2.3 Rate, Distance, Time & Speed Word Problems

2.3 Rate, Distance, Time & Speed Word Problems

Motion and speed problems form a core component of the AFQT Arithmetic Reasoning subtest. These word problems require setting up equations based on the relationships between distance ($d$), rate or speed ($r$), and time ($t$).

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The Fundamental Distance Formula Triangle ($d = rt$)

All motion problems rely on the primary distance equation: $d = r \cdot t$

By rearranging the algebraic variables, you derive three working formulas:

$\text{Distance: } d = r \cdot t$ $\text{Rate (Speed): } r = \frac{d}{t}$ $\text{Time: } t = \frac{d}{r}$

CRITICAL RULE: Units MUST be consistent across all variables! If rate is given in miles per hour and time in minutes, you MUST convert time to hours ($t = \frac{\text{minutes}}{60}$) before multiplying!

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Opposite-Direction Motion (Closing / Separating Distance)

When two vehicles or objects move in opposite directions (either traveling toward each other to meet, or departing from the same point in opposite directions), their speeds add together to determine the rate of distance change.

Combined Rate Formula for Opposite Directions

$r_{\text{combined}} = r_1 + r_2$ $\text{Total Distance} = (r_1 + r_2) \cdot t$ $\text{Time to Meet / Separate} = \frac{\text{Initial Distance}}{r_1 + r_2}$

Example: Two military convoys start $300$ miles apart and drive directly toward each other. Convoy A travels at $45 \text{ mph}$ and Convoy B travels at $55 \text{ mph}$. $\text{Combined Rate} = 45 + 55 = 100 \text{ mph}$ $\text{Time to meet} = \frac{300 \text{ miles}}{100 \text{ mph}} = 3 \text{ hours}$

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Same-Direction Catch-Up Problems

When two objects move in the same direction and the faster object attempts to catch up to the slower object (which had a head start), their speeds subtract to determine the relative rate of closure.

Relative Speed Formula for Same Direction

$r_{\text{relative}} = r_{\text{faster}} - r_{\text{slower}}$ $\text{Head Start Distance} = r_{\text{slower}} \cdot t_{\text{head start}}$ $\text{Time to Catch Up} = \frac{\text{Head Start Distance}}{r_{\text{faster}} - r_{\text{slower}}}$

Alternatively, set distance equations equal ($d_1 = d_2$): $r_{\text{slower}} \cdot (t + t_{\text{head start}}) = r_{\text{faster}} \cdot t$

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Round-Trip and Average Speed Calculations

One of the most frequently missed question types on the AFQT is finding the average speed for a round trip or multi-leg journey.

The Arithmetic Mean Trap

If a vehicle travels to a destination at $30 \text{ mph}$ and returns along the exact same route at $60 \text{ mph}$, the average speed is NOT $\frac{30 + 60}{2} = 45 \text{ mph}$! Because the vehicle spends more time traveling at the slower speed, $45 \text{ mph}$ is mathematically incorrect.

The Harmonic Average Speed Formula

$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{d_1 + d_2}{t_1 + t_2}$

For a round trip of distance $d$ each way at speeds $r_1$ and $r_2$: $R_{\text{avg}} = \frac{2 \cdot r_1 \cdot r_2}{r_1 + r_2}$

Validation using example ($30 \text{ mph}$ and $60 \text{ mph}$): $R_{\text{avg}} = \frac{2(30)(60)}{30 + 60} = \frac{3600}{90} = 40 \text{ mph}$

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Current and Wind Resistance Problems

Naval vessels traveling in water currents or aircraft flying through wind experience rate adjustments based on direction.

• Downstream / Tailwind (With the current/wind): Current speed adds to still-water/air speed. $r_{\text{effective}} = r_{\text{vehicle}} + c$
• Upstream / Headwind (Against the current/wind): Current speed subtracts from still-water/air speed. $r_{\text{effective}} = r_{\text{vehicle}} - c$

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Step-by-Step Worked AFQT Examples

Worked Example 1: Opposite Direction Motion

Problem: Two naval cutters depart from the same port at the same time, one heading east at $22 \text{ knots}$ and the other heading west at $18 \text{ knots}$. How many hours will it take for them to be $200$ nautical miles apart?

• Step 1: Identify opposite directions $\implies$ add rates. $r_{\text{combined}} = 22 + 18 = 40 \text{ knots (nautical miles/hour)}$
• Step 2: Use $t = \frac{d}{r}$ to find time. $t = \frac{200 \text{ nautical miles}}{40 \text{ knots}} = 5 \text{ hours}$

Worked Example 2: Same Direction Catch-Up

Problem: A supply truck leaves an airbase traveling at $40 \text{ mph}$. Two hours later, a fast courier jeep leaves the same airbase along the same route traveling at $60 \text{ mph}$. How long will it take the courier jeep to catch up to the supply truck?

• Step 1: Calculate head start distance of the supply truck. $\text{Head Start Distance} = 40 \text{ mph} \times 2 \text{ hours} = 80 \text{ miles}$
• Step 2: Determine relative closing speed. $r_{\text{relative}} = 60 \text{ mph} - 40 \text{ mph} = 20 \text{ mph}$
• Step 3: Divide head start distance by relative closing speed. $\text{Catch-up Time} = \frac{80 \text{ miles}}{20 \text{ mph}} = 4 \text{ hours}$

Worked Example 3: Upstream vs Downstream River Motion

Problem: A patrol boat travels at $15 \text{ mph}$ in still water. It travels $36$ miles downstream with a $3 \text{ mph}$ current and then returns $36$ miles upstream against the same current. What is the total time for the round trip?

• Step 1: Calculate downstream rate and time. $r_{\text{downstream}} = 15 + 3 = 18 \text{ mph}$ $t_{\text{downstream}} = \frac{36 \text{ miles}}{18 \text{ mph}} = 2 \text{ hours}$
• Step 2: Calculate upstream rate and time. $r_{\text{upstream}} = 15 - 3 = 12 \text{ mph}$ $t_{\text{upstream}} = \frac{36 \text{ miles}}{12 \text{ mph}} = 3 \text{ hours}$
• Step 3: Sum downstream and upstream times. $\text{Total Trip Time} = 2 + 3 = 5 \text{ hours}$

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Common AFQT Traps & Shortcut Strategies

1. The Minute-Hour Conversion Trap: If a problem states a speed of $60 \text{ mph}$ and a duration of $45 \text{ minutes}$, do NOT multiply $60 \times 45 = 2,700$! Convert $45 \text{ minutes}$ to $\frac{45}{60} = 0.75 \text{ hours}$, yielding $60 \times 0.75 = 45 \text{ miles}$.
2. The Average Speed Distractor: On round trip average speed problems, test creators will almost always include the simple average as Option A or B. Always use $\frac{\text{Total Distance}}{\text{Total Time}}$!