Math 25 Activity 9: Distance, Rate, and Time
This week’s activity focuses on application problems that involve distance, rate, and time. First, we
will look at some concepts that appear within application problems. Then, we will walk through an
application problem one step at a time. At the end, your group will be given time to work on a
couple of problems to practice the skills.
A formula you will need for this activity is
to practice using the formula on straight forward applications.
In your group, answer the following questions using
or
for short. We want
:
1. If a car is travelling at the constant rate of 60 miles per hour and traveled 330 miles, then how
long did this take assuming there weren’t any stops? Change your answer, if needed, to hours and
minutes.
2. If a bicyclist rode to work in 45 minutes and work was a straight shot 9 miles from home, then
how fast was the bicyclist riding? Change your answer, if needed, to answer in miles per hour.
3. A hiker was walking up a mountain trail at a steady rate of 2.8 kilometers per hour. How far did
the hiker get before resting after walking 3 hours?
If you didn’t notice by the way the questions are worded, it is important to think about units of
measurement when working with distance, rate, and time.
Another skill we want to practice in addition to using the
formula, is noticing in a word
problem if we are describing situations where distances are being added together, or, if distances
are being set equal to each other. Being able to tell the difference will help us when we are setting
up an equation in a word problem.
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4. In your group, decide if these situations are describing when distances are equal or when
distances are needing to be added together by placing an X in the appropriate box. You are not
trying to solve anything- the situations are written with details left out.
Distances
traveled will
be equal
Situation
Distances
traveled will be
added together
A man drives to the mountains in 2 hours going slow through a
rainstorm, and the return trip only takes 1 hour.
A hiker starts at a trailhead before a cyclist, but the cyclist
eventually passes the hiker.
Two trains are moving apart at difference speeds and after four
hours they are 300 miles apart
A woman walks from her home to her office in 1 hour on
Monday, but rides her scooter to work in 15 minutes on Tuesday.
Two cars are 400 miles apart and traveling towards each other at
different speeds.
Now that you have practiced using the formula and practiced noticing how distances may be
described, let’s look at an application problem in full. We will walk you through one problem step
by step and then have you practice a couple of problems using what you know and have learned.
Two cars leave towns 900 kilometers apart at the same time and travel towards each other. One
car’s rate is 20 kilometers per hour more than the other’s. If they meet in 5 hours, what is the rate
of the faster car?
The first thing we want to do is understand some of the components. We like to list the variables
we will use, what we do know, and what we don’t know. It is easiest if we use a table to help
organize.
Here is what the table looks like before we have filled out any information from the problem:
Rate (km per hour)
Time (hours)
Distance (km)
Faster Car
Slower Car
Now that we have a plan to organize our information, let’s put what we know on the table. The only
thing we know for certain is that they both traveled for 5 hours.
Rate (km per hour)
Faster Car
Slower Car
Time (hours)
5
5
Distance (km)
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Next, we want to decide what the variable will be and where to place it on the table. It can be
difficult to identify the variable, because it is not always the answer to the question we are trying to
solve. In this problem, the phrase “one car’s rate is 20 kilometers per hour more than the other’s”
indicates that we can use x for one rate and x + 20 for the other.
5. Which rate is faster in this application problem: x or x+20?
Since x + 20 is faster than x, we will put the x + 20 in the rate box for the faster car and x for the rate
of the slower car.
Faster Car
Slower Car
Rate (km per hour)
x + 20
x
Now we can use the formula
Faster Car
Slower Car
Rate (km per hour)
x + 20
x
Time (hours)
5
5
Distance (km)
to fill out the distance parts of the table.
Time (hours)
5
5
Distance (km)
5(x + 20)
5x
The table is completely filled out. The next step is to build the equation so we can solve for the
variable.
6. Refer to the part of the activity when we discussed situations dealing with distances. Will these
distances be added together or set equal to each other?
Since these cars are traveling towards each other until they meet, and covering 900 kilometers, that
means we will add their distances together and set it equal to 900. Our new equation is:
7. Solve the equation for x.
What is the variable x representing?
Does solving for x answer the original question?
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You should have gotten x = 80. Since x represents the speed of the slower car, and we need to
figure out the faster speed, our last step is to figure out the faster car’s speed: x + 20 = 80 + 20 =
100. The final answer to the problem is that the faster car is traveling at 100 kilometers per hour.
Practice solving distance, rate, and time application problems in your group:
A long distance runner starts at the beginning of a trail and runs at a rate of 6 miles per hour. One
hour later, a cyclist starts at the beginning of the trail and travels at a rate of 14 miles per hour.
What is the amount of time that the cyclist travels before overtaking the runner? Do not do any
rounding.
Two trains leave the station at the same time, one heading east and the other west. The
eastbound train travels at 80 miles per hour. The westbound train travels at 60 miles per hour.
How long will it take for the two trains to be 196 miles apart? Do not do any rounding.
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