Distance, rate, time word problems

For distance word problems, it is important to remember the formula for speed:

Definition:   \(Speed = \Large \frac{{Distance}}{{Time}}\)

We can use this definition to solve different types of problems.  Let’s jump straight to an example:

Example:  Huong drove to the ferry office and back.  The trip there took three hours and the trip back took four hours.  She averaged 10 mph faster on the trip there than on the return trip.  What was Huong’s average speed on the outbound trip?

Solution:  First make sure you understand which of the two trips was the outbound trip.  The outbound trip was the first trip Houng made, when she was travelling toward the ferry office.  This trip took three hours, and the speed of this trip is the speed we are concerned with in this problem.  We will denote Huong’s speed on the outbound trip as \(S\). Now we can set up an equation to model the outbound trip.

\(S = \Large \frac{{Distance  Travelled}}{{3  hours}}\)

On the return trip, Houng averaged 10 mph less than on the outbound trip.  But the distance travelled remained the same.  So we can write, for the return trip:

\(S - 10 = \Large \frac{{Distance  Travelled}}{{4  hours}}\)

Now we have two equations, but it appears that we have two variables.  But let’s simplify both of these equations by multiplying them by values that will eliminate the denominators and clean things up.  We have

\(\left( {3  hours} \right)\left( S \right) = Distance  Travelled\)


\(\left( {4  hours} \right)\left( {S - 10} \right) = Distance  Travelled\)

Do you see where this is going?  We have two equations which represent the distance travelled.  But since the distance travelled is the same in both cases, we can conclude that the left sides of the equations equal each other as well!  That is,

\(\left( {3  hours} \right)\left( S \right) = (4  hours)(S - 10)\)

Now we have an equation with only one unknown value, the outbound speed!  This is what we want to find anyway.  Let’s solve for  \(S\).

\({\text{}}3S = 4S - 40\)

\(S = 40\) miles per hour


Example:  An Air Force plane left Singapore and flew toward the maintenance facility at an average speed of 150 mph.  A cargo plane left some time later flying in the same direction at an average speed of 180 mph.  After flying for five hours the cargo plane caught up with the Air Force plane.  How long did the Air Force plane fly before the cargo plane caught up?

Solution:  For the Air Force Plane, we have

\(Speed = \Large \frac{{Distance}}{{Time}}\)

so that

\(Distance = Speed \cdot Time\)

\(Distance = 150  mph \cdot x  hours\)

For the cargo plane, we have

\(Distance = Speed \cdot Time\)

\(Distance = 180  mph \cdot 5  hrs\)

Since at the moment they caught up, their distance travelled would be equal, we can conclude that

\(150x = 180 \cdot 5\)

\(150x = 900\)

\(x = 6  hours\)


That is, the Air Force Plane flew for six hours before the cargo plane caught up.


Below you can download some free math worksheets and practice.

28316 x

This free worksheet contains 10 assignments each with 24 questions with answers.

Example of one question:


Watch bellow how to solve this example:


24686 x

This free worksheet contains 10 assignments each with 24 questions with answers.

Example of one question:


Watch bellow how to solve this example:


15987 x

This free worksheet contains 10 assignments each with 24 questions with answers.

Example of one question:


Watch bellow how to solve this example:


Facebook PageYouTube Channel

Algebra and Pre-Algebra

Beginning Algebra
Adding and subtracting integer numbers
Dividing integer numbers
Multiplying integer numbers
Sets of numbers
Order of operations
The Distributive Property
Verbal expressions
Beginning Trigonometry
Finding angles
Finding missing sides of triangles
Finding sine, cosine, tangent
Absolute value equations
Distance, rate, time word problems
Mixture word problems
Work word problems
One step equations
Multi step equations
Graphing exponential functions
Operations and scientific notation
Properties of exponents
Writing scientific notation
By grouping
Common factor only
Special cases
Linear Equations and Inequalities
Plotting points
Graphing absolute value equations
Percent of change
Markup, discount, and tax
Adding and subtracting
Quadratic Functions
Completing the square by finding the constant
Solving equations by completing the square
Solving equations by factoring
Solving equations by taking square roots
Solving equations with The Quadratic Formula
Understanding the discriminant
Absolute value inequalities
Graphing Single Variable Inequalities
Radical Expressions
Adding and subtracting
Simplifying single radicals
The Distance Formula
The Midpoint Formula
Rational Expressions
Adding and subtracting
Multiplying and dividing
Simplifying and excluded values
Systems of Equations and Inequalities
Graphing systems of inequalities
Solving by elimination
Solving by graphing
Solving by substitution
Word problems