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## 8th grade

### Course: 8th grade > Unit 3

Lesson 10: Comparing linear functions- Comparing linear functions: equation vs. graph
- Comparing linear functions: same rate of change
- Comparing linear functions: faster rate of change
- Compare linear functions
- Comparing linear functions word problem: climb
- Comparing linear functions word problem: walk
- Comparing linear functions word problem: work
- Comparing linear functions word problems

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# Comparing linear functions word problem: climb

Sal is given the formula and a table of values that represent two people climbing a wall, and is asked to determine which one started out higher. Created by Sal Khan.

## Want to join the conversation?

- it was hard i didn´t get it??(5 votes)
- The goal here is to find who started off higher than the other.

Finding Alyssa's height is fairly straightforward.

a=1/3t+5

We substitute "0" for "t".

a=1/3(0)+5

This removes 1/3 making:

a=5

Alyssa started off at 5 feet up the wall.

t|Nick's height

6|6

8|7

10|8

Every 2 seconds Nick goes up one foot.

Nick is moving at 1/2 a foot per second.

We divide one of the x value of 6 by 2 giving us 3.

To find Nick's height we can make an equation

Nick's height=1/2(x)+3

if x=6

Nick's height=1/2(6)+3

6/2=3

3+3=6

We have proved that b=3.

Nick's height when he started was 3 feet.

Alyssa's height when she started was 5 feet.

Alyssa started off higher than Nick.(7 votes)

- can't you also use y=mx+b(1 vote)
- That is exactly what he is doing, he just substituted n for y and t for x. The variables make the problem a little bit easier to understand.(5 votes)

- What would Alyssa's height be?(2 votes)
- 5ft, he said at the beginning of the video(3 votes)

- Do we know the formula of the slope?(0 votes)
- If you went thru the KA lesson sequence, then yes you should know the formula for slope. You should also know how to find the slope using the slope-intercept form of the equation. In case you need to review those lessons, see the lessons for linear equations at this link: https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-linear-equations-functions(4 votes)

- W(t)=23t, i do not get this. really confused(1 vote)
- W is the function of t, in this case, 23 is equal to W, if you give a input:

for example, 2, then this function would be:

W(2)=23(2)

And the output of this function would be 46

Remember, a function can only have one output. This is how I think of it, hope it helps!(1 vote)

- Compare features of two real-world relationships that can be modeled by linear functions, where the functions are represented in different ways.(1 vote)
- On the previouse exercise there was a question:

P(x)=-2x+10 and Q(x)=-11/5x+10 Which linear function has a greater rate of change?

The answer was P(x) since the slope -2>-11/5

But function Q(x) decreases by 11 when x increses by 5 and function P(x) decreases by 10 when x increses by 5. It is clear that Q(x) steeper than P(x) so shouldn't Q(x) has a greater rate of change than P(x)?(2 votes)- No it isn't clear. smh watch the video again its right there.(0 votes)

- I don't get the equation. Why the 1/3 times t?(1 vote)
- 1/3t just equals it since Alyssa climbs 1/3 ft a second. And t= alyssa(1 vote)

- Which 6 is he using for n? Is he using the same 6 as time or is he using the 6 for Nick's height? Can someone please break this down for me?(0 votes)
- at2:17, isn't nick higher than allyssa(1 vote)

## Video transcript

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.