9.2 Speed, Distance, Time & Combined Work

Speed, Distance, Time & Combined Work

Core Concept: Rate problems evaluate your ability to link physical dimensions across time. Whether tracking two express trains closing a distance gap or calculating how rapidly three pipes fill a storage cistern, the mathematical architecture is identical: rates are reciprocal quantities that aggregate additively or subtractively under unified units.

Most candidate errors in this domain stem not from difficult algebra, but from inconsistent dimensional units—such as multiplying kilometres per hour by minutes, or taking the simple arithmetic mean of two speeds across equal distances. Mastering the conversion factors and reciprocal rate models guarantees quick, reliable solutions.

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The Kinematic Triangle & Dimensional Integrity

The fundamental relationship connecting uniform motion is:

$\text{Distance} = \text{Speed} \times \text{Time} \quad (D = S \cdot T)$ $\text{Speed} = \frac{\text{Distance}}{\text{Time}} \quad \left(S = \frac{D}{T}\right)$ $\text{Time} = \frac{\text{Distance}}{\text{Speed}} \quad \left(T = \frac{D}{S}\right)$

          /\ 
         /  \ 
        / D  \ 
       /------\ 
      / S | T  \ 
     /__________\ 
   D = S × T  |  S = D / T  |  T = D / S

Dimensional Consistency Rule

Every variable in a kinematic formula must share compatible units:

• If Distance is in kilometres (km) and Time is in hours (h), Speed must be in km/h.
• If Distance is in metres (m) and Time is in seconds (s), Speed must be in m/s.

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High-Speed Unit Conversions (The 3.6 Factor)

Converting between $\text{km/h}$ and $\text{m/s}$ is tested heavily on selective entry exams because train and vehicle lengths are specified in metres while speeds are given in kilometres per hour.

The Mathematical Derivation

$1\text{ km/h} = \frac{1,000\text{ metres}}{3,600\text{ seconds}} = \frac{10}{36}\text{ m/s} = \frac{5}{18}\text{ m/s} = \frac{1}{3.6}\text{ m/s}$

$\mathbf{\text{km/h} \xrightarrow{\quad \div 3.6 \quad} \text{m/s}}$ $\mathbf{\text{m/s} \xrightarrow{\quad \times 3.6 \quad} \text{km/h}}$

High-Frequency Speed Benchmarks

Memorise these common multiples to avoid dividing decimals during the exam:

Rapid Minute-to-Fractional-Hour Conversions

Never perform long division when converting minutes to hours. Use standard rational fractions:

$$10\text{ min} = \frac{1}{6}\text{ hr} \qquad 12\text{ min} = \frac{1}{5}\text{ hr} \qquad 15\text{ min} = \frac{1}{4}\text{ hr} \qquad 20\text{ min} = \frac{1}{3}\text{ hr} \qquad 24\text{ min} = \frac{2}{5}\text{ hr}$$ $$30\text{ min} = \frac{1}{2}\text{ hr} \qquad 36\text{ min} = \frac{3}{5}\text{ hr} \qquad 40\text{ min} = \frac{2}{3}\text{ hr} \qquad 45\text{ min} = \frac{3}{4}\text{ hr} \qquad 48\text{ min} = \frac{4}{5}\text{ hr}$$

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The Average Speed Trap: Total Distance over Total Time

The most pervasive error on selective school examinations is assuming that average speed across a multi-leg journey equals the simple arithmetic mean of the individual speeds.

$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{D_1 + D_2 + \dots + D_n}{T_1 + T_2 + \dots + T_n}$

The Equal Distance Case (The Harmonic Mean)

When an object covers two equal distances ($d_1 = d_2 = d$) at speeds $S_1$ and $S_2$:

$T_1 = \frac{d}{S_1}, \qquad T_2 = \frac{d}{S_2}$ $S_{\text{avg}} = \frac{2d}{T_1 + T_2} = \frac{2d}{\frac{d}{S_1} + \frac{d}{S_2}} = \frac{2}{\frac{1}{S_1} + \frac{1}{S_2}} = \frac{2 S_1 S_2}{S_1 + S_2}$

Worked Example: The Round-Trip Commute

A commuter travels $120\text{ km}$ to a regional office at an average speed of $60\text{ km/h}$, and returns along the identical $120\text{ km}$ route at an average speed of $90\text{ km/h}$. What is the overall average speed for the entire $240\text{ km}$ round trip?

• Step 1 (Calculate time for Leg 1): $T_1 = \frac{120\text{ km}}{60\text{ km/h}} = 2.0\text{ hours}$
• Step 2 (Calculate time for Leg 2): $T_2 = \frac{120\text{ km}}{90\text{ km/h}} = \frac{4}{3}\text{ hours} = 1\text{ hr } 20\text{ min}$
• Step 3 (Calculate aggregate metrics): $\text{Total Distance} = 120 + 120 = 240\text{ km}$ $\text{Total Time} = 2 + \frac{4}{3} = \frac{10}{3}\text{ hours}$
• Step 4 (Compute True Average Speed): $S_{\text{avg}} = \frac{240}{\frac{10}{3}} = 240 \times \frac{3}{10} = 72\text{ km/h}$

Exam Trap: The arithmetic mean of $60$ and $90$ is $\frac{60 + 90}{2} = 75\text{ km/h}$. Notice that $72 < 75$. Because the commuter spends more time travelling at the slower speed ($2$ hours vs $1.33$ hours), the slower speed is weighted more heavily in the true average. Selecting $75\text{ km/h}$ is a fatal distractor trap.

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Relative Speed: Opposing vs. Same-Direction Motion

When two moving objects are evaluated relative to one another, their individual velocities combine according to their directional vectors.

Opposing Directions (Moving Towards Each Other):
   [Object 1] ───► S1          S2 ◄─── [Object 2]
   Closing Speed: S_rel = S1 + S2
   Time to Collision/Meeting: T = Initial Separation ÷ (S1 + S2)

Same Direction (Overtaking / Chasing):
   [Chaser] ───► S1            [Lead] ───► S2  (where S1 > S2)
   Closing Speed: S_rel = S1 - S2
   Time to Catch Up: T = Head Start Separation ÷ (S1 - S2)

Worked Example: Pursuit and Interception

A cyclist departs a checkpoint at a steady pace of $18\text{ km/h}$. Exactly $15\text{ minutes}$ later, a motorcyclist sets off from the same checkpoint along the identical route at a constant speed of $45\text{ km/h}$ in pursuit. How many minutes will it take the motorcyclist to catch up to the cyclist?

• Step 1 (Calculate initial separation distance): $\text{Cyclist's Head Start Time} = 15\text{ min} = \frac{15}{60}\text{ hr} = 0.25\text{ hr}$ $\text{Head Start Distance} = 18\text{ km/h} \times 0.25\text{ hr} = 4.5\text{ km}$
• Step 2 (Determine relative closing speed): $S_{\text{rel}} = 45\text{ km/h} - 18\text{ km/h} = 27\text{ km/h}$
• Step 3 (Calculate time to overtake): $T = \frac{\text{Separation Distance}}{S_{\text{rel}}} = \frac{4.5\text{ km}}{27\text{ km/h}} = \frac{1}{6}\text{ hour}$
• Step 4 (Convert fractional hours to minutes): $\text{Time in minutes} = \frac{1}{6} \times 60\text{ minutes} = 10\text{ minutes}$

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Train Crossing Geometry

Train problems require candidates to account for the physical length of the moving vehicle. A train has not "passed" an object until its very last carriage has cleared the boundary.

Worked Example: Clearing a Tunnel

A high-speed regional passenger train measuring $150\text{ metres}$ in length travels at a constant speed of $90\text{ km/h}$. It enters a straight railway tunnel that is $600\text{ metres}$ long. How many seconds elapse from the instant the nose of the train enters the tunnel until the rear carriage completely exits the tunnel?

• Step 1 (Convert speed to m/s): $S = 90\text{ km/h} = \frac{90}{3.6} = 25\text{ m/s}$
• Step 2 (Determine total clearance distance): $D = L_{\text{train}} + L_{\text{tunnel}} = 150\text{ m} + 600\text{ m} = 750\text{ metres}$
• Step 3 (Calculate time in seconds): $T = \frac{D}{S} = \frac{750\text{ m}}{25\text{ m/s}} = 30\text{ seconds}$

Exam Trap: Candidates who divide only the tunnel length ($600 / 25 = 24\text{ s}$) forget that the train's entire $150\text{ m}$ body must also exit the tunnel opening.

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Combined Work & Fluid Reservoir Problems

Combined work problems function identically to kinematic problems: work done ($W$) replaces distance ($D$), production rate ($R$) replaces speed ($S$), and time taken ($T$) remains time ($T$).

$\text{Work} = \text{Rate} \times \text{Time} \quad (W = R \cdot T)$ $\text{If a worker or pipe completes } 1\text{ entire job in } t\text{ hours, their rate is: } R = \frac{1}{t}$

The Cooperative Rate Principle

When multiple entities work simultaneously toward the same outcome, their individual rates add together:

$R_{\text{total}} = R_1 + R_2 + \dots + R_n = \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n}$ $\text{Total Time Taken } T_{\text{total}} = \frac{1}{R_{\text{total}}}$

For two entities working together: $T_{\text{total}} = \frac{1}{\frac{1}{a} + \frac{1}{b}} = \frac{a \cdot b}{a + b} \quad \left(\mathbf{\frac{\text{Product}}{\text{Sum}}}\right)$

Opposing Rates (Inlet Pipes vs. Drainage Outlets)

If an entity empties, removes, or undermines progress (such as a leak or drainage valve), its rate is subtracted:

$R_{\text{net}} = R_{\text{inlet 1}} + R_{\text{inlet 2}} - R_{\text{drain}} = \frac{1}{t_{\text{in 1}}} + \frac{1}{t_{\text{in 2}}} - \frac{1}{t_{\text{drain}}}$

Worked Example: Reservoir with Competing Pipes

An industrial reservoir has two inlet pipes and one drainage valve. Inlet Pipe A can fill the reservoir alone in $6\text{ hours}$, while Inlet Pipe B can fill it alone in $8\text{ hours}$. When full, the drainage valve can empty the entire reservoir in $12\text{ hours}$. If all three are opened simultaneously from empty, how long will it take to fill the reservoir completely?

• Step 1 (Express individual hourly rates): $R_A = +\frac{1}{6}, \quad R_B = +\frac{1}{8}, \quad R_{\text{drain}} = -\frac{1}{12}$
• Step 2 (Find net rate using common denominator 24): $R_{\text{net}} = \frac{4}{24} + \frac{3}{24} - \frac{2}{24} = \frac{4 + 3 - 2}{24} = \frac{5}{24}\text{ of the tank per hour}$
• Step 3 (Invert net rate to find total time): $T = \frac{1}{R_{\text{net}}} = \frac{24}{5}\text{ hours} = 4\frac{4}{5}\text{ hours} = 4\text{ hours } 48\text{ minutes} = 4.8\text{ hours}$

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Kinematics & Rates Formula Summary Table

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High-Yield Traps in Speed, Distance & Work

1. Adding Completion Times Directly

Untrained candidates see that Worker A takes 4 hours and Worker B takes 6 hours, and conclude they will take $4 + 6 = 10$ hours together. Two workers cooperating must finish faster than the fastest individual worker ($T < 4\text{ hours}$). The correct calculation is $\frac{4 \times 6}{4 + 6} = \frac{24}{10} = 2.4\text{ hours}$.

2. The Arithmetic Mean Fallacy in Average Speeds

Whenever an exam item states "A car travels at 40 km/h for the first half of the distance and 60 km/h for the second half", examiners know candidates will rush to pick $50\text{ km/h}$. Remember: the car spends more time at the slower speed, pulling the true average down to $48\text{ km/h}$.

3. Neglecting Train Length in Signal vs. Platform Crossings

When a train passes a person or signal pole, distance is simply the train's own length. But when a train crosses a bridge or tunnel, distance is train length plus bridge length. Forgetting the train's length produces an under-calculated time that matches an attractive distractor.