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# Comparing linear functions word problem: climb

## Videóátirat

Nick and Alyssa are racing up a wall. Alyssa's height on the wall is given by the equation. a is equal to 1/3 t plus 5-- so it sounds like they're wall climbers of some kind-- where a is Alyssa's height in feet after climbing 4t seconds. Nick started racing at the same time as Alyssa and is also climbing at a constant speed. His height is shown in the following table. So this is t in seconds, time in seconds. This is height in feet. Who started out higher, Alyssa or Nick? So to figure out their starting position, we just need to figure out what was their height at time equals 0. That's when this whole race started. For Alyssa, this is pretty straightforward. When time is equal to 0, you have 1/3 times 0 plus 5. Well, that's just going to be 5 feet. So Alyssa's starting position is at 5 feet when time is equal to 0. Now, let's think about Nick's height at time equals 0, and there's a couple of ways that we can go about doing this. One is just to back up, to kind go backwards on this chart. So let me show you what I'm talking about. So if this is time, and this is, let's say, n for Nick's height because we have a for Alyssa's height. So I'm going to make a little table here. We already know that at time 6 seconds, or after 6 seconds, he's 6 feet in the air or along the wall. At time 8, he is 7 feet in the air or along the wall. And at time 10, he's gotten to a height of 8 feet. So what's happening here? Every time 2 seconds goes by, he increases in 1 foot. You have another 2 seconds, he increases in height by 1 foot. So you could go backwards. If we take away 2 seconds to 4 seconds, he will decrease in height by 1 foot. If we go back another 2 seconds, he will decrease in height by another 1 foot. The reason why we can say this is because we know he's climbing at a constant speed. So if we decrease by another 2 seconds to our starting time, then we know that he would have been 1 foot lower, so he would have been at 3 feet. So just like that, we now know at time equals 0, Nick's height is 3 feet in the air. So Alyssa started out higher than Nick. So this right over here would be the correct answer. Now, the other way to do it is set up an equation just like we had for Alyssa and substitute for time equals 0. And the way to do that is to recognize that Nick's height, as a function of time, is also going to be a linear equation. Because they're both climbing. They're both climbing at constant speeds, or we know that Nick is climbing at a constant speed. So Nick's height, as a function of time, is going to look like-- Nick's height is going to be some slope, some rate of change, essentially his height per second at times time plus his initial position. So how can we solve for m, the slope, and his initial position? Well, the slope is just his rate of change of height, so it's literally how much does his height change per unit time? So m right over here, m is just going to be for a unit time, for change in time, how much is his height changing? And his height is-- we used the letter n. So we already know that when time increases by 2, his height increases by 1 foot. So we know that m is equal to 1/2. He increases 1/2 feet per second. And you see that there because it takes him 2 seconds to go 1 foot. So we can fill in m here. So we know now that n is equal to 1/2 t plus b. Now, to solve for b, you could just substitute one of these points. All of these points must satisfy this equation right over here. So we could use the point 6. So if we put a 6 in here, so when time is 6, we know that n is 6. So you have 6 is equal to 1/2 times 6 plus b. Or you get 6 is equal to 3 plus b. Subtract 3 from both sides, you get b is equal to 3. So there you have it. You get Nick's equation, or Nick's height as a function of time. Nick as a function of time is going to be equal to 1/2 t plus 3. So now we have an equation just like Alyssa's. And we can say, well, when time is equal to 0, he's at a height of 3, which is lower than Alyssa's initial height.