Apply the uniform-motion strategy for two vehicles using a rate–time–distance table.

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When a uniform motion problem involves two travelers or vehicles, the general formula-based strategy is enhanced with visual and organizational tools. The adapted seven-step process is:

1. **Read** the problem and make sure all the words and ideas are understood.
   - **Draw a diagram** to illustrate the situation (e.g., sketch the route with labeled arrows for each traveler).
   - **Create a table** with columns for Rate, Time, and Distance, and one row for each scenario.
   - Fill in the information that is given.
2. **Identify** what you are looking for.
3. **Name** the unknown by choosing a variable, then complete the table. Use variable expressions to represent unknown quantities in each row, and multiply rate by time to fill in the distance column (since $$d = rt$$).
4. **Translate** into an equation by restating the relationship between the two distances in one sentence, then converting that sentence into an equation.
5. **Solve** the equation using standard algebra techniques.
6. **Check** the answer in the original problem to make sure it is reasonable.
7. **Answer** the question with a complete sentence.

The key adaptations for distance, rate, and time problems are the diagram and the rate–time–distance table, which organize the two-scenario information visually and make it straightforward to identify the equation connecting the distances.

Problem-Solving Strategy for Distance, Rate, and Time Applications

A logistics manager is training a new dispatcher on the company's standard 7-step protocol for solving distance, rate, and time problems for delivery routes. Match each step of the problem-solving strategy with the specific action required to complete it.

A logistics dispatcher is using the company's 7-step strategy to coordinate the meeting point of two delivery trucks. When completing the 'Distance' column of the organizational table in Step 3, which formula must be recalled to correctly calculate the distance expressions?

A logistics supervisor is training a new dispatcher to use a standard 7-step strategy for calculating delivery schedules. True or False: This strategy recommends that the dispatcher should draw a diagram and create a rate-time-distance table during the very first step ('Read') of the process.

A logistics supervisor is training a new dispatcher to use a standard 7-step problem-solving strategy for calculating delivery schedules. According to the 'Translate' step (Step 4) of this process, what specific action must the dispatcher take to organize their thoughts immediately before converting the information into a mathematical equation?

Procedural Step for Translating Motion Scenarios

Based on the 7-step problem-solving strategy for distance, rate, and time applications, identify which specific step the dispatcher is performing and state the criteria used to evaluate the answer in this step.

Evaluating Delivery Route Feasibility

A logistics supervisor is training a new dispatcher on the company's 7-step strategy for solving distance, rate, and time problems involving two delivery vehicles. To help visualize the route and the relative directions of the vehicles, the dispatcher is instructed in Step 1 to draw a(n) ____.

Imagine you are a training supervisor at a logistics company. A new dispatcher is learning how to calculate delivery schedules for routes involving two separate vehicles. To ensure consistency and accuracy, the company uses a standardized 7-step problem-solving strategy for distance, rate, and time applications. Write a short essay that lists each of the seven steps of this strategy in the correct sequence. In your description of the first step, identify the two specific visual or organizational tools the dispatcher is required to create and briefly state why these tools are important for organizing information in problems involving two different scenarios.

Training Manual: The 7-Step Motion Strategy

A logistics supervisor is training a team on the company's 7-step strategy for solving distance, rate, and time problems involving two delivery vehicles. According to the strategy, what is the primary purpose of the 'key adaptations'—specifically the diagram and the rate–time–distance table?

A logistics coordinator is training a new dispatcher on the company's 7-step strategy for solving distance, rate, and time problems for delivery routes. According to the instructions for Step 6 ('Check'), where should the dispatcher verify their calculated answer to ensure it is accurate and reasonable?

Apply the distance, rate, and time problem-solving strategy to find two unknown uniform motion speeds when both travelers cover an identical route and distance but travel at different uniform rates and times.

Problem: Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis's speed is $$7$$ miles per hour faster than Wayne's speed, so it takes Wayne $$2$$ hours to ride to the beach while it takes Dennis $$1.5$$ hours for the ride. Find the speed of both bikers.

1. Read and draw: Sketch the situation. Both bikers travel the exact same route from the park to the beach, meaning they cover the exact same distance. Note the specific times and relative speeds. Create a rate–time–distance table.
2. Identify what you are looking for: you are asked to find the speed (rate) of both bikers.
3. Name the unknowns using a single variable. Let $$r$$ = Wayne's speed. Because Dennis’s speed is $$7$$ mph faster, we represent that as $$r + 7$$. Fill this information into the chart and multiply the rate by the time to get the respective distance expression:

| Biker | Rate (mph) | Time (hrs) | Distance (miles) |
|---|---|---|---|
| Dennis | $$r + 7$$ | $$1.5$$ | $$1.5(r + 7)$$ |
| Wayne | $$r$$ | $$2$$ | $$2r$$ |

4. Translate into an equation. Because both bikers leave from Riverside Park and travel to the beach, they travel the same distance. The equation to model this situation comes from equating their two distance expressions: 
$$1.5(r + 7) = 2r$$

5. Solve the equation using standard algebra techniques:
- Distribute the $$1.5$$ factor: $$1.5r + 10.5 = 2r$$
- Subtract $$1.5r$$ from both sides to combine variables: $$10.5 = 0.5r$$
- Divide both sides by $$0.5$$ (which is the same as multiplying by $$2$$): $$21 = r$$

Wayne's speed is $$21$$ mph. Now find Dennis's speed using his specific expression, $$r + 7$$: 
$$21 + 7 = 28$$

6. Check the results and verify they make mathematical sense in context:
- For Dennis, traveling at $$28$$ mph for $$1.5$$ hours implies $$28(1.5) = 42$$ miles.
- For Wayne, traveling at $$21$$ mph for $$2$$ hours implies $$21(2) = 42$$ miles.
The distances are truly equal. $$\checkmark$$

7. Answer the question with a complete sentence: Wayne rode at $$21$$ mph and Dennis rode at $$28$$ mph.

This application uses the principle that parallel trips covering the same route have equal distances. By defining the faster traveler's speed as an additive shift from the slower traveler's speed and multiplying by their specific times ($$D=rt$$), an intuitive linear equation can be set up directly from equating those two distance outputs without ever needing to know the actual distance in miles beforehand.

Solving a Uniform Motion Application: Finding Biking Speeds Using Equal Distances

Apply the distance, rate, and time problem-solving strategy to find an unknown speed when the distances for two segments are known and the difference in their travel times is given, producing a rational equation that often leads to a quadratic equation.

**Problem:** Hamilton rode his bike downhill $$12$$ miles to the ocean and then rode uphill $$12$$ miles to return home. His uphill speed was $$8$$ mph slower than his downhill speed. It took him $$2$$ hours longer to get home than it took him to get to the ocean. Find his downhill speed.

1. **Read and draw:** Sketch the route with an arrow downhill ($$12$$ miles) and an arrow uphill ($$12$$ miles). Create a rate–time–distance table.
2. **Identify:** Hamilton's downhill speed.
3. **Name:** Let $$h$$ = Hamilton's downhill speed in mph. His uphill speed is $$h - 8$$. The distance in both directions is $$12$$ miles. Using $$t = \frac{D}{r}$$, 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 longer than the downhill time:
$$\frac{12}{h - 8} = \frac{12}{h} + 2$$
5. **Solve:** Multiply both sides by the LCD, $$h(h - 8)$$, to clear the fractions:
$$h(h - 8) \cdot \frac{12}{h - 8} = h(h - 8) \left( \frac{12}{h} + 2 \right)$$
$$12h = 12(h - 8) + 2h(h - 8)$$
$$12h = 12h - 96 + 2h^2 - 16h$$
$$0 = 2h^2 - 16h - 96$$
Divide by $$2$$:
$$0 = h^2 - 8h - 48$$
Factor:
$$0 = (h - 12)(h + 4)$$
So, $$h = 12$$ or $$h = -4$$. Discard the negative speed since it is not physically meaningful.
6. **Check:** Is $$12$$ mph a reasonable downhill biking speed? Yes. Downhill time is $$\frac{12}{12} = 1$$ hour. Uphill speed is $$12 - 8 = 4$$ mph, so uphill time is $$\frac{12}{4} = 3$$ hours. The uphill time ($$3$$ hours) is $$2$$ hours more than the downhill time ($$1$$ hour). $$\checkmark$$
7. **Answer:** Hamilton's downhill speed is $$12$$ mph.

This example illustrates the **known-distances-with-time-difference** scenario. Because individual times are unknown, each is represented as a rational expression. Equating one time to the other plus the given difference produces a **rational equation**, which, after clearing denominators, can be solved using standard quadratic techniques.

Finding Speed Using Known Distances and Time Difference

Apply the distance, rate, and time problem-solving strategy to find unknown speeds when a trip is divided into segments with a known total time.

Problem: Joon drove $$4$$ hours to his home, driving $$208$$ miles on the interstate and $$40$$ miles on country roads. If he drove $$15$$ mph faster on the interstate than on the country roads, what was his rate on the country roads?

1. Read and draw: Sketch the two segments of the trip: $$208$$ miles on the interstate and $$40$$ miles on country roads. Create a rate-time-distance table.
2. Identify: Joon's driving rate on the country roads.
3. Name: Let $$r$$ = Joon's rate on country roads. His interstate rate is $$r + 15$$. The time spent on the interstate is $$\frac{208}{r + 15}$$ and the time spent on country roads is $$\frac{40}{r}$$.
4. Translate: The total driving time is $$4$$ hours, meaning the sum of the times is $$4$$:
$$\frac{208}{r + 15} + \frac{40}{r} = 4$$
5. Solve: Multiply both sides by the least common denominator, $$r(r + 15)$$:
$$208r + 40(r + 15) = 4r(r + 15)$$
$$208r + 40r + 600 = 4r^2 + 60r$$
$$248r + 600 = 4r^2 + 60r$$
$$0 = 4r^2 - 188r - 600$$
Divide by $$4$$:
$$0 = r^2 - 47r - 150$$
$$0 = (r - 50)(r + 3)$$
The solutions are $$r = 50$$ and $$r = -3$$. Speed cannot be negative, so $$r = 50$$ mph.
6. Check: Interstate time: $$\frac{208}{50 + 15} = \frac{208}{65} = 3.2$$ hours. Country road time: $$\frac{40}{50} = 0.8$$ hours. Total time: $$3.2 + 0.8 = 4$$ hours. $$\checkmark$$
7. Answer: Joon's rate on the country roads was $$50$$ mph.

Solving a Uniform Motion Application: Driving Rates on Different Roads

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 two unknown uniform speeds for consecutive trip segments when the segment distances add up to a known total.

**Problem:** Phuong left home on his bicycle at 10:00. He rode on the flat street until 11:15, then rode uphill until 11:45. He rode a total of $$31$$ miles. His speed riding uphill was $$0.6$$ times his speed on the flat street. Find his speed biking uphill and on the flat street.

1. **Read and draw:** Sketch the route from home with two segments: flat street (10:00 to 11:15) and uphill (11:15 to 11:45). The total distance is $$31$$ miles. Create a rate–time–distance table.
2. **Identify:** Phuong's biking speed on the flat street and his biking speed uphill.
3. **Name:** Let $$r$$ = the speed on the flat street in mph. Because his uphill speed was $$0.6$$ times as fast, it equals $$0.6r$$. Convert the clock times to elapsed times: riding on the flat street lasts from 10:00 to 11:15, which is $$1.25$$ hours; riding uphill lasts from 11:15 to 11:45, which is $$0.5$$ hours. Multiply rate by time to fill in the distance column:

| | Rate (mph) | Time (hrs) | Distance (miles) |
|---|---|---|---|
| **Flat** | $$r$$ | $$1.25$$ | $$1.25r$$ |
| **Uphill** | $$0.6r$$ | $$0.5$$ | $$0.5(0.6r)$$ |
| **Total** | | | $$31$$ |

4. **Translate:** The flat distance plus the uphill distance equals the total of $$31$$ miles:
$$1.25r + 0.5(0.6r) = 31$$
5. **Solve:** Multiply $$0.5 \cdot 0.6 = 0.3$$:
$$1.25r + 0.3r = 31$$
Combine like terms: $$1.55r = 31$$. Divide both sides by $$1.55$$:
$$r = 20$$
The speed on the flat street is $$20$$ mph. The uphill speed is $$0.6 \cdot 20 = 12$$ mph.
6. **Check:** Flat street: $$20 \cdot 1.25 = 25$$ miles. Uphill: $$12 \cdot 0.5 = 6$$ miles. Total distance: $$25 + 6 = 31$$ miles. $$\checkmark$$
7. **Answer:** Phuong's speed biking on the flat street was $$20$$ mph and his speed biking uphill was $$12$$ mph.

Finding Flat and Uphill Biking Speeds Using a Known Total Distance

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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