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Algebra / Solving Equations and Inequalities

Distance, Rate, and Time Problems

Every motion word problem runs on one formula, d = rt, and one setup tool, a small distance-rate-time chart. Fill the chart with one row per moving object, put the known quantities in, write the unknown quantities as expressions in a single variable, and one relationship between the distances gives you the equation. This lesson works through the four scenarios that cover almost every motion problem you'll see.

Practice Problems
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Motion problems intimidate people out of proportion to how hard they are. There is exactly one formula, , and one organizing tool, a three-column chart. The Word Problems five-step method still applies; this lesson just adds the chart, which makes the translation step almost mechanical.

The whole trick is filling the chart with one row per moving thing, using a single variable, and then finding the one sentence in the problem that relates the distances. That sentence becomes the equation, and the equation is linear.

The Distance, Rate, and Time Formula

Distance, Rate, and Time Problems — key formula
Key formula

is distance, is rate (speed), and is time. A car at mph for hours covers miles. Every motion problem applies this formula once per moving object, then links the objects with a single equation about their distances.

The Distance-Rate-Time Chart

Set up a table with one row per object:

ObjectRateTimeDistance
Object 1
Object 2

Fill in every rate and time you are given. Name one unknown with a variable and write the others in terms of it. The Distance column is always rate times time — you never enter a distance directly, you compute it. The equation comes from a relationship between the two Distance entries.

Scenario 1: Moving Toward Each Other (or Apart)

Two objects start at different points and move toward each other, or start at the same point and move in opposite directions. Their distances add to the total gap.

If they start at the same time, both travel for the same .

Scenario 2: Same Direction, One Catches the Other

A faster object leaves after a slower one and catches up. At the moment it catches up, the two distances are equal.

The times differ by the head start: if the slow object left hours earlier, .

Scenario 3: Round Trip

An object travels out at one speed and back over the same route at another speed. The two distances are equal (it is the same route), and the equation usually comes from a stated total time.

Scenario 4: Current or Wind

An object with still-water (or no-wind) speed travels in a moving medium of speed :

The two legs cover the same distance, so , or the distances are each set equal to a known value.

Worked Example A: Toward Each Other

Two cars leave towns miles apart at the same time, driving toward each other. One averages mph, the other mph. After how many hours do they meet?

Chart:

CarRateTimeDistance
A
B

Equation — the distances add to :

Answer: they meet after hours. Check: . ✓

Worked Example B: One Catches the Other

A freight train leaves a station traveling mph. Two hours later a passenger train leaves the same station on the same track at mph. How long does the passenger train travel before it catches the freight train?

Chart — let be the passenger train’s time; the freight train has been going hours:

TrainRateTimeDistance
Freight
Passenger

Equation — the distances are equal when it catches up:

Answer: the passenger train travels hours. Check: miles and miles. ✓

Worked Example C: A Round Trip

A boater travels upstream at mph and returns downstream at mph over the same stretch of river. The round trip takes hours. How far upstream did the boater go?

Chart — let be the one-way distance:

LegRateTimeDistance
Upstream
Downstream

Equation — the times add to :

Multiply every term by the LCD, :

Answer: the boater went miles upstream. Check: hours. ✓

Worked Example D: A Current Problem

A plane flies miles with a tailwind in hours and makes the return trip against the same wind in hours. Find the plane’s speed in still air and the wind speed.

Chart — let be the still-air speed and the wind speed:

TripRateTimeDistance
With wind
Against wind

Two equations — each distance is :

Add the equations:

Answer: the plane’s still-air speed is mph and the wind speed is mph.

Common Mistakes to Avoid

  • Entering a distance directly. The Distance column is always rate times time. If you know a distance, use it in the equation, not as a chart entry you also compute.
  • Giving both objects the same time when one has a head start. Different departure times mean different time expressions, like and .
  • Adding distances when they should be equal, or vice versa. Toward-each-other and opposite-directions add; same route (round trip, catch-up) sets equal.
  • Mixing units. Convert minutes to hours and feet to miles before building the chart.
  • Adding rates in a current problem. The effective rate is or ; you cannot just average the two trip speeds.
  • Forgetting to answer the actual question. If the problem asks for total distance or the meeting point, solving for is only step one.

Where This Shows Up Later

  • Work Problems. Same chart idea, with “rate of work” instead of speed and “job done” instead of distance.
  • Systems of Equations. Current and wind problems with two unknowns (still speed and current) are solved cleanly as a two-equation system, as in Worked Example D.
  • Rational Equations. Round-trip problems that give a total time lead to an equation with the variable in a denominator.
  • Quadratic Applications. A few motion problems — a boat whose speed relates to the current in a nonlinear way — end in a quadratic.

Practice Problems

Work each problem yourself before opening the answer.

Problem 1. Two hikers start miles apart and walk toward each other, one at mph and the other at mph. After how many hours do they meet?

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Answer: hours.

Problem 2. A car and a truck leave the same point at the same time in opposite directions. The car goes mph and the truck mph. After how many hours are they miles apart?

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Answer: hours.

Problem 3. A runner leaves a park at mph. Half an hour later a cyclist leaves the same park on the same path at mph. How long does the cyclist ride before catching the runner?

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Let be the cyclist’s time; the runner has been going .

Answer: hour, or minutes.

Problem 4. A boat travels miles downstream in hours and the same miles upstream in hours. Find the boat’s speed in still water and the current’s speed.

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Adding: , so mph and mph.

Problem 5. Two trains leave stations miles apart at the same time, heading toward each other. One goes mph faster than the other. They meet in hours. Find each speed.

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Let the slower speed be ; the faster is .

Answer: mph and mph.

Problem 6. A jogger runs out at mph and walks back over the same route at mph. The whole trip takes hours. How far out did the jogger go?

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Multiply by : , so and miles.

Problem 7. A plane flies miles with the wind in hours and returns against the wind in hours. Find the plane’s still-air speed and the wind speed.

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Adding: , so mph and mph.

Problem 8. A family drives to a lake at mph and returns on the same road at mph. The return trip takes minutes longer. How far is the lake?

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minutes is hour.

Multiply by : , so miles.

Problem 9. Car A leaves at noon going mph. Car B leaves the same point at 1:00 p.m. going mph in the same direction. At what time does Car B catch Car A?

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Let be Car B’s travel time; Car A’s is .

Answer: Car B catches Car A hours after 1:00 p.m., at 4:00 p.m.

Problem 10. Two cyclists miles apart ride toward each other. One rides twice as fast as the other. They meet in hours. Find each speed.

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Let the slower speed be ; the faster is .

Answer: mph and mph.

Problem 11. A boat’s speed in still water is mph. It travels miles downstream and miles back in a total of hours. Find the speed of the current.

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Let be the current speed.

Multiply by : , so , giving , , mph.

Problem 12. Maria leaves home at mph walking to work. Her brother leaves the same house minutes later, riding a scooter at mph on the same route. How far from home does he catch her?

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minutes is hour. Let be the brother’s time; Maria’s is .

Distance: miles from home.

Quick Reference

ScenarioDistance relationship
Toward each other / opposite directions
Same direction, faster catches slower
Round trip with known total time
With current / windrate (downstream) or (upstream)
Every object, always

The setup routine is the Word Problems five-step method with a chart bolted on. Work Problems reuse the chart with rates of work, and round-trip problems that lead to a variable in the denominator connect to Rational Expressions. See Applications of Linear Equations for geometry and interest problems, and the full Algebra lessons for everything else.

Frequently Asked Questions

What is the formula for distance, rate, and time problems?+

: distance equals rate multiplied by time. Rearranged, and . Every motion word problem is built from this one relationship applied to each moving object.

How do I set up a distance-rate-time chart?+

Make one row for each moving object and columns for rate, time, and distance. Fill in the two quantities you know or can name with a variable, then compute the third using . The equation comes from a sentence relating the distances (they are equal, they add to a total, or one exceeds the other).

When do the two distances add and when are they equal?+

If two objects move toward each other or in opposite directions, their distances add to the total gap. If they cover the same route (a round trip, or a slower object caught by a faster one), the two distance expressions are set equal to each other.

Why do my units have to match?+

only balances when rate and time use consistent units. A rate in miles per hour must pair with a time in hours; mixing in minutes or feet gives a numerically wrong answer even with perfect algebra. Convert everything to one system before setting up the chart.

How do I handle a head start or a delayed departure?+

Give the two objects different time expressions. If one leaves 2 hours earlier and travels for hours, the other travels for hours. Both still use with their own time.

What does 'still water' or 'no wind' mean in a boat or plane problem?+

It is the object's own speed without help or resistance. Going downstream or with the wind, you add the current or wind speed to it; going upstream or against the wind, you subtract it. The two legs then have different effective rates.

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