Solve Uniform Motion Applications: example 3.49

Christopher and his parents live 115 miles apart. They met at a restaurant between their homes to celebrate his mother’s birthday. Christopher drove 1.5 hours while his parents drove 1 hour to get to the restaurant. Christopher’s average speed was 10 miles per hour faster than his parents’ average speed. What were the average speeds of Christopher and of his parents as they drove to the restaurant?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

1
Solution
Step 1. Read the problem. Make sure all the words and ideas are understood.
- Draw a diagram to illustrate what it happening. Below shows a sketch of what is happening in the example.
- Create a table to organize the information.
- Label the columns rate, time, distance.
- List the two scenarios.
- Write in the information you know.
Step 2. Identify what we are looking for.
- We are asked to find the average speeds of Christopher and his parents.
Step 3. Name what we are looking for. Choose a variable to represent that quantity.
- Complete the chart.
- Use variable expressions to represent that quantity in each row.
- We are looking for their average speeds. Let’s let r represent the average speed of the parents. Since the Christopher’s speed is 10 mph faster, we represent that as $r + 10 .$
Fill in the speeds into the chart.

Multiply the rate times the time to get the distance.

Step 4. Translate into an equation.
- Restate the problem in one sentence with all the important information.
- Then, translate the sentence into an equation.
- Again, we need to identify a relationship between the distances in order to write an equation. Look at the diagram we created above and notice the relationship between the distance Christopher traveled and the distance his parents traveled.
The distance Christopher travelled plus the distance his parents travel must add up to 115 miles. So we write:

Step 5. Solve the equation using good algebra techniques.

Now solve this equation. $\begin{matrix} 1.5 (r + 10) + r & = & 115 \\ 1.5 r + 15 + r & = & 115 \\ 2.5 r + 15 & = & 115 \\ 2.5 r & = & 100 \\ r & = & 40 \end{matrix}$

So the parents' speed was 40 mph.

Christopher's speed is $r + 10$ . $\begin{matrix} r + 10 \\ 40 + 10 \\ 50 \end{matrix}$

Christopher's speed was 50 mph.

Table 3.12

Step 6. Check the answer in the problem and make sure it makes sense.

Christopher drove $\begin{matrix} 50 mph (1.5 hours) & = & 75 miles \end{matrix}$

His parents drove $\begin{matrix} 40 mph (1 hours) & = & \underset{_______}{40 miles} \\ 115 miles \end{matrix}$

Table 3.13

Step 7. Answer the question with a complete sentence. Christopher's speed was 50 mph.
His parents' speed was 40 mph.

Table 3.14

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Carina is driving from her home in Anaheim to Berkeley on the same day her brother is driving from Berkeley to Anaheim, so they decide to meet for lunch along the way in Buttonwillow. The distance from Anaheim to Berkeley is 410 miles. It takes Carina 3 hours to get to Buttonwillow, while her brother drives 4 hours to get there. The average speed Carina’s brother drove was 15 miles per hour faster than Carina’s average speed. Find Carina’s and her brother’s average speeds.

How did it go?

Now solve this equation.	$\begin{matrix} 1.5 (r + 10) + r & = & 115 \\ 1.5 r + 15 + r & = & 115 \\ 2.5 r + 15 & = & 115 \\ 2.5 r & = & 100 \\ r & = & 40 \end{matrix}$
	So the parents' speed was 40 mph.
Christopher's speed is $r + 10$ .	$\begin{matrix} r + 10 \\ 40 + 10 \\ 50 \end{matrix}$
	Christopher's speed was 50 mph.

Christopher drove	$\begin{matrix} 50 mph (1.5 hours) & = & 75 miles \end{matrix}$
His parents drove	$\begin{matrix} 40 mph (1 hours) & = & \underset{_______}{40 miles} \\ 115 miles \end{matrix}$

Solution

Your turn