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**Uniform motion problems** are a category of word problems in which two or more objects travel at constant (uniform) speeds, and the goal is to find an unknown rate, time, or distance by comparing the two travel scenarios. Typical setups include vehicles going the same route at different speeds, travelers heading in opposite directions, or one person catching up to another. Because each object's speed does not change during its trip, the distance formula $$d = rt$$ can be applied separately to each traveler, and the resulting distance expressions are then related through a condition stated in the problem — for example, the distances are equal, or they add up to a known total. These problems require more algebra than single-object distance problems because they involve expressing two rates or two times in terms of a single variable and then setting up an equation that links the two distance expressions.

Uniform Motion Problems

Apply the distance, rate, and time problem-solving strategy to find two unknown speeds when both trips cover the same distance but use different modes of travel, requiring a unit conversion from minutes to hours and clearing fractions to solve.

**Problem:** When Naoko walks to school, it takes her $$30$$ minutes. If she rides her bike, it takes her $$15$$ minutes. Her biking speed is $$3$$ miles per hour faster than her walking speed. What are her walking speed and her biking speed?

1. **Read and draw:** Sketch the route from home to school. The walking path ($$30$$ minutes) and the biking path ($$15$$ minutes, $$3$$ mph faster) cover the same distance. Create a rate–time–distance table.
2. **Identify:** The walking speed and the biking speed.
3. **Name:** Let $$r$$ = walking speed in mph. Then the biking speed is $$r + 3$$. Because the rate is in miles per hour, convert the times into hours: $$30$$ minutes is $$\frac{30}{60}$$ or $$\frac{1}{2}$$ hour, and $$15$$ minutes is $$\frac{15}{60}$$ or $$\frac{1}{4}$$ hour. Multiply rate by time to fill in the distance column:

| | Rate (mph) | Time (hrs) | Distance (miles) |
|---|---|---|---|
| Walk | $$r$$ | $$\frac{1}{2}$$ | $$\frac{1}{2}r$$ |
| Bike | $$r + 3$$ | $$\frac{1}{4}$$ | $$\frac{1}{4}(r + 3)$$ |

4. **Translate:** The distance from home to school is the same whether Naoko walks or bikes:

$$\frac{1}{2}r = \frac{1}{4}(r + 3)$$

5. **Solve:** Clear the fractions by multiplying both sides by the LCD of $$2$$ and $$4$$, which is $$4$$:

$$4 \cdot \frac{1}{2}r = 4 \cdot \frac{1}{4}(r + 3)$$

$$2r = r + 3$$

Subtract $$r$$ from both sides: $$r = 3$$. The walking speed is $$3$$ mph. Her biking speed is $$3 + 3 = 6$$ mph.

6. **Check:** Walk: $$3 \cdot 0.5 = 1.5$$ miles. Bike: $$6 \cdot 0.25 = 1.5$$ miles. Both distances are $$1.5$$ miles. $$\checkmark$$

7. **Answer:** Naoko's walking speed is $$3$$ mph and her biking speed is $$6$$ mph.

This example extends the equal-distance uniform motion scenario by requiring a time conversion before setup and solving an algebraic equation with fractional coefficients.

Finding Walking and Biking Speeds Using Equal Distances

Apply the distance, rate, and time problem-solving strategy to find two unknown speeds when a single traveler covers two consecutive segments at different uniform rates, with the segment distances adding up to a known total and the faster rate expressed as a multiple of the slower rate.

**Problem:** Hamilton drives from his home in Orange County to Las Vegas, a total distance of $$255$$ miles. He departs at 2:00 PM and drives on congested city freeways. At 4:00 PM the congestion clears and he drives through the desert at a speed $$1.75$$ times his city speed. He arrives in Las Vegas at 6:30 PM. What was his speed during each part of the trip?

1. **Read and draw:** Sketch the route from Home to Las Vegas with two arrows — one labeled "city driving" (2:00 PM to 4:00 PM) and one labeled "desert driving" (4:00 PM to 6:30 PM). The total distance between the two endpoints is $$255$$ miles. Create a rate–time–distance table.

2. **Identify:** The driving speed for the city segment and the desert segment.

3. **Name:** Let $$r$$ = the city driving speed in mph. Because the desert speed is $$1.75$$ times as fast, it equals $$1.75r$$. Convert the clock times to elapsed times: city driving lasts from 2:00 PM to 4:00 PM, which is $$2$$ hours; desert driving lasts from 4:00 PM to 6:30 PM, which is $$2.5$$ hours. Multiply rate by time to fill in the distance column:

| | Rate (mph) | Time (hrs) | Distance (miles) |
|---|---|---|---|
| City | $$r$$ | $$2$$ | $$2r$$ |
| Desert | $$1.75r$$ | $$2.5$$ | $$2.5(1.75r)$$ |
| | | | $$255$$ |

4. **Translate:** The city distance plus the desert distance equals the total of $$255$$ miles:

$$2r + 2.5(1.75r) = 255$$

5. **Solve:** Multiply $$2.5 \times 1.75 = 4.375$$:

$$2r + 4.375r = 255$$

Combine like terms: $$6.375r = 255$$. Divide both sides by $$6.375$$:

$$r = 40$$

The city speed is $$40$$ mph. The desert speed is $$1.75 \times 40 = 70$$ mph.

6. **Check:** City: $$40 \times 2 = 80$$ miles. Desert: $$70 \times 2.5 = 175$$ miles. Total: $$80 + 175 = 255$$ miles. $$\checkmark$$

7. **Answer:** Hamilton drove $$40$$ mph in the city and $$70$$ mph in the desert.

This example demonstrates the consecutive-segment scenario: a single traveler covers two parts of a trip at different uniform speeds, and the individual segment distances add up to a known total. Unlike the equal-distance and opposite-direction examples, here the two speeds are related by a **multiplicative factor** ($$1.75r$$ rather than $$r + 12$$), the elapsed times for each segment are determined by converting separate clock-time intervals to durations, and multiplying the decimal coefficients produces the term $$4.375r$$. Combining like terms with decimal arithmetic yields $$6.375r = 255$$, which is solved by a single division.

Finding Speeds for Consecutive Trip Segments Using a Known Total Distance

In many real-world uniform motion applications, a current — such as a river's water current or the prevailing wind — affects the actual speed at which a vehicle travels. Cross-country airplane flights in the United States, for instance, generally take longer heading west than east because of prevailing wind currents.

To analyze these situations, let $$b$$ represent the speed of the vehicle in still conditions (e.g., a boat's speed in still water or a plane's speed in calm air), and let $$c$$ represent the speed of the current.

- **Downstream (with the current):** The current pushes the vehicle in the same direction it is already moving, so the actual speed is **faster** than the still-water speed. The effective rate is $$b + c$$.
- **Upstream (against the current):** The current opposes the vehicle's motion, so the actual speed is **slower** than the still-water speed. The effective rate is $$b - c$$.

These adjusted rates replace the simple rate $$r$$ in the distance formula $$d = rt$$. For example, if a boat moves at $$b$$ mph in still water and the river flows at $$c$$ mph, the distance traveled downstream in time $$t$$ is $$d = (b + c)t$$, while the distance traveled upstream in the same time is $$d = (b - c)t$$. Recognizing which direction aligns with or opposes the current is the essential first step before setting up the rate–time–distance table for these problems.

Effect of Currents on Actual Speed in Uniform Motion

Apply the distance, rate, and time problem-solving strategy to find an unknown speed when a single traveler covers two consecutive segments at different uniform rates, with the individual segment distances and the total time all known, producing a rational equation that leads to a quadratic.

**Problem:** Jazmine trained for $$3$$ hours on Saturday. She ran $$8$$ miles and then biked $$24$$ miles. Her biking speed is $$4$$ mph faster than her running speed. What is her running speed?

1. **Read and draw:** Sketch the route with two arrows — one labeled "run" ($$8$$ miles) and one labeled "bike" ($$24$$ miles). The total time for both segments is $$3$$ hours. Create a rate–time–distance table.

2. **Identify:** Jazmine's running speed.

3. **Name:** Let $$r$$ = Jazmine's running speed in mph. Then her biking speed is $$r + 4$$. Because $$D = r \cdot t$$, solving for time gives $$t = \frac{D}{r}$$. Divide each distance by its rate to fill in the time column:

| | Rate (mph) | Time (hrs) | Distance (miles) |
|---|---|---|---|
| Run | $$r$$ | $$\frac{8}{r}$$ | $$8$$ |
| Bike | $$r + 4$$ | $$\frac{24}{r + 4}$$ | $$24$$ |

4. **Translate:** The running time plus the biking time equals $$3$$ hours:

$$\frac{8}{r} + \frac{24}{r + 4} = 3$$

5. **Solve:** Multiply both sides by the LCD, $$r(r + 4)$$:

$$r(r + 4) \cdot \frac{8}{r} + r(r + 4) \cdot \frac{24}{r + 4} = 3 \cdot r(r + 4)$$

Cancel matching factors: $$8(r + 4) + 24r = 3r(r + 4)$$. Distribute: $$8r + 32 + 24r = 3r^2 + 12r$$. Combine like terms on the left: $$32r + 32 = 3r^2 + 12r$$. Rearrange to standard form:

$$0 = 3r^2 - 20r - 32$$

Factor: $$0 = (3r + 4)(r - 8)$$. Apply the Zero Product Property:

$$(3r + 4) = 0 \Rightarrow r = -\frac{4}{3}$$ or $$(r - 8) = 0 \Rightarrow r = 8$$

A negative speed is not meaningful, so discard $$r = -\frac{4}{3}$$.

6. **Check:** Running at $$8$$ mph: $$\frac{8}{8} = 1$$ hour. Biking at $$12$$ mph: $$\frac{24}{12} = 2$$ hours. Total: $$1 + 2 = 3$$ hours. $$\checkmark$$

7. **Answer:** Jazmine's running speed is $$8$$ mph.

This example demonstrates the **known-distances-with-total-time** scenario. Because the individual distances are given but the individual times are unknown, each time is expressed as a rational expression using $$t = \frac{D}{r}$$. Adding these rational expressions and setting their sum equal to the known total time produces a **rational equation**. Unlike the airplane headwind/tailwind example — where clearing the LCD yielded a linear equation — here the LCD is $$r(r + 4)$$ and clearing it produces a **quadratic** equation, $$3r^2 - 20r - 32 = 0$$. The quadratic is solved by factoring and applying the Zero Product Property, and the negative root is discarded because speed must be positive.

Finding Running Speed Using Known Segment Distances and Total Time

In the context of fleet management and logistics, what is the defining characteristic of 'uniform motion' in a travel scenario?

A logistics coordinator is calculating the fuel needs for a long-haul truck traveling at a constant speed. To find the total distance covered, the coordinator must multiply the truck's rate of speed by the ____ spent on the road.

In the context of fleet management and logistics, match each uniform motion term to its corresponding real-world description.

A logistics coordinator is modeling a travel scenario where two vehicles are moving at constant speeds. To solve this uniform motion problem algebraically, arrange the following steps in the correct standard sequence.

In a fleet logistics context, when modeling a 'catch-up' uniform motion problem where one vehicle is sent to reach another vehicle traveling on the same route, what is the primary mathematical relationship established between their distances ($d_1$ and $d_2$) to solve the equation?

Distance Relationships in Catch-Up Scenarios

In a logistics planning model, a travel scenario is classified as a 'uniform motion' problem only if the vehicle's rate of speed remains constant throughout the duration of the trip being analyzed.

In professional logistics analysis, solving uniform motion problems requires identifying how the distances of two different vehicles relate to one another. Based on the provided scenarios, state the two primary mathematical ways that the distance traveled by the first vehicle (d1) and the distance traveled by the second vehicle (d2) are typically related to form an algebraic equation.

Mathematical Modeling of Fleet Scenarios

In the field of logistics and transportation planning, 'uniform motion problems' are used to model scenarios where vehicles travel at constant speeds. Identify and describe the two primary travel setups used in these problems (specifically 'moving in opposite directions' and 'catch-up/same route' scenarios). For each setup, state the mathematical relationship between the distances traveled by the vehicles that is used to form an algebraic equation.

Categorizing Distance Relationships in Logistics Modeling

In a logistics planning model, when two delivery trucks depart from a central distribution center at the same time and travel in opposite directions, which algebraic relationship is used to determine the total distance between the two vehicles?

A logistics coordinator is using the standard formula for uniform motion to calculate the distance of a local delivery. If the delivery truck's rate is recorded in miles per hour, but the travel time is recorded in minutes, what foundational step must be taken to ensure the calculation is mathematically accurate?

Apply the seven-step problem-solving strategy to determine the time required for two vehicles traveling in opposite directions to reach a specific separation distance.

Problem: Two truck drivers depart simultaneously from an interstate rest area. The westbound truck moves at $$70$$ mph, while the eastbound truck maintains an average speed of $$60$$ mph. How many hours will elapse before they are separated by $$325$$ miles?

1. Read and draw: Sketch a diagram showing the rest stop with arrows indicating the westbound ($$70$$ mph) and eastbound ($$60$$ mph) trucks. The total separation is $$325$$ miles. Create a table with columns for Rate, Time, and Distance.
2. Identify: We need to find the travel time until the trucks are $$325$$ miles apart.
3. Name: Let $$t$$ equal the travel time in hours for both trucks. Multiply each vehicle's rate by $$t$$ to determine their individual distances. The westbound distance is $$70t$$ and the eastbound distance is $$60t$$.
4. Translate: Since the trucks are moving in opposite directions from the same starting point, the sum of their individual distances equals the total separation distance: $$70t + 60t = 325$$.
5. Solve: Combine the like terms to get $$130t = 325$$. Dividing both sides by $$130$$ yields $$t = 2.5$$.
6. Check: The westbound truck travels $$70 \cdot 2.5 = 175$$ miles. The eastbound truck travels $$60 \cdot 2.5 = 150$$ miles. The total distance is $$175 + 150 = 325$$ miles. $$\checkmark$$
7. Answer: The trucks will be $$325$$ miles apart after $$2.5$$ hours.

Solving a Uniform Motion Application: Finding Travel Time for Trucks Moving in Opposite Directions

Apply the distance, rate, and time problem-solving strategy to find an unknown speed on a round trip where both legs cover the same distance, the return speed is slower by a fixed amount, and the time difference between the two legs is known — producing a rational equation that leads to a quadratic.

**Problem:** Hamilton rode his bike downhill $$12$$ miles on the river trail from his house to the ocean and then rode uphill to return home. His uphill speed was $$8$$ miles per hour slower than his downhill speed. It took him $$2$$ hours longer to get home than it took him to get to the ocean. Find Hamilton's downhill speed.

1. **Read and draw:** Sketch two parallel paths — one labeled "downhill" ($$12$$ miles) and one labeled "uphill" ($$12$$ miles, $$8$$ mph slower, $$2$$ hours longer). Create a rate–time–distance table.
2. **Identify:** Hamilton's downhill speed.
3. **Name:** Let $$h$$ = Hamilton's downhill speed in mph. Then his uphill speed is $$h - 8$$. Because $$D = r \cdot t$$, solving for time gives $$t = \frac{D}{r}$$. Divide each distance by its rate to fill in the time column:

| | Rate (mph) | Time (hrs) | Distance (miles) |
|---|---|---|---|
| Downhill | $$h$$ | $$\frac{12}{h}$$ | $$12$$ |
| Uphill | $$h - 8$$ | $$\frac{12}{h - 8}$$ | $$12$$ |

4. **Translate:** The uphill time is $$2$$ hours more than the downhill time:
$$\frac{12}{h - 8} = \frac{12}{h} + 2$$
5. **Solve:** Multiply both sides by the LCD, $$h(h - 8)$$:
$$h(h - 8) \cdot \frac{12}{h - 8} = h(h - 8) \cdot \left(\frac{12}{h} + 2\right)$$
Cancel matching factors on the left; distribute on the right:
$$12h = 12(h - 8) + 2h(h - 8)$$
Expand: $$12h = 12h - 96 + 2h^2 - 16h$$. Collect all terms on one side:
$$0 = 2h^2 - 16h - 96$$
Factor out the GCF of $$2$$: $$0 = 2(h^2 - 8h - 48)$$. Factor the trinomial:
$$0 = 2(h - 12)(h + 4)$$
Apply the Zero Product Property:
$$h - 12 = 0 \Rightarrow h = 12 \quad \text{or} \quad h + 4 = 0 \Rightarrow h = -4$$
A negative speed is not meaningful, so discard $$h = -4$$.
6. **Check:** Downhill at $$12$$ mph: $$\frac{12}{12} = 1$$ hour. Uphill at $$12 - 8 = 4$$ mph: $$\frac{12}{4} = 3$$ hours. The uphill time is $$2$$ hours more than the downhill time. $$\checkmark$$
7. **Answer:** Hamilton's downhill speed is $$12$$ mph.

This example demonstrates the **equal-distance round trip with known time difference** scenario. Unlike the Jazmine running/biking example — where two different distances are covered and the total time is known — here both legs of the trip cover the same $$12$$ miles, and the linking condition is a **time difference** rather than a time sum. Expressing each time as $$t = \frac{D}{r}$$ produces two rational expressions, and setting one equal to the other plus $$2$$ yields a **rational equation**. Multiplying both sides by the LCD $$h(h - 8)$$ clears the denominators and produces the quadratic $$2h^2 - 16h - 96 = 0$$. After factoring out the GCF and factoring the trinomial, the Zero Product Property gives two solutions, and the negative root is discarded because speed must be positive.

Example 7.49: Finding Downhill Speed on a Round Trip Using a Known Time Difference

Apply the distance, rate, and time problem-solving strategy to find unknown speeds when different modes of transportation are used and a time difference is known.

**Problem:** Kayla rode her bike $$75$$ miles home from college one weekend and then rode the bus back to college. It took her $$2$$ hours less to ride back to college on the bus than it took her to ride home on her bike, and the average speed of the bus was $$10$$ miles per hour faster than Kayla’s biking speed. Find Kayla’s biking speed.

1. **Read and draw:** Sketch the journey with $$75$$ miles by bike and $$75$$ miles by bus. Create a rate-time-distance table.
2. **Identify:** Kayla's biking speed.
3. **Name:** Let $$r$$ = biking speed in mph. The bus speed is $$r + 10$$. The distance for both trips is $$75$$ miles. Because $$t = \frac{D}{r}$$, the biking time is $$\frac{75}{r}$$ and the bus time is $$\frac{75}{r + 10}$$.

| | Rate (mph) | Time (hrs) | Distance (miles) |
|---|---|---|---|
| Bike | $$r$$ | $$\frac{75}{r}$$ | $$75$$ |
| Bus | $$r + 10$$ | $$\frac{75}{r + 10}$$ | $$75$$ |

4. **Translate:** The bus ride took $$2$$ hours less than the bike ride, so the biking time minus $$2$$ equals the bus time:
$$\frac{75}{r} - 2 = \frac{75}{r + 10}$$
5. **Solve:** Multiply both sides by the least common denominator, $$r(r + 10)$$:
$$r(r + 10) \left(\frac{75}{r} - 2\right) = r(r + 10) \left(\frac{75}{r + 10}\right)$$
$$75(r + 10) - 2r(r + 10) = 75r$$
$$75r + 750 - 2r^2 - 20r = 75r$$
$$-2r^2 - 20r + 750 = 0$$
Divide by $$-2$$:
$$r^2 + 10r - 375 = 0$$
Factor the trinomial:
$$(r + 25)(r - 15) = 0$$
The solutions are $$r = -25$$ and $$r = 15$$. Since speed cannot be negative, discard $$r = -25$$. Therefore, $$r = 15$$ mph.
6. **Check:** Biking time: $$\frac{75}{15} = 5$$ hours. Bus time: $$\frac{75}{15 + 10} = \frac{75}{25} = 3$$ hours. The bus ride was $$5 - 3 = 2$$ hours shorter. $$\checkmark$$
7. **Answer:** Kayla's biking speed was $$15$$ mph.

Try It 7.97: Solving a Uniform Motion Application for Biking and Bus Speeds

Apply the distance, rate, and time problem-solving strategy to find an unknown jogging rate when two segments of a trip have different distances and speeds, and their time difference is known.

**Problem:** Victoria jogs $$12$$ miles to the park along a flat trail and then returns by jogging on an $$20$$ mile hilly trail. She jogs $$1$$ mile per hour slower on the hilly trail than on the flat trail, and her return trip takes her two hours longer. Find her rate of jogging on the flat trail.

1. **Read and draw:** Sketch the trip with a $$12$$-mile flat trail and a $$20$$-mile hilly trail. Create a rate-time-distance table.
2. **Identify:** Victoria's jogging rate on the flat trail.
3. **Name:** Let $$r$$ = jogging rate on the flat trail in mph. Her rate on the hilly trail is $$r - 1$$. Because $$t = \frac{D}{r}$$, the time on the flat trail is $$\frac{12}{r}$$ and the time on the hilly trail is $$\frac{20}{r - 1}$$.

| | Rate (mph) | Time (hrs) | Distance (miles) |
|---|---|---|---|
| Flat trail | $$r$$ | $$\frac{12}{r}$$ | $$12$$ |
| Hilly trail | $$r - 1$$ | $$\frac{20}{r - 1}$$ | $$20$$ |

4. **Translate:** The hilly trail took $$2$$ hours longer than the flat trail, so the flat trail time plus $$2$$ equals the hilly trail time:
$$\frac{12}{r} + 2 = \frac{20}{r - 1}$$
5. **Solve:** Multiply both sides by the least common denominator, $$r(r - 1)$$:
$$r(r - 1) \left(\frac{12}{r} + 2\right) = r(r - 1) \left(\frac{20}{r - 1}\right)$$
$$12(r - 1) + 2r(r - 1) = 20r$$
$$12r - 12 + 2r^2 - 2r = 20r$$
$$2r^2 + 10r - 12 = 20r$$
$$2r^2 - 10r - 12 = 0$$
Divide by $$2$$:
$$r^2 - 5r - 6 = 0$$
Factor the trinomial:
$$(r - 6)(r + 1) = 0$$
The solutions are $$r = 6$$ and $$r = -1$$. Since speed cannot be negative, discard $$r = -1$$. Therefore, $$r = 6$$ mph.
6. **Check:** Flat trail time: $$\frac{12}{6} = 2$$ hours. Hilly trail time: $$\frac{20}{6 - 1} = \frac{20}{5} = 4$$ hours. The return trip took $$4 - 2 = 2$$ hours longer. $$\checkmark$$
7. **Answer:** Victoria's rate of jogging on the flat trail was $$6$$ mph.

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