Round-Trip Problems: Same Distance, Different Speeds

Distance & Rate 9th-10th Grade
Problem
Brantley and Shawna walk to Grandma's house at 4 mi/hr and ride back at 8 mi/hr on the same route. Walking takes 1 hour longer than riding. How long did it take them to walk?

What This Problem Teaches

  • How to use the distance formula (distance = speed × time) when the same distance is traveled at different speeds
  • Setting up equations based on time relationships between two parts of a journey
  • Recognizing that round-trip problems rely on the fact that both legs cover identical distances
  • Translating word phrases like "takes 1 hour longer" into algebraic expressions
  • Solving linear equations that arise from real-world motion scenarios

Picture This

Let's visualize what's happening in this round-trip journey:

Brantley and Shawna walk to Grandma's house at 4 mi/hr and ride back at 8 mi/hr on the same route. Walking takes 1...

The key insight is that both legs of the journey cover exactly the same distance, but at different speeds and for different amounts of time. The walking leg takes 1 hour longer than the riding leg.

Solution: Method 1 — Time Relationship Approach

Since we're asked for the walking time, let's make that our variable and work from there.

Step 1 — Define the variable

Let t = time to walk to Grandma's house (in hours). This is what we want to find.

Step 2 — Express the riding time

We're told "walking takes 1 hour longer than riding." If walking takes t hours, then riding takes (t - 1) hours.

Step 3 — Set up the distance equation

Both legs cover the same distance. Using distance = speed × time:

Walking distance = Riding distance
4 × t = 8 × (t - 1)

Step 4 — Solve the equation

Expand the right side:

4t = 8(t - 1)
4t = 8t - 8

Collect like terms:

4t - 8t = -8
-4t = -8
t = 2

Therefore, it took 2 hours to walk to Grandma's house.

Solution: Method 2 — Distance as the Primary Variable

Instead of starting with time, let's work with the distance and see how the times relate.

Step 1 — Define the distance variable

Let d = distance to Grandma's house (in miles).

Step 2 — Express both times in terms of distance

Using time = distance ÷ speed:

  • Walking time = d ÷ 4 hours
  • Riding time = d ÷ 8 hours

Step 3 — Set up the time relationship equation

Walking takes 1 hour longer than riding:

Walking time = Riding time + 1
d/4 = d/8 + 1

Step 4 — Solve for the distance

Multiply everything by 8 to clear fractions:

8 × (d/4) = 8 × (d/8) + 8 × 1
2d = d + 8
d = 8

Step 5 — Find the walking time

Now that we know the distance is 8 miles:

Walking time = 8 ÷ 4 = 2 hours
It took 2 hours to walk to Grandma's house.

Verification

Let's check our answer by substituting back into the original conditions.

Check the times:
• Walking time: 2 hours
• Riding time: 2 - 1 = 1 hour
• Time difference: 2 - 1 = 1 hour ✓
Check the distances:
• Walking distance: 4 mi/hr × 2 hr = 8 miles
• Riding distance: 8 mi/hr × 1 hr = 8 miles ✓
• Both distances are equal, as required.

Our answer satisfies both conditions: the walking takes exactly 1 hour longer than the riding, and both legs cover the same 8-mile distance.

Watch Out For These

✗ Mistake 1: Averaging the speeds
Some students think: "The average speed is (4 + 8) ÷ 2 = 6 mi/hr, so I can work with that." This is wrong because time is not evenly split between the two speeds. The slower leg takes longer, so you can't just average the rates.
✗ Mistake 2: Mixing up which time is longer
Writing "riding time = walking time + 1" instead of "walking time = riding time + 1." Since walking is slower, it must take longer. Always double-check that your equation reflects the logical relationship.
✗ Mistake 3: Setting up the equation backwards
Writing "8t = 4(t - 1)" instead of "4t = 8(t - 1)." Remember: distance = speed × time. The walking equation is 4 × time, not 8 × time.

The General Pattern

This problem belongs to a family called "round-trip with different rates." The general structure is:

If speed₁ × time₁ = speed₂ × time₂ (same distance)
and time₁ = time₂ + k (time difference of k)
then speed₁ × (time₂ + k) = speed₂ × time₂

Solving this general equation:

speed₁ × time₂ + speed₁ × k = speed₂ × time₂
speed₁ × k = (speed₂ - speed₁) × time₂
time₂ = (speed₁ × k) ÷ (speed₂ - speed₁)

In our case: speed₁ = 4, speed₂ = 8, k = 1, so time₂ = (4 × 1) ÷ (8 - 4) = 1 hour for riding, and time₁ = 2 hours for walking.

Important limitation: This pattern only works when the slower method takes longer. If the problem were reversed (riding takes longer than walking), the relationship would be impossible given realistic speeds.

How to Spot This Problem Type

Round-trip problems with different speeds typically include these telltale phrases:

  • "same route" or "same path" — signals that both distances are equal
  • "takes __ longer than" or "takes __ more time" — gives you the time relationship
  • Two different speeds mentioned for the same journey
  • "round trip" or "there and back" — confirms you're dealing with equal distances

Variations you might see include walking vs. biking, driving vs. flying, upstream vs. downstream, or any scenario where the same distance is covered at two different rates.

Don't confuse with: Average speed problems, where you're asked to find the overall speed for the entire trip. Those require the total distance divided by total time, not the individual leg analysis we used here.

Where This Shows Up in Real Life

This type of calculation appears in several practical contexts:

  • Commute planning: Walking to the bus stop vs. getting a ride back, with different time constraints for each direction.
  • Delivery logistics: Trucks traveling empty (faster) vs. loaded (slower) on return trips, with fuel and time costs depending on the leg durations.
  • Exercise planning: Running uphill vs. jogging back down, where the terrain creates natural speed differences but the distance remains constant.

What If?

1
Different Speed Ratio
They walk to Grandma's house at 3 mi/hr and ride back at 12 mi/hr on the same route. Walking takes 3 hours longer than riding. How long did it take them to walk?
Step 1 — Define the variable

Let t = time to walk (in hours).

Step 2 — Express riding time

Since walking takes 3 hours longer than riding: riding time = t - 3 hours.

Step 3 — Set up distance equation

Walking distance = Riding distance
3t = 12(t - 3)

Step 4 — Solve

3t = 12t - 36
-9t = -36
t = 4

Step 5 — Verify

Walking: 4 hours, distance = 3 × 4 = 12 miles
Riding: 1 hour, distance = 12 × 1 = 12 miles ✓
Time difference: 4 - 1 = 3 hours ✓

Answer: 4 hours

2
Find the Unknown Speed
They walk to Grandma's house at 5 mi/hr and ride back at an unknown speed. Walking takes 1.5 hours longer than riding, and the distance is 10 miles each way. What is the riding speed?
Step 1 — Find walking time

Distance = 10 miles, speed = 5 mi/hr
Walking time = 10 ÷ 5 = 2 hours

Step 2 — Find riding time

Walking takes 1.5 hours longer than riding
Riding time = 2 - 1.5 = 0.5 hours

Step 3 — Calculate riding speed

Speed = distance ÷ time
Riding speed = 10 ÷ 0.5 = 20 mi/hr

Step 4 — Verify

Walking: 2 hours at 5 mi/hr = 10 miles ✓
Riding: 0.5 hours at 20 mi/hr = 10 miles ✓
Time difference: 2 - 0.5 = 1.5 hours ✓

Answer: 20 mi/hr

3
Total Round-Trip Time
They walk at 4 mi/hr and ride back at 10 mi/hr. The walking leg takes 2 hours longer than the riding leg. What is the total time for the complete round trip?
Step 1 — Set up variables

Let t = walking time, so riding time = t - 2

Step 2 — Distance equation

4t = 10(t - 2)
4t = 10t - 20
-6t = -20
t = 20/6 = 10/3 hours

Step 3 — Find individual times

Walking time: 10/3 hours
Riding time: 10/3 - 2 = 10/3 - 6/3 = 4/3 hours

Step 4 — Calculate total time

Total = walking + riding = 10/3 + 4/3 = 14/3 hours
14/3 = 4⅔ hours = 4 hours 40 minutes

Step 5 — Verify

Walking: (10/3) × 4 = 40/3 miles
Riding: (4/3) × 10 = 40/3 miles ✓
Time difference: 10/3 - 4/3 = 6/3 = 2 hours ✓

Answer: 4⅔ hours (4 hours 40 minutes)

4
Three-Leg Journey with a Stop
Walk to Grandma's at 3 mi/hr, stay for 1 hour, then ride back at 9 mi/hr. The total trip (including the 1-hour visit) takes 6 hours. How long did the walking take?
Step 1 — Define variables

Let d = distance to Grandma's house
Walking time = d/3, riding time = d/9

Step 2 — Set up total time equation

Total time = walking + visit + riding = 6 hours
d/3 + 1 + d/9 = 6

Step 3 — Solve for distance

Subtract the 1-hour visit: d/3 + d/9 = 5
Common denominator: 3d/9 + d/9 = 5
4d/9 = 5
d = 45/4 = 11.25 miles

Step 4 — Calculate walking time

Walking time = distance ÷ speed = 11.25 ÷ 3 = 3.75 hours
3.75 hours = 3 hours 45 minutes

Step 5 — Verify

Walking: 3.75 hours
Visit: 1 hour
Riding: 11.25 ÷ 9 = 1.25 hours
Total: 3.75 + 1 + 1.25 = 6 hours ✓

Answer: 3.75 hours (3 hours 45 minutes)

Frequently Asked Questions

How do you solve round-trip problems with different speeds? +
Use the fact that both legs cover the same distance. Set up the equation: distance = speed × time for each leg, then use the given time relationship. In this problem, if walking takes time t, then riding takes time (t-1), so 4t = 8(t-1), giving us t = 2 hours.
Why do you set the distances equal in round-trip problems? +
Because it's the same route traveled in both directions. The distance formula is distance = speed × time, so if the walking distance equals the riding distance, you can write one equation relating the two different speeds and times.
What's the most common mistake in round-trip speed problems? +
Trying to average the speeds instead of using the time relationship. You cannot simply average 4 mi/hr and 8 mi/hr to get 6 mi/hr. The speeds and times are related through the distance equation, not through arithmetic averaging.
DN

Dr. Neven Jurkovic

Mathematics educator with 15+ years of experience helping students master algebra and problem-solving strategies.

NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

Contact

This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-09-02

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