Distance, Rate, and Time Applications An athlete in training...

Distance, Rate, and Time Applications An athlete in training rides his bike $20 \mathrm{mi}$ and then immediately follows with a 10 -mi run. The total workout takes him 2.5 hr. He also knows that he bikes about twice as fast as he runs. Determine his biking speed and his running speed. (PICTURE NOT COPY)

Distance, Rate, and Time Applications
An athlete in training rides his bike $20 \mathrm{mi}$ and then immediately follows with a 10 -mi run. The total workout takes him 2.5 hr. He also knows that he bikes about twice as fast as he runs. Determine his biking speed and his running speed.
(PICTURE NOT COPY)

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Key Concepts

Distance-Rate-Time Relationship

This concept revolves around the formula distance = rate × time, which is central to problems involving moving objects. It allows one to relate the distance traveled by an object to its speed and the time taken, enabling the analysis and solution of problems where these quantities interplay.

Systems of Equations

In many word problems, multiple relationships between variables exist that can be modeled using equations. Solving a system of equations involves finding values for the unknowns that satisfy all equations simultaneously, which is particularly useful when different segments of a journey or process are described with interrelated conditions.

Ratio Relationships

Ratios provide a way to express a relationship between two quantities. In problems where one rate or speed is a multiple of another, this concept allows one to set up a proportional relationship between the variables, thereby reducing the number of unknowns and simplifying the problem-solving process.

Transcript

00:02 All right, so what we need to do is solve for how fast were we biking and how fast we're running.

00:08 If we go bike for 20 miles and then immediately follow that with a 10 -mile run.

00:12 And we know the whole thing takes 2 .5 hours.

00:15 And we also know that we basically bike twice as fast as we can run.

00:18 So we need to set up our equation.

00:20 So what we do know is that our bike trip for 20 miles and our running for 10 miles needs to equal the 2 .5 hours total.

00:32 And if we're going to talk about rates, we're going to be adding these together.

00:35 Our rate for biking and a rate for running.

00:40 So we know that we're going to bike twice as fast as we can run.

00:44 So let's call our running x, which means our rate for our biking then is going to be 2x.

00:50 Now we need to go ahead and solve.

00:52 In order to get rid of our variable down in the denominator, we're going to have to find our gcf.

00:58 So we got 2x and x.

00:59 So our gcf then is going to be 2x.

01:03 So we're going to multiply everything by 2x so we can get rid of the x in the variable down there in the denominator.

01:10 So we're going to multiply 2x here.

01:11 And we're going to multiply by 2x here.

01:13 And we're going to multiply by 2x here...

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