t + ½at²: s = 10(20) + ½(1.0)(20²) = 200 + 200 = 400 m ✔
When landing height ≠ launch height, substitute y = Δh into the vertical equation to get a quadratic in t. Discard the negative root (physically meaningless — occurs before launch). The impact speed is greater than v₀ because the projectile descends below the launch elevation, gaining speed.
Problem
PROBLEM 2 (Projectile — Elevated Landing): A projectile is fired at v₀ = 50 m/s at θ = 30° from the top of a 20 m high tower. Find (a) the horizontal distance from the base of the tower to the impact point and (b) the speed at impact. Use g = 9.81 m/s².
Solution
Set origin at launch point, upward positive. v₀ₓ = 50cos30° = 43.30 m/s v₀ᵧ = 50sin30° = 25.00 m/s (a) At impact, y = −20 m (20 m below launch point): y = v₀ᵧt − ½gt² −20 = 25t − ½(9.81)t² 4.905t² − 25t − 20 = 0 t = [25 ± √(625 + 4 × 4.905 × 20)] / (2 × 4.905) t = [25 ± √(625 + 392.4)] / 9.81 t = [25 ± √1017.4] / 9.81 = [25 ± 31.90] / 9.81 Take positive root: t = 56.90/9.81 = 5.80 s x = v₀ₓ · t = 43.30 × 5.80 = 251.1 m (b) vₓ = 43.30 m/s (constant) vᵧ = v₀ᵧ − gt = 25 − 9.81(5.80) = 25 − 56.9 = −31.9 m/s v = √(43.30² + 31.9²) = √(1874.9 + 1017.6) = √2892.5 = 53.8 m/s
Notice that aₙ >> aₜ even at half speed. The centripetal acceleration governs the structural integrity of spinning components. Always state the direction of each component: aₙ toward center, aₜ tangential (opposing motion during deceleration).
Problem
PROBLEM 3 (Rotational — Deceleration to Rest): A flywheel decelerates uniformly from 600 rpm to rest in 40 complete revolutions. Find (a) the angular deceleration in rad/s², (b) the time to stop, and (c) the tangential and normal accelerations of a point on the rim (r = 0.4 m) at the instant when ω = 300 rpm.
Solution
Convert: ω₀ = 600 × 2π/60 = 62.83 rad/s; ω = 0 θ = 40 × 2π = 251.33 rad (a) ω² = ω₀² + 2αθ 0 = (62.83)² + 2α(251.33) α = −3947.4 / 502.65 = −7.854 rad/s² Deceleration = 7.854 rad/s² (b) ω = ω₀ + αt 0 = 62.83 + (−7.854)t t = 62.83/7.854 = 8.0 s (c) At ω = 300 rpm = 31.42 rad/s, r = 0.4 m: aₜ = rα = 0.4 × 7.854 = 3.14 m/s² (magnitude; directed opposing rotation) aₙ = rω² = 0.4 × (31.42)² = 0.4 × 987.2 = 394.9 m/s² (toward axis) Total a = √(3.14² + 394.9²) = √(9.86 + 155,925) ≈ 394.9 m/s²
For non-constant acceleration given as f(t), integrate term by term. Apply initial conditions at t = 0 to evaluate the constants of integration. This type of problem appears in the Board Exam when 'classical' kinematics does not apply.
Problem
PROBLEM 4 (Variable Acceleration — Integration): A particle moves along a straight line with acceleration a = (12t² − 6) m/s². At t = 0: v = 4 m/s, s = 2 m. Find the position and velocity at t = 2 s.
Solution
v(t) = v₀ + ∫₀ᵗ a dt = 4 + ∫₀² (12t² − 6) dt = 4 + [4t³ − 6t]₀² = 4 + (32 − 12) − 0 = 4 + 20 = 24 m/s s(t) = s₀ + ∫₀ᵗ v(t) dt v(t) = 4 + 4t³ − 6t s = 2 + ∫₀² (4 + 4t³ − 6t) dt = 2 + [4t + t⁴ − 3t²]₀² = 2 + (8 + 16 − 12) − 0 = 2 + 12 = 14 m
The gear ratio N_A/N_B = ω_B/ω_A — more teeth on the driven gear means lower speed. For tangential (linear) velocity at the pitch circle: v = r_A × ω_A = r_B × ω_B — the pitch-point velocity is the same. Angular acceleration scales inversely with the gear ratio, just as angular velocity does.
Problem
PROBLEM 5 (Combined — Gear Train): In a gear drive, Gear A (N_A = 80 teeth) drives Gear B (N_B = 200 teeth). Gear A runs at 1500 rpm. Find: (a) the speed of Gear B in rpm, (b) the angular accelerations if Gear A accelerates at 5 rad/s², and (c) the centripetal acceleration of a tooth tip on Gear B, at a pitch radius of 100 mm, at steady-state speed.
Solution
Gear ratio: ω_A/ω_B = N_B/N_A (inverse ratio of teeth) (a) ω_A = 1500 rpm; N_B/N_A = 200/80 = 2.5 ω_B = ω_A / 2.5 = 1500/2.5 = 600 rpm In rad/s: ω_B = 600 × 2π/60 = 62.83 rad/s (b) For meshing gears: r_A × α_A = r_B × α_B (Pitch radii are proportional to tooth count: r_A/r_B = N_A/N_B = 80/200 = 0.4) α_B = α_A × (r_A/r_B) = 5 × (0.4) = 2.0 rad/s² (c) At r_B = 0.1 m, ω_B = 62.83 rad/s: aₙ = r_B × ω_B² = 0.1 × (62.83)² = 0.1 × 3947.4 = 394.7 m/s²
Exam Preparation Tips
- MASTER THE THREE-EQUATION TRIO: Write all five quantities {v₀, v, a, s, t} and circle the three given. Choose the one equation that contains exactly those three knowns plus the one unknown. Never memorize all equations blindly — understand the derivation.
- ALWAYS DEFINE A POSITIVE DIRECTION FIRST: Write it in the margin. Gravity is then either +g or −g depending on your choice. Changing sign convention mid-problem is the #1 source of errors.
- RANGE FORMULA CHECK: Before applying R = v₀²sin2θ/g, verify that launch and landing elevations are the same. If they differ, use the parametric equations y = v₀sinθ·t − ½gt² and solve the quadratic for t.
- RPM-TO-RAD/S CONVERSION IS MANDATORY: ω (rad/s) = 2πN/60. Do this as the very first step in every rotational problem. The rotational kinematic equations do not work with rpm.
- CENTRIPETAL ACCELERATION IS ALWAYS INWARD: aₙ = v²/r = rω² is directed toward the center of curvature. It does not depend on whether the speed is changing — it exists even at constant speed. Only aₜ = rα requires changing speed.
- USE THE CHECK EQUATION: After solving, verify your answer with the unused fourth equation. On board exams, if two methods agree, confidence in the answer is high.
- QUADRATIC IN PROJECTILE PROBLEMS: When t appears as t², form the standard quadratic At² + Bt + C = 0 and use the quadratic formula. Discard the negative root unless the problem explicitly involves motion before t = 0.
- COMPLEMENTARY ANGLES GIVE EQUAL RANGE: θ and (90° − θ) give the same R when launched from level ground. If you get two angle solutions to a range problem, both are physically valid — report both unless the problem specifies 'minimum angle' or 'maximum angle'.
- TREAT HORIZONTAL AND VERTICAL INDEPENDENTLY: In projectile problems, list horizontal quantities separately from vertical quantities. The only shared quantity is time t.
- FOR VARIABLE ACCELERATION: If a = f(t) → integrate with respect to t. If a = f(s) or a = f(v) → use the chain-rule form a = v(dv/ds) or a = v(dv/ds) and separate variables before integrating.
- KNOW THE UNITS: rad, rad/s, rad/s² for angular; m, m/s, m/s² for linear. 1 rev = 2π rad. Errors from wrong units are common and easily avoided.
- EXAM TIME MANAGEMENT: Kinematics problems are usually solvable within 3–5 minutes if equations are recalled correctly. If a problem takes more than 5 minutes, mark it and move on — come back after completing faster items.
In summary
Kinematics is the language of motion. Every dynamic analysis — whether in structural engineering (impact loads), geotechnical engineering (pile driving), transportation engineering (stopping distances), or mechanical systems (rotating machinery) — begins with a kinematic description of how bodies move in space and time. For the PRC Civil Engineer Licensure Examination, success in this chapter rests on three pillars: (1) disciplined sign convention, applied from the first line of every solution; (2) correct formula selection — particularly distinguishing sin2θ from sin²θ in projectile problems and remembering to convert rpm to rad/s before any rotational calculation; and (3) verification — every answer should be cross-checked using the unused fourth equation or an alternative method. The three domains of kinematics (rectilinear, projectile, rotational) share the same mathematical structure, making it efficient to master rectilinear motion first and then transfer that understanding directly to rotational kinematics. Projectile motion, while two-dimensional, reduces to two independent one-dimensional problems linked only by time. Commit the governing equations to memory, practice the worked problems until solution pathways become automatic, and apply the exam tips above. With consistent practice at this level, kinematics problems should be among the most reliably solved items on your board exam.
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