Using the Distance Formula to Solve Word Problems

Site: Saylor Academy
Course: MA007: Algebra
Book: Using the Distance Formula to Solve Word Problems
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Date: Tuesday, 15 July 2025, 7:45 AM

Description

Use the Distance, Rate, and Time Formula

One formula you will use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant rate. Rate is an equivalent word for "speed". The basic idea of rate may already familiar to you. Do you know what distance you travel if you drive at a steady rate of 60 miles per hour for 2 hours? (This might happen if you use your car's cruise control while driving on the highway.) If you said 120 miles, you already know how to use this formula!

 Distance, Rate, and Time

For an object moving at a uniform (constant) rate, the distance traveled, the elapsed time, and the rate are related by the formula:

\(\begin{aligned} d=r t \quad \text { where } \quad d &=\text { distance } \\ & r=\text { rate } \\ & t=\text { time } \end{aligned}\)

We will use the Strategy for Solving Applications that we used earlier in this chapter. When our problem requires a formula, we change Step 4. In place of writing a sentence, we write the appropriate formula. We write the revised steps here for reference.

HOW TO

Solve an application (with a formula).
  1. Step 1. Read the problem. Make sure all the words and ideas are understood.
  2. Step 2. Identify what we are looking for.
  3. Step 3. Name what we are looking for. Choose a variable to represent that quantity.
  4. Step 4. Translate into an equation. Write the appropriate formula for the situation. Substitute in the given information.
  5. Step 5. Solve the equation using good algebra techniques.
  6. Step 6. Check the answer in the problem and make sure it makes sense.
  7. Step 7. Answer the question with a complete sentence.

You may want to create a mini-chart to summarize the information in the problem. See the chart in this first example.

Example 2.58

Jamal rides his bike at a uniform rate of 12 miles per hour for \(3\frac{1}{2}\) hours. What distance has he traveled?

Solution
Step 1. Read the problem.  
Step 2. Identify what you are looking for. distance traveled
Step 3. Name. Choose a variable to represent it. Let \(d = distance\).
Step 4. Translate: Write the appropriate formula. \(d=rt\)
  \(d=?\)
\(r=12 \mathrm{mph}\)
\(t=3 \frac{1}{2}\) hours
Substitute in the given information. \(d=12 \cdot 3 \frac{1}{2}\)
Step 5. Solve the equation. \( d\)=42 miles
Step 6. Check  
Does 42 miles make sense?  
Jamal rides:  
 
Step 7. Answer the question with a complete sentence. Jamal rode 42 miles.
Try It 2.115

Lindsay drove for \(5 \frac{1}{2}\) hours at 60 miles per hour. How much distance did she travel?

Try It 2.116

Trinh walked for \(2 \frac{1}{3}\) hours at 3 miles per hour. How far did she walk?

Example 2.59

Rey is planning to drive from his house in San Diego to visit his grandmother in Sacramento, a distance of 520 miles. If he can drive at a steady rate of 65 miles per hour, how many hours will the trip take?

Solution
Step 1. Read the problem.  
Step 2. Identify what you are looking for. How many hours (time)
Step 3. Name.
Choose a variable to represent it.
Let \(t = time\)..
 

\(d=520 \, \text{miles}\)
\(r=65 \, \text{mph}\)
\(t=? \, \text{hours}\)

Step 4. Translate.
Write the appropriate formula.
\(d=rt\)
Substitute in the given information. \(520=65t\)
Step 5. Solve the equation. \(t=8\)
Step 6. Check. Substitute the numbers into
the formula and make sure the result is a
true statement.
 
  \(\begin{aligned} d &=r t \\ 520 & \stackrel{?}{=} 65 \cdot 8 \\ 520 &=520 \text{✓} \end{aligned}\)  
Step 7. Answer the question with a complete sentence. Rey's trip will take 8 hours.
Try It 2.117

Lee wants to drive from Phoenix to his brother's apartment in San Francisco, a distance of 770 miles. If he drives at a steady rate of 70 miles per hour, how many hours will the trip take?

Try It 2.118

Yesenia is 168 miles from Chicago. If she needs to be in Chicago in 3 hours, at what rate does she need to drive?


Source: OpenStax, https://openstax.org/books/elementary-algebra/pages/2-6-solve-a-formula-for-a-specific-variable
Creative Commons License This work is licensed under a Creative Commons Attribution 4.0 License.

Solve a Formula for a Specific Variable

You are probably familiar with some geometry formulas. A formula is a mathematical description of the relationship between variables. Formulas are also used in the sciences, such as chemistry, physics, and biology. In medicine they are used for calculations for dispensing medicine or determining body mass index. Spreadsheet programs rely on formulas to make calculations. It is important to be familiar with formulas and be able to manipulate them easily.

In Example 2.58 and Example 2.59, we used the formula \(d=rt\). This formula gives the value of \(d\), distance, when you substitute in the values of \(r \text{and }t\), the rate and time. But in Example 2.59, we had to find the value of \(t\). We substituted in values of \(d \text{and }r\) and then used algebra to solve for \(t\). If you had to do this often, you might wonder why there is not a formula that gives the value of \(t\) when you substitute in the values of \(d\text{and }r\). We can make a formula like this by solving the formula \(d=rt\) for \(t\).

To solve a formula for a specific variable means to isolate that variable on one side of the equals sign with a coefficient of 1. All other variables and constants are on the other side of the equals sign. To see how to solve a formula for a specific variable, we will start with the distance, rate and time formula.

Example 2.60

Solve the formula \(d=rt\) for \(t\):

  1. when \(d=520\) and \(r=65\)
  2. in general
Solution

We will write the solutions side-by-side to demonstrate that solving a formula in general uses the same steps as when we have numbers to substitute.

  1. when \(d=520\) and \(r=65\)
  1. in general
Write the formula. \(d=rt\)
Write the formula. \(d=rt\)
Substitute. \(520=65t\)
Divide, to isolate \(t\).
\(\frac{520}{65}= \frac{65t}{65}\)
Divide, to isolate \(t\).
\(\frac{d}{r} = \frac{rt}{t}\)
Simplify. \(8=t\)
Simplify. \(\frac{d}{r} = t\)

We say the formula \(t=dr\) is solved for \(t\).

Try It 2.119

Solve the formula \(d=rt\) for \(r\):

  1. when \(d=180\) and \(t=4\)
  2. in general
Try It 2.120

Solve the formula \(d=rt\) for \(r\)

  1. when \(d=780\) and \(t=12 \)
  2. in general

Example 2.61

Solve the formula \(A=\frac{1}{2}bh\) for \(h\):

  1. when \(A=90\) and \(b=15 \)
  2. in general
Solution
  1. when \(A=\frac{1}{2}bh\) and \(b=15\)
  1. in general
Write the formula. \(A=\frac{1}{2}bh\)
Write the formula. \(A=\frac{1}{2}bh\)
Substitute. \(90=\frac{1}{2} \cdot 15 \cdot h\)
Clear the fractions.
\(2 \cdot 90 = 2 \cdot \frac{1}{2}15h\)
Clear the fractions.
\(2 \cdot A = 2 \cdot \frac{1}{2}bh\)
Simplify. \(180=h\)
Simplify. \(2A = bh\)
Solve for \(h\)
\(12=h\) Solve for \(h\)
 \(\frac{2A}{b}=h\)

We can now find the height of a triangle, if we know the area and the base, by using the formula \(h=\frac{2A}{b}\).

Try It 2.121

Use the formula \(A=\frac{1}{2}bh\) to solve for \(h\):

  1. when \(A=170\) and \(b=17\)
  2. in general
Try It 2.122

Use the formula \(A=\frac{1}{2}bh\) to solve for \(b\):

  1. when \(A=62\) and \(h=31 \)
  2. in general

The formula \(I=Prt\) is used to calculate simple interest, \(I\), for a principal, \(P\), invested at rate, \(r\), for \(t\) years.

Example 2.62

Solve the formula \(I=Prt\) to find the principal, \(P\):

  1. when \(I=$5,600\), \(r=4%\), \(t=7\) years 
  2. in general
Solution
  1. \(I=$5,600\), \(r=4%\), \(t=7\) years
  1. in general
Write the formula. \(I=Prt\)
Write the formula. \(I=Prt\)
Substitute. \(5600=P(0.04)(7)\)
Simplify. \(5600=P(28)\) Simplify. \(I=P(rt)\)
Divide, to isolate P. \(\frac{5600}{0.28}=\frac{P(0.28)}{0.28}\)
Divide, to isolate P.
\(\frac{I}{rt}=\frac{P(rt)}{rt}\)
Simplify
\(20,000=P\)
Simplify. \(\frac{I}{rt}=P\)
The principal is
\( \ $20,000\)
\(P=\frac{I}{rt}\)

Try It 2.123

Use the formula \(I=Prt\) to find the principal, \(P\):

  1. when \(I=$2,160\), \(r=6%\), \(t=3\) years 
  2. in general
Try It 2.124

Use the formula \(I=Prt\) to find the principal, \(P\):

  1. when \(I=$5,400\), \(r=12%\), \(t=5\) years
  2. in general

Later in this class, and in future algebra classes, you'll encounter equations that relate two variables, usually \(x\) and \(y\). You might be given an equation that is solved for \(y\) and need to solve it for \(x\), or vice versa. In the following example, we're given an equation with both \(x\) and \(y\) on the same side and we'll solve it for \(y\).

Example 2.63

Solve the formula \(3x+2y=18\) for \(y\):

  1. when \(x=4\)
  2. in general
Solution
  1. when \(x=4\)
  1. in general

\(3x+2y=18\)

\(3x+2y=18\)
Substitute. \(3(4)+2y=18\)
Subtract to isolate the y-term.
\(12-12+2y=18-12\)
Subtract to isolate the y-term.
\(3x-3x+2y=18-3x\)
Divide. \(\frac{2y}{2}=\frac{6}{2}\)
Divide. \(\frac{2y}{2}=\frac{18}{2}-\frac{3x}{2}\)
Simplify.
\(y=3\) Simplify.
\(y=-\frac{3x}{2}+9\)
Try It 2.125

Solve the formula \(3x+4y=10\) for \(y\):

  1. when \(x=\frac{14}{3}\)
  2. in general
Try It 2.126

Solve the formula \(5x+2y=18\) for \(y\):

  1. when \(x=4\)
  2. in general

In Examples 1.60 through 1.64 we used the numbers in part ⓐ as a guide to solving in general in part ⓑ. Now we will solve a formula in general without using numbers as a guide.

Example 2.64

Solve the formula \(P=a+b+c\) for \(a\).

Solution
We will isolate \(a\) on one side of the equation. \(P=a+b+c\)
Both \(b\) and \(c\) are added to \(a\), so we subtract them from both sides of the equation. \(P - b - c =a+b+c - b - c\)
Simplify.

\(P -b - c =a\)

\(a=P -b -c\)


Try It 2.127

Solve the formula \(P=a+b+c\) for \(b\).

Try It 2.128

Solve the formula \(P=a+b+c\) for \(c\).

Example 2.65

Solve the formula \(6x+5y=13\) for \(y\).

Solution
\(6x+5y=13\)
Subtract \(6x\) from both sides to isolate the term with \(y\).
\(6x+5y-6x=13-6x\)
Simplify. \(5y=13-6x\)
Divide by 5 to make the coefficient 1. \(\frac{5y}{5}=\frac{13-6x}{5}\)
Simplify. \(y=\frac{13-6x}{5}\)


The fraction is simplified. We cannot divide \(13−6x\) by 5.

Try It 2.129

Solve the formula \(4x+7y=9\) for \(y\).

Try It 2.130

Solve the formula \(5x+8y=1\) for \(y\).