Direct Variation - Problem 2

Imagine a biker who's cruising out on flat land, that biker pretty much maintains a constant speed or a constant rate of moving. So with this kind of problem is going to be a way to apply direct variation to a real world situation.

A biker's distance travelled on a flat terrain varies directly with his travelling time. If the biker has gone 90 miles after two and a half hours, how long will it take him to go 306 miles?

So what I'm going to do is to set up an equation that looks like y equals kx and then I'm going to plug in 306 as my y value, my dependent variable and solve for x. Before I can do that I need to find out what k is. K is a lot like slope and in order to find slope I'm going to use the points (0,0) because direct variation always includes the points 0,0 and then let's see. He's gone 2 and a half hours that's my independent variable and his result was 90 miles.

Let's go a head and find the slope or we are going to call that k, the constant variation here. Y change on top of my x change 2.5 take away 0. You notice that I used 2.5 instead of 2 and a half just because I'm going to this on my calculator 90 divided by 2.5 gives me his k value which in this case is 36.

If you wanted to interpret this in the real world situation this would mean 36 miles per hour. Let's go ahead and plug that into our formula here so that you can have the equation here for this problem.
36x distance travelled equals 36 times this time and now we are ready to solve this last piece. How long will it take him to go 306 miles? So 306 is going to be my y value, that's my dependent variable. 306 equals 36 times what x number solve for x by dividing both sides by 36 and you get x is 8 and a half hours.

So direct variation which looks always like y equals kx does actually have some applications to the real world. A lot of the times in your homework or maybe even on your test you are going to see these similar problems of this that involves distance, rate and time.