1Cademy - Solving a Uniform Motion Application: Finding Biking Speeds Using Equal Distances

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Problem-Solving Strategy for Distance, Rate, and Time Applications

Example

Solving a Uniform Motion Application: Finding Biking Speeds Using Equal Distances

Apply the distance, rate, and time problem-solving strategy to find two unknown uniform motion speeds when both travelers cover an identical route and distance but travel at different uniform rates and times.

Problem: Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis's speed is $7$ miles per hour faster than Wayne's speed, so it takes Wayne $2$ hours to ride to the beach while it takes Dennis $1.5$ hours for the ride. Find the speed of both bikers.

Read and draw: Sketch the situation. Both bikers travel the exact same route from the park to the beach, meaning they cover the exact same distance. Note the specific times and relative speeds. Create a rate–time–distance table.
Identify what you are looking for: you are asked to find the speed (rate) of both bikers.
Name the unknowns using a single variable. Let $r$ = Wayne's speed. Because Dennis’s speed is $7$ mph faster, we represent that as $r + 7$ . Fill this information into the chart and multiply the rate by the time to get the respective distance expression:

Biker	Rate (mph)	Time (hrs)	Distance (miles)
Dennis	$r + 7$	$1.5$	$1.5(r + 7)$
Wayne	$r$	$2$	$2r$

Translate into an equation. Because both bikers leave from Riverside Park and travel to the beach, they travel the same distance. The equation to model this situation comes from equating their two distance expressions: $1.5(r + 7) = 2r$
Solve the equation using standard algebra techniques:

Distribute the $1.5$ factor: $1.5r + 10.5 = 2r$
Subtract $1.5r$ from both sides to combine variables: $10.5 = 0.5r$
Divide both sides by $0.5$ (which is the same as multiplying by $2$ ): $21 = r$

Wayne's speed is $21$ mph. Now find Dennis's speed using his specific expression, $r + 7$ : $21 + 7 = 28$

Check the results and verify they make mathematical sense in context:

For Dennis, traveling at $28$ mph for $1.5$ hours implies $28(1.5) = 42$ miles.
For Wayne, traveling at $21$ mph for $2$ hours implies $21(2) = 42$ miles. The distances are truly equal. $\checkmark$

Answer the question with a complete sentence: Wayne rode at $21$ mph and Dennis rode at $28$ mph.

This application uses the principle that parallel trips covering the same route have equal distances. By defining the faster traveler's speed as an additive shift from the slower traveler's speed and multiplying by their specific times ( $D=rt$ ), an intuitive linear equation can be set up directly from equating those two distance outputs without ever needing to know the actual distance in miles beforehand.