5.1 Motion, Work Rate & Fluid Flow Word Problems

5.1 Motion, Work Rate & Fluid Flow Word Problems

Core Principle: In military field operations, calculating logistical transit intervals, coordinating multi-column tactical movements, scheduling naval interdictions, and projecting combat engineering completion timelines require absolute mathematical accuracy under pressure. On the AFP Service Aptitude Test (AFPSAT), calculators are strictly barred. Officer candidates must rapidly construct linear algebraic models and execute mental arithmetic using structured rate-time templates.

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1. Uniform Motion Fundamentals ($D = R \times T$)

All classical motion problems on the AFPSAT evaluate the relationship between three core variables: Distance ($D$), Rate or Speed ($R$), and Elapsed Time ($T$).

$\text{Distance} = \text{Rate} \times \text{Time} \quad (D = R \cdot T)$ $\text{Rate} = \frac{\text{Distance}}{\text{Time}} \quad \left(R = \frac{D}{T}\right) \qquad \text{Time} = \frac{\text{Distance}}{\text{Rate}} \quad \left(T = \frac{D}{R}\right)$

Unit Consistency and Dimensional Alignment

A frequent source of test errors is mismatched physical units. Examinees must verify dimensional alignment prior to calculating:

• Speed in kilometers per hour (km/h) with time in hours ($h$) and distance in kilometers ($km$).
• Speed in meters per second (m/s) with time in seconds ($s$) and distance in meters ($m$). $\text{Conversion Factor: } 1 \text{ km/h} = \frac{1,000 \text{ m}}{3,600 \text{ s}} = \frac{5}{18} \text{ m/s} \quad \Longleftrightarrow \quad 1 \text{ m/s} = \frac{18}{5} \text{ km/h} = 3.6 \text{ km/h}$
• Naval Velocity: In maritime interdiction scenarios involving the Philippine Navy or Coast Guard, speed is expressed in knots (1 knot = 1 nautical mile per hour = 1.852 km/h).

+--------------------------------------------------------------------------+
|                       RATE-TIME-DISTANCE TABULAR SETUP                   |
+---------------------+-------------------+----------------+---------------+
| Entity / Scenario   | Rate (R)          | Time (T)       | Distance (D)  |
+---------------------+-------------------+----------------+---------------+
| Element A (Lead)    | R_A               | T_A            | D_A = R_A·T_A |
| Element B (Chase)   | R_B               | T_B            | D_B = R_B·T_B |
+---------------------+-------------------+----------------+---------------+
| System Condition    | Closing: R_B - R_A| Elapsed Time   | D_B - D_A = 0 |
|                     | Opposing: R_A+R_B | Meeting Time   | D_A+D_B = Tot |
+---------------------+-------------------+----------------+---------------+

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2. Core Motion Problem Archetypes

Every motion word problem on military entrance examinations falls into one of four structural archetypes. Identifying the archetype immediately dictates the setup equation.

Archetype A: Opposite Direction / Converging Motion (Approaching Objects)

When two bodies move directly toward each other from a fixed initial separation distance ($D_{\text{total}}$), or depart simultaneously in opposite directions from a single central point:

• Their effective closing rate is the sum of their individual speeds: $R_{\text{relative}} = R_1 + R_2$.
• The total distance traversed collectively equals the sum of the individual distances: $D_{\text{total}} = D_1 + D_2 = (R_1 \cdot T) + (R_2 \cdot T) = (R_1 + R_2) \cdot T$.
• The time elapsed until convergence is: $T_{\text{meet}} = \frac{D_{\text{total}}}{R_1 + R_2}$

Worked Tactical Scenario: A Philippine Army logistics convoy departs Camp O'Donnell traveling south along the expressway toward Fort Magsaysay at an average sustained speed of $45 \text{ km/h}$. At the identical moment, an armored reconnaissance patrol leaves Fort Magsaysay heading north toward Camp O'Donnell at $55 \text{ km/h}$. The two military installations are separated by a highway distance of $250 \text{ km}$. How long after departure do the two military units meet, and how far has the logistics convoy traveled from Camp O'Donnell?

1. Isolate Knowns: $R_1 = 45 \text{ km/h}$, $R_2 = 55 \text{ km/h}$, $D_{\text{total}} = 250 \text{ km}$.
2. Compute Relative Speed: $R_{\text{rel}} = 45 + 55 = 100 \text{ km/h}$.
3. Solve for Meeting Time: $T = \frac{250 \text{ km}}{100 \text{ km/h}} = 2.5 \text{ hours}$ (2 hours and 30 minutes).
4. Calculate Convoy Distance: $D_{\text{convoy}} = R_1 \cdot T = 45 \text{ km/h} \times 2.5 \text{ h} = 112.5 \text{ km}$ from Camp O'Donnell.

Archetype B: Same Direction Motion (Catch-Up & Tactical Interception)

In an overtaking or interception problem, a faster pursuit craft pursues a slower vehicle that has already departed or holds an initial positional lead ($D_{\text{lead}}$):

• The net closing speed is the absolute difference between their rates: $R_{\text{closing}} = R_{\text{pursuit}} - R_{\text{target}}$, where $R_{\text{pursuit}} > R_{\text{target}}$.
• If the target has a head start of $t_{\text{headstart}}$ hours, the initial gap is $D_{\text{lead}} = R_{\text{target}} \times t_{\text{headstart}}$.
• The time required for the pursuit unit to eliminate the lead distance is: $T_{\text{intercept}} = \frac{D_{\text{lead}}}{R_{\text{pursuit}} - R_{\text{target}}}$

Worked Naval Interdiction Scenario: A suspicious, unflagged maritime vessel is detected fleeing a coastal anchorage in the Sulu Sea at a speed of $12 \text{ knots}$. Two hours later, a Philippine Navy Littoral Combat Vessel departs the same base in direct pursuit along the identical maritime corridor at $20 \text{ knots}$. How many hours will the naval vessel take to overtake the target, and at what distance from the base will the interdiction occur?

1. Calculate Target's Lead Distance: $D_{\text{lead}} = 12 \text{ knots} \times 2 \text{ hours} = 24 \text{ nautical miles (NM)}$.
2. Compute Closing Rate: $R_{\text{closing}} = 20 - 12 = 8 \text{ knots}$.
3. Solve Intercept Time: $T_{\text{intercept}} = \frac{24 \text{ NM}}{8 \text{ knots}} = 3 \text{ hours}$.
4. Determine Interception Range from Base: $D = R_{\text{pursuit}} \times T_{\text{intercept}} = 20 \text{ knots} \times 3 \text{ hours} = 60 \text{ nautical miles}$. Cross-Verification: The target traveled for $2 + 3 = 5 \text{ hours}$ at $12 \text{ knots}$: $12 \times 5 = 60 \text{ NM}$. The positions coincide exactly.

Archetype C: The Average Speed Trap and Harmonic Mean

A notorious trap on standardized military aptitude examinations is calculating average speed for a round trip over identical outbound and inbound distances.

The Trap: If a vehicle travels from Point A to Point B at $R_1$ and returns along the identical route from Point B to Point A at $R_2$, the average speed is never the simple arithmetic mean $\frac{R_1 + R_2}{2}$.

$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$

Because the vehicle travels at a lower speed over one leg, it spends significantly more time moving at that slower speed. Time acts as a weighting factor that pulls the true average speed toward the lower velocity.

Algebraic Derivation of the Harmonic Mean:

Let $d$ represent the one-way distance between endpoints.

• Total round-trip distance $= d + d = 2d$.
• Outbound transit time: $T_1 = \frac{d}{R_1}$.
• Inbound transit time: $T_2 = \frac{d}{R_2}$.
• Total round-trip time: $T_{\text{total}} = T_1 + T_2 = \frac{d}{R_1} + \frac{d}{R_2} = d \left(\frac{R_1 + R_2}{R_1 R_2}\right)$.

Substituting into the average speed definition: $\text{Average Speed} = \frac{2d}{d \left(\frac{R_1 + R_2}{R_1 R_2}\right)} = \frac{2 \cdot R_1 \cdot R_2}{R_1 + R_2}$

Numerical Proof: A military dispatch motorcycle travels outbound through an urban zone at $30 \text{ km/h}$ and returns along the identical road during off-peak hours at $60 \text{ km/h}$.

• Arithmetic Mean (WRONG): $\frac{30 + 60}{2} = 45 \text{ km/h}$.
• Harmonic Mean (CORRECT): $\frac{2 \times 30 \times 60}{30 + 60} = \frac{3,600}{90} = 40 \text{ km/h}$.
• Verification via arbitrary distance ($d = 60 \text{ km}$): Outbound time $= 60 / 30 = 2 \text{ h}$. Inbound time $= 60 / 60 = 1 \text{ h}$. Total distance $= 120 \text{ km}$. Total time $= 3 \text{ h}$. Average speed $= 120 / 3 = 40 \text{ km/h}$.

Archetype D: River Currents and Headwind/Tailwind Problems

When a vehicle operates within a moving medium (water current for boats, wind for aircraft), the medium directly modifies the craft's effective ground speed ($R_{\text{effective}}$):

• Let $R_v = \text{craft's velocity in still water or calm air}$.
• Let $R_m = \text{velocity of the moving fluid medium (current or wind)}$.

$\text{Downstream / With Tailwind: } R_{\text{effective}} = R_v + R_m$ $\text{Upstream / Against Headwind: } R_{\text{effective}} = R_v - R_m$

Adding and subtracting these two fundamental equations yields direct formulas to isolate craft speed and current speed: $R_v = \frac{R_{\text{down}} + R_{\text{up}}}{2} \qquad R_m = \frac{R_{\text{down}} - R_{\text{up}}}{2}$

Worked Aerial Navigation Example: A Philippine Air Force C-295 transport aircraft flies a $720 \text{ km}$ route from Villamor Air Base to Palawan in $2 \text{ hours}$ assisted by a tailwind. On the return flight along the exact reverse heading, facing an identical opposing headwind, the flight takes $2.4 \text{ hours}$. What is the aircraft's airspeed in calm conditions, and what is the velocity of the prevailing wind?

1. Compute Effective Speeds:
   • Tailwind Ground Speed: $R_{\text{down}} = \frac{720 \text{ km}}{2 \text{ h}} = 360 \text{ km/h}$.
   • Headwind Ground Speed: $R_{\text{up}} = \frac{720 \text{ km}}{2.4 \text{ h}} = 300 \text{ km/h}$.
2. Isolate Calm Airspeed ($R_v$): $R_v = \frac{360 + 300}{2} = \frac{660}{2} = 330 \text{ km/h}$
3. Isolate Wind Velocity ($R_m$): $R_m = \frac{360 - 300}{2} = \frac{60}{2} = 30 \text{ km/h}$

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3. Work Rate Problems and Collaborative Production

Work problems apply the uniform motion equation to tasks and civil-military engineering projects. Rather than traversing a spatial distance ($D$), workers generate units of finished work ($W$).

$\text{Work Done} = \text{Rate} \times \text{Time} \quad (W = R \cdot T)$

The Reciprocal Rule of Work

If a single military engineering unit can complete an entire project (defined as $W = 1$ complete job) in $T$ days working alone, its daily work rate is the reciprocal of time: $R = \frac{1}{T} \quad \text{of the total job completed per day}$

Combined Collaborative Rate Formula

When multiple entities work concurrently without interfering with one another, their production rates are additive: $R_{\text{combined}} = R_1 + R_2 + R_3 + \dots + R_n = \frac{1}{T_1} + \frac{1}{T_2} + \frac{1}{T_3} + \dots + \frac{1}{T_n}$

For two workers with individual completion times $T_1$ and $T_2$: $\frac{1}{T_{\text{together}}} = \frac{1}{T_1} + \frac{1}{T_2} = \frac{T_1 + T_2}{T_1 \cdot T_2} \quad \implies \quad T_{\text{together}} = \frac{T_1 \cdot T_2}{T_1 + T_2} \quad \text{("Product over Sum" Rule)}$

Workers Joining or Leaving Midway Through Tasks

When participants work for varying time intervals, set up an equation where the sum of the fractional work completed equals $1$: $\sum (R_i \cdot t_i) = (R_1 \cdot t_1) + (R_2 \cdot t_2) + \dots + (R_n \cdot t_n) = 1$

Worked Engineering Scenario: Detachment Alpha can construct a tactical perimeter trench in $12 \text{ days}$. Detachment Bravo can construct the same trench in $18 \text{ days}$. Detachment Alpha works on the fortification alone for $4 \text{ days}$, after which Detachment Bravo is ordered to assist. How many additional days will both detachments take working together to finish the trench?

1. Identify Individual Rates: $R_A = \frac{1}{12}$ job/day, $R_B = \frac{1}{18}$ job/day.
2. Calculate Work Accomplished Solo by Alpha: $W_{\text{initial}} = R_A \times 4 = \frac{4}{12} = \frac{1}{3} \text{ of the trench completed}$
3. Determine Remaining Balance: $W_{\text{remaining}} = 1 - \frac{1}{3} = \frac{2}{3} \text{ of the trench}$
4. Compute Combined Daily Rate: $R_{\text{combined}} = \frac{1}{12} + \frac{1}{18} = \frac{3}{36} + \frac{2}{36} = \frac{5}{36} \text{ job/day}$
5. Solve for Collaborative Time ($t$): $t = \frac{W_{\text{remaining}}}{R_{\text{combined}}} = \frac{2/3}{5/36} = \frac{2}{3} \times \frac{36}{5} = \frac{24}{5} = 4.8 \text{ days}$ Both detachments require an additional $4.8 \text{ days}$ (4 days, 19 hours, and 12 minutes) to complete the project.

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4. Fluid Flow: Pipes, Cisterns & Reservoirs

Fluid flow problems model storage reservoirs, fuel bladders, and naval ballast systems. They follow the exact logic of work problems with one critical distinction: flow directions have opposing algebraic signs.

• Inlet Conduits (Filling): Add volume to the system; treated as positive rates ($+R_i = +\frac{1}{T_{\text{fill}}}$).
• Drainage Conduits / Leaks (Emptying): Remove volume from the system; treated as negative rates ($-R_d = -\frac{1}{T_{\text{drain}}}$).

$\text{Net Rate} = \sum R_{\text{inlets}} - \sum R_{\text{outlets}} = \left(\frac{1}{T_{\text{in},1}} + \frac{1}{T_{\text{in},2}}\right) - \left(\frac{1}{T_{\text{out},1}} + \frac{1}{T_{\text{leak}}}\right)$

Worked Fluid Mechanics Example: An expeditionary field hospital maintains a 12,000-liter potable water bladder. Main Supply Line 1 can fill the empty bladder in $6 \text{ hours}$. Secondary Line 2 can fill it in $8 \text{ hours}$. An automated purification distribution valve empties a full bladder in $12 \text{ hours}$. If both supply lines and the distribution valve are opened simultaneously when the bladder is completely empty, how many hours will it take to fill the bladder completely?

1. Establish Individual Rates: $R_1 = +\frac{1}{6}$, $R_2 = +\frac{1}{8}$, $R_{\text{drain}} = -\frac{1}{12}$.
2. Calculate Net Rate using Least Common Denominator (LCD = 24): $R_{\text{net}} = \frac{1}{6} + \frac{1}{8} - \frac{1}{12} = \frac{4}{24} + \frac{3}{24} - \frac{2}{24} = \frac{5}{24} \text{ bladder per hour}$
3. Invert Net Rate to Find Total Time: $T = \frac{1}{R_{\text{net}}} = \frac{24}{5} = 4.8 \text{ hours} = 4 \text{ hours and } 48 \text{ minutes}$

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Comprehensive Word Problem Archetype Matrix

The following master reference table summarizes the governing equations, algebraic setups, and military operational contexts tested on the AFPSAT: