The distance dd a car travels varies directly as the time tt. If the car travels 180km180\operatorname{\mathrm{km}} in 33 hours, find how far it will travel in 55 hours.

300km⁡300\operatorname{\mathrm{km}}

The distance d d a car travels varies directly as the time t t . If the car travels 180 k m 180\operatorname{\mathrm{km}} in 3 3 hours, find how far it will travel in 5 5 hours.

here we have an application problem involving direct variation. We have here that the distance d of a car travels that a car travels varies directly as the time. If the car travels a 180 kilometers in three hours, find how far the car will travel in five hours. Okay. So to solve this problem, we need to recognize this is a direct variation problem. Now I can see that the distance that the car travels is varying directly with time. Now since this is a direct variation problem, I know that there's some constant k that's going to go here. I don't really know what that constant is at this moment, so I'm just gonna go ahead and put k. So we now have this equation, and what I need to do is find my constant. That's the goal for the initial parts of the problem. Now what I can see is that the car travels 180 kilometers. Kilometers is a distance, and it does this in three hours. Obviously, hours is a time. So what that means is I can take the distance and replace it with 180 kilometers. And then we have this constant, and then I can take the time and replace it with three hours. So let's put an h there for hours. Now what I'm going to do is solve for this constant. So I'm gonna go ahead and divide three hours on both sides of the equal sign. And notice that the three hours will cancel, leaving me with just the constant k. Now don't worry too much about these kind of letters that you see. Just focus on the numbers that are being divided here because that's what we do for direct variation problems. So focus on 180 divided by three. Now we should know that that's 60. Right? So 180 divided by three is 60. And then what I can do is recognize that since we're dividing kilometers by hours, it's going to be kilometers per hour. So this right here would be our constant k. So it's gonna be just 60 or 60 kilometers per hour. Now that we found our constant, well, what I'm gonna do is rewrite this equation. So we'll have that distance is equal to our constant, which we just figured out is 60, and it's going to be multiplied by time. Now what do we do from here? Well, we're asked to figure out how far this car will travel in five hours. So because of that, what I can do is take the time and replace it with five hours. So our distance is going to be 60 times five. And 60 times five comes out to 300, and that's going to be in kilometers. So 300 kilometers would be the solution to this problem. Now there is one more thing I want to mention here. For this application problem, you can actually recognize something about what we got for our constant. Since we got this in kilometers per hour, we should know kilometers per hour, miles per hour, meters per second. That's all just speed. Right? So this constant that we were finding was actually the speed that the car was moving. So you could put an s for speed, or you could keep it as just some constant k, depending on how you wish to solve this problem. But as you can see, when it comes to solving these direct variation problems where we have application, it really just is the same process for any type of direct variation that we see. We just need to look through what our problem's telling us and set up our equation to get our answer. So I hope you found this video helpful, and let's move on.

The distance dd a car travels varies directly as the time tt. If the car travels 180km⁡180\operatorname{\mathrm{km}} in 33 hours, find how far it will travel in 55 hours.

300km⁡300\operatorname{\mathrm{km}}

60km⁡60\operatorname{\mathrm{km}}

180km⁡180\operatorname{\mathrm{km}}

900km⁡900\operatorname{\mathrm{km}}

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Direct & Inverse Variation

Prealgebra, Beginning Algebra, & Intermediate Algebra

16. Relations and Functions

Direct & Inverse Variation

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The distance $d$ a car travels varies directly as the time $t$ . If the car travels $180 k m$ in $3$ hours, find how far it will travel in $5$ hours.

$60 k m$

$180 k m$

$300 k m$

$900 k m$

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The distance dd a car travels varies directly as the time tt. If the car travels 180km180\(\operatorname{\mathrm{km}\)} in 33 hours, find how far it will travel in 55 hours.

다음 시청

The distance $d$ a car travels varies directly as the time $t$ . If the car travels $180 k m$ in $3$ hours, find how far it will travel in $5$ hours.