Solve Distance, Rate, and Time Using Two Different Speeds
David gets to work in 20 minutes when he drives his car. Riding his bike (by the same route) takes him 45 minutes. His average driving speed is 4.5 mph greater than his average speed on his bike. How far does he travel for work?
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Setup: same route means same distance, so (1/3)(b + 4.5) = (3/4)b, where b is the bike speed in mph.
Result: bike 3.6 mph, car 8.1 mph, distance 2.7 miles.
One-line check:8.1 × 1/3 = 2.7 and 3.6 × 3/4 = 2.7. ✓
If your answer was 162 or 0.045, jump to the pitfalls section below.
What You Will Learn
- The "same trip" principle. When two trips cover the same route, the distances are equal. That single fact turns a word problem into an equation.
- Unit discipline. Why minutes must become hours before you multiply by a speed in mph, and what goes wrong when they don't.
- Describing two related speeds with one variable. "4.5 mph greater" becomes
b + 4.5. - Inverse proportionality. For a fixed distance, speed and time trade off exactly, which gives a second, faster solution path.
- A reusable formula. One expression solves every problem of this shape, so you never have to rebuild it from scratch.
Before You Tackle This
Three small skills are all you need. If any feel shaky, a quick review will save you time:
- The relationship
distance = rate × timeand its rearrangements. - Converting minutes to hours by dividing by 60.
- Solving a linear equation with fractions, for example by multiplying both sides by a common denominator.
Solution: Method 1 — Set the Two Distances Equal
The key insight is that David travels the same route either way. The car and the bike cover an identical distance, and only the speed and the time differ. That lets us write the distance twice and make the two expressions equal.
Step 1 — Convert the times to hours
Speeds are in miles per hour, so the times must be in hours too:
45 min = 45/60 = 3/4 hour
Step 2 — Name the unknown speed
We know how the two speeds relate to each other but not their actual values, so one variable can describe both. Let b be David's biking speed in mph. The car is 4.5 mph faster, so its speed is b + 4.5.
Step 3 — Write distance = rate × time for each trip
Bike: d = b × (3/4)
Step 4 — Set the distances equal and solve
Both expressions equal the same d, so they equal each other. The denominators are 3 and 4, so multiply every term by 12 to clear the fractions:
4(b + 4.5) = 9b
4b + 18 = 9b
18 = 5b
b = 3.6
So David bikes at 3.6 mph and drives at 3.6 + 4.5 = 8.1 mph.
Step 5 — Answer the actual question
The problem asks for the distance, not a speed. This is the step students most often forget. Use either trip:
Solution: Method 2 — The Speed-Ratio Approach
This method uses no equation with fractions at all. Over a fixed distance, speed and time are inversely proportional: a trip that takes half as long must be twice as fast. So the time ratio tells us the speed ratio directly.
Step 1 — Compare the times
Car time : bike time = 20 : 45 = 4 : 9. Minutes work fine here because only the ratio matters.
Step 2 — Flip it to get the speed ratio
Speed is inversely proportional to time, so car speed : bike speed = 9 : 4. Think of the car as going 9 "parts" per hour and the bike as going 4 "parts" per hour.
Step 3 — Use the speed difference to size one part
The car is 9 − 4 = 5 parts faster than the bike, and the problem says that gap is 4.5 mph:
1 part = 0.9 mph
Bike: 4 × 0.9 = 3.6 mph. Car: 9 × 0.9 = 8.1 mph.
Step 4 — Convert speed and time to distance
Same answer, with the algebra replaced by a bit of proportional reasoning.
A third route, for completeness: let the unknown be the distance d itself. Then speed = distance ÷ time, so the car goes d ÷ (1/3) = 3d mph and the bike goes d ÷ (3/4) = 4d/3 mph. Their difference is 4.5: 3d − 4d/3 = 5d/3 = 4.5, so d = 2.7. Choosing what to call x is a real decision. Here it is a matter of taste, but in harder problems the right choice can make the algebra much shorter.
The Answer
David travels 2.7 miles to work (one way).
Biking speed: 3.6 mph | Driving speed: 8.1 mph
Verification
Check the answer against both trips and against the speed condition in the original problem:
| Trip | Speed | Time | Distance = speed × time |
|---|---|---|---|
| Car | 8.1 mph | 20 min = 1/3 h | 8.1 × 1/3 = 2.7 mi ✓ |
| Bike | 3.6 mph | 45 min = 3/4 h | 3.6 × 3/4 = 2.7 mi ✓ |
Both trips give 2.7 miles, and the speed difference is 8.1 − 3.6 = 4.5 mph, exactly as the problem states. ✓
Sanity Check
Is a 2.7-mile commute with an 8.1 mph "driving" average believable? It is slow for open road, but fine for city streets with stoplights, and a 3.6 mph bike speed is leisurely. The numbers are textbook-friendly rather than realistic, but they are internally consistent.
A stronger check is the proportion. The bike trip takes 45 ÷ 20 = 2.25 times as long as the car trip, so the car should be 2.25 times as fast. And 8.1 ÷ 3.6 = 2.25. ✓
It is also worth knowing the boundaries. If the car were only barely faster, the two times would be nearly equal. A 25-minute gap in travel time can only come from a sizable speed gap, and here the car's speed is more than double the bike's.
What Trips People Up
✗ Skipping the final conversion: 20(b + 4.5) = 45b → b = 3.6 → d = 45 × 3.6 = 162
This one is subtle, because it gets the speed right. Using minutes on both sides scales both expressions by the same factor, so b = 3.6 still comes out correctly. But the "distance" 162 is in mph·minutes, which is not miles. Dividing by 60 rescues it (162 ÷ 60 = 2.7), but the safer habit is to convert to hours before writing any equation.
✗ Mixing units: 20(b + 4.5) = (3/4)b
Here one time is in minutes and the other in hours. The equation becomes 19.25b = −90, a negative speed. A negative or absurd result is a strong signal that the units are inconsistent. Always ask whether every time in the equation is in the same unit as the speed.
✗ 4.5 mph × 25 min = 4.5 × (25/60) = 1.875 miles
It is tempting to link the 4.5 mph speed gap to the 25-minute time gap. But those two gaps belong to different quantities: the speed gap is between two speeds, while the time gap is between two durations over the same distance. The relationship between them is multiplicative (speed × time = distance), not additive, so subtracting or multiplying the differences does not work.
The Tell-Tale Signs
This problem type is easy to spot once you know the clues:
- "By the same route" / "the same trip" / "to and from" signals that the distances are equal.
- Two different times for one journey, with the speeds related by "greater than" or "times as fast as."
- No distance given even though the question asks for it. The distance is the shared quantity you use to connect the two trips.
- Times given in minutes but speeds in mph is a warning to convert units.
The same structure appears in problems like "A plane flies with the wind in 3 hours and against it in 4 hours..." and "Two cyclists ride the same loop at different speeds..." In each, the equal-distance principle is the tool.
The General Formula
Suppose a trip takes t₁ hours at the faster speed and t₂ hours at the slower speed, and the speeds differ by Δ. With slow speed s, we have (s + Δ)t₁ = s·t₂. Solving for s gives s = Δt₁ / (t₂ − t₁), and multiplying by t₂ gives the distance:
Plug in this problem's values: Δ = 4.5, t₁ = 1/3, t₂ = 3/4.
= 4.5 × (1/4) / (5/12)
= 4.5 × 0.6 = 2.7 miles
Use with care: this formula only works when both trips cover the same distance and the times are in hours. Understanding why it works matters more than memorizing it, because variations (a headwind, a stop partway) break it immediately, while the equal-distance idea still carries you through.
Beyond the Textbook
- Commute planning. Transit apps compare modes of travel over one route exactly this way: same distance, different speeds, different times.
- Logistics. Delivery companies compare truck and rail transit times over the same corridor to estimate effective speeds.
- Physics. Equal displacement at two different velocities is the basis for many kinematics problems, from relative motion to Doppler-style timing.
Push Further
Try each variation yourself first, then open the solution to check your work. The sequence builds in difficulty.
David's trip times are unchanged (20 minutes by car, 45 minutes by bike), but now his driving speed is 13.5 mph greater than his biking speed. How far is his route?
20 min = 1/3 h and 45 min = 3/4 h.
Let b be the bike speed. The car speed is b + 13.5. Equal distances give (1/3)(b + 13.5) = (3/4)b.
Multiply by 12: 4b + 54 = 9b, so 5b = 54 and b = 10.8 mph. The car goes 24.3 mph.
d = 10.8 × 3/4 = 8.1, so the route is 8.1 miles.
Car: 24.3 × 1/3 = 8.1 ✓. The speed gap tripled and so did the distance (2.7 × 3 = 8.1). With the times fixed, distance is proportional to the speed gap.
A different commuter's route is 6 miles. Driving takes 20 minutes and biking takes 45 minutes. By how many mph is the driving speed greater than the biking speed?
20 min = 1/3 h, 45 min = 3/4 h.
Now the distance is known, so each speed comes directly from division.
Car: 6 ÷ (1/3) = 18 mph. Bike: 6 ÷ (3/4) = 8 mph.
18 − 8 = 10, so the car is 10 mph faster.
Check the ratio: 18 ÷ 8 = 2.25 = 45 ÷ 20 ✓. And 18 × 1/3 = 6, 8 × 3/4 = 6 ✓.
Using David's original numbers (a 2.7-mile route, biking at 3.6 mph, driving at 8.1 mph), suppose he could instead take a bus that travels twice as fast as his bike. How many minutes would the bus ride take, and how many minutes faster or slower is it than driving?
Twice the bike speed: 2 × 3.6 = 7.2 mph.
time = distance ÷ speed = 2.7 ÷ 7.2 = 0.375 hours.
0.375 × 60 = 22.5 minutes.
The car takes 20 minutes, so the bus is 22.5 − 20 = 2.5 minutes slower. The bus takes 22.5 minutes, 2.5 minutes longer than driving.
Twice the bike speed means half the bike time: 45 ÷ 2 = 22.5 min ✓. Also the bus (7.2 mph) is slightly slower than the car (8.1 mph), consistent with its slightly longer time.
David bikes to work (2.7 miles) in 45 minutes. On the way home he bikes the same route, mostly downhill, in only 27 minutes. What is his average speed for the entire round trip? (Is it the average of his two speeds?)
Going: 2.7 ÷ 0.75 = 3.6 mph. Returning: 27 min = 0.45 h, so 2.7 ÷ 0.45 = 6 mph.
Averaging the two speeds gives (3.6 + 6) ÷ 2 = 4.8 mph. This treats both speeds as if he rode at each for equal time, but he spent more time at the slower speed.
Distance: 2.7 + 2.7 = 5.4 miles. Time: 45 + 27 = 72 min = 1.2 hours.
5.4 ÷ 1.2 = 4.5, so his average speed is 4.5 mph, lower than 4.8 mph.
4.5 × 1.2 = 5.4 miles ✓. The answer lies between 3.6 and 6, and closer to the slower speed, as it should since he spent more time at the slower speed.
Frequently Asked Questions
How do you solve a distance-rate-time problem when the same trip is made at two different speeds?
Because the route is the same, the distance is the same for both trips, so you can write distance = rate × time for each trip and set the two expressions equal. In this example, a car trip of 1/3 hour at speed b + 4.5 and a bike trip of 3/4 hour at speed b give (1/3)(b + 4.5) = (3/4)b. Multiplying by 12 gives 4b + 18 = 9b, so b = 3.6 mph and the distance is (3/4)(3.6) = 2.7 miles.
Why do I need to convert minutes to hours in a rate problem?
Speeds in miles per hour only produce miles when they are multiplied by time in hours. If you multiply mph by minutes, the result is in the wrong units. In this problem, 20 minutes is 20/60 = 1/3 hour and 45 minutes is 45/60 = 3/4 hour. Skipping that conversion would give a distance 60 times too large (162 instead of 2.7).
What is the relationship between speed and time when the distance is fixed?
Speed and time are inversely proportional: the trip that takes less time must be proportionally faster. In this example, a 20-minute trip and a 45-minute trip over the same route have a time ratio of 4 : 9, so the speed ratio is 9 : 4. If the faster speed is 9 parts and the slower is 4 parts, their difference is 5 parts. Setting 5 parts equal to the 4.5 mph difference gives 1 part = 0.9 mph, so the speeds are 8.1 mph and 3.6 mph.
2026-07-20