Lesson Objectives
• Demonstrate an understanding of how to translate phrases into algebraic expressions and equations
• Demonstrate an understanding of the six-step method used for solving applications of linear equations
• Learn how to solve motion word problems

## How to Solve Motion Word Problems

In our last lesson, we reviewed the six-step method used to solve a word problem that involves a linear equation in one variable.

### Six-step method for Solving Word Problems with Linear Equations in One Variable

1. Read the problem and determine what you are asked to find
2. Assign a variable to represent the unknown
• If more than one unknown exists, we express the other unknowns in terms of this variable
3. Write out an equation that describes the given situation
4. Solve the equation
5. State the answer using a nice clear sentence
6. Check the result
• We need to make sure the answer is reasonable. In other words, if asked how many students were on a bus, the answer shouldn't be (-4) as we can't have a negative amount of students on a bus.

### Solving Motion Word Problems » d = r x t

At this point, most of us have seen motion word problems. These problems use the distance formula (d = r x t) in order to gain a solution. The distance formula:
d » distance traveled
r » rate of speed
t » time traveled
This formula is one of the most intuitive. If we take a road trip and travel at an average rate of speed of 60 miles per hour for 12 hours, we can calculate our distance traveled as:
d = r • t
d = 60 • 12
d = 720
So in this example, we would have traveled a total distance of 720 miles. Let's look at an example.
Example 1: Solve each word problem.
Mishel can get to work in 15 minutes when she takes the bus. When she rides her bike to work, it takes 45 minutes. Her average speed when riding her bike is 10 miles per hour slower than the speed of the bus. What is the average speed of the bus?
Step 1) After reading the problem, it is clear that we need to find the average speed of the bus in miles per hour
Step 2) In this case, we have quite a few things that are unknown. We are given the time it takes to get to work by bus (15 minutes) and by bike (45 minutes). We are not given a rate for either scenario. We are just told that whatever the rate is for the bus, the bike is 10 miles per hour slower. If we let a variable like x be equal to the speed of the bus, then (x - 10) would be the speed of the bike.
let x = speed of the bus in miles per hour
then: x - 10 = speed of the bike in miles per hour
It is also important to note that for this type of problem we assume the distance to work is always the same. This may not be very realistic as a bike path in the real world may not be exactly the same as the path taken for a car. Here we will just say the distance to work is always the same.
Step 3) To make an equation, let's look at our information in a table. Remember our distance formula is: d = r x t. This means we can multiply rate of speed by time traveled to obtain a distance.
d r t
Bike0.75(x - 10)x - 100.75
Bus0.25xx0.25
First and foremost, looking at the table might cause some confusion. Why is time set to 0.75 for the bike and 0.25 for the bus? We did this because we are working with a speed that is in miles per hour. Therefore, we converted our time into hours. 45 minutes is the same as 3/4 of an hour or 0.75 hours. Similarly, 15 minutes is 1/4 of an hour or 0.25 hours. Now that we have modeled our situation, we can set up an equation. We know the distances are equal for the two scenarios. Let's set those equal to each other and solve:
0.75(x - 10) = 0.25x
Step 4) Solve the equation:
0.75x - 7.5 = 0.25x
Multiply by 100 to clear the decimals:
75x - 750 = 25x
75x - 25x = 750
50x = 750
x = 15
Step 5) Since x represented the speed of the bus in miles per hour, this tells us the bus travels at an average speed of 15 miles per hour and the bike will travel at (15 - 10 = 5) 5 miles per hour. Let's state our answer as:
The bus travels at an average speed of 15 miles per hour, while Mishel's bike travels at an average speed of 5 miles per hour.
Step 6) To check our answer, let's see if our answer matches up with the information given.
The distance traveled to work should be the same in each case:
Bike:
0.75 hours • 5 miles per hour
Bus:
0.25 hours • 15 miles per hour
0.75 • 5 = 0.25 • 15
3.75 = 3.75
Example 2: Solve each word problem.
Beth made a trip to Jamestown and back. The trip there took three hours and the trip back took five hours. What was Beth's average speed on the trip there if she averaged 45 mph on the return trip?
1) The main objective here is to determine Beth's average speed on the trip to Jamestown.
2) We can assign a variable like x to represent her average speed on the trip to Jamestown.
let x = the average speed of Beth in miles per hour on her trip to Jamestown
3) We write our equation. Let's first think about the information using a table. This can sometimes help to organize our thoughts:
Trip Rate (mph) Time (hours) Distance (miles)
There x33x
Back 455225
If we look at our table, we see a column for the rate in miles per hour, the time in hours, and the distance in miles. We know that she travels to and from the same location. Although in our everyday life we may choose a different route, we will assume that Beth travels the same route and therefore has the same distance to Jamestown and back from Jamestown. This means the distance there (3x) can be set equal to the distance coming back (225).
This sets up our equation as:
3x = 225 4) Let's solve our equation:
3x = 225
3/3 x = 225/3
x = 75
5) Since x = 75, this tells us Beth has traveled at an average speed of 75 mph on her trip there. Let's create a nice clean sentence:
Beth drove at an average speed of 75 mph on her trip to Jamestown.
6) Check
Let's check the distance to Jamestown on each leg of the trip.
Trip There: distance (225) = rate (75) x time (3)
225 = 75 x 3
225 = 225
Trip Back: distance (225) = rate (45) x time (5)
225 = 45 x 5
225 = 225
Our solution is correct. Beth drove at an average speed of 75 mph on her trip to Jamestown.

#### Skills Check:

Example #1

Solve each word problem.

Bill left home and drove toward the ferry office. Micaela left one hour later driving 10 miles per hour faster in an effort to catch up to him. After four hours Micaela finally caught up. What was Bill's average speed?

A
15 mph
B
48 mph
C
40 mph
D
25 mph
E
17 mph

Example #2

Solve each word problem.

A freight train made a trip to St. Paul and back. The trip there took 15 hours and the trip back took 6 hours. It averaged 45 miles per hour faster on the return trip than on the outbound trip. Find the freight train's average speed on the outbound trip.

A
25 mph
B
10 mph
C
12 mph
D
18 mph
E
30 mph

Example #3

Solve each word problem.

Alberto left home three hours before Ashley. They drove in opposite directions. Ashley drove at 30 miles per hour for three hours. After this time they were 480 miles apart. Find Alberto's speed.

A
30 mph
B
50 mph
C
65 mph
D
20 mph
E
17 mph         