Middle School Math

[loxxii]

Jim and Mike live 20 meters away from each other. They want to meet up somewhere between their houses. They leave at the same time and Jim walks at 2 m/s and Mike walks at 2.5 m/s. How far from Mike's house will they meet and how long did each of them walk?

I know one way to find this answer, but I cant think of a more simple way to explain it to a remedial math student. Can you help?

[jay_shark]

The total distance that Jim walks divided by the total distance that Mike walks should be in the ratio 2:2.5 .

So Jim walked a total distance of 2/(2+ 2.5)*20 = (2/4.5)*20

Speed = Distance/Time .
2=[2/4.5*20]/T

Solve for T and we get T= 20/4.5 s

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Solution 2 :

Total distance traveled by Jim and Mike is 20m .

Distance jim travels as a function of time(s) is 2t
Distance Mike travels as a function of time(s) is 2.5t

So 2t + 2.5t = 20
Solve for t and we get 4.5t=20 and t =20/4.5s

Now we use that Speed = distance/time
2=distance/{20/4.5)

Solve for distance and we get distance = 2/4.5*20

[SamIAm]

Ugh. If I was explaining to a remedial middle school student, I wouldn't define a function of time and solve for t.

Instead, I'd point out that every second they get 4.5 meters closer. They start 20 meters apart. How long till they're together?

[jay_shark]

You need a good understanding of fractions to do this .

Jim walked 2/(2+2.5) of the total distance (20m) .

so we solve 2/4.5 = x/20 ; x=8.888888

I don't think it gets much easier than this .

[Enrique]

Well the technique of "every second they get 4.5 meters closer" from IamSam is easier to understand.

Then the answer for how long is just 20 / 4.5 = 40/9 = 4.444

And distance is just that times 2, so 8.888.

Seems easier for a remedial student to understand, and you don't deal with fractions.

[madnak]

[ QUOTE ]
You need a good understanding of fractions to do this.

[/ QUOTE ]

I think this is true. I also think the answer should be presented as 11 1/9m, not 11.11111...m. C'mon, decimals will just confuse him.

If you can talk him up to the "five out of nine" concept, the rest should be (relatively) easy. You can point out simple things like "if they're going the same speed they'll meet in the middle. So what will happen if Mike is going faster than Jim?" But getting the fractions right seems more difficult than getting the basic concept.

Also, remember we're solving for distance from Mike's house. You guys are solving for distance from Jim's house.

[KikoSanchez]

Yes, I think it is easiest to keep variables out. Just basic division and multiplication should do:

2 + 2.5 = 4.5 they complete 4.5m/s
20 / 4.5 = 4.44s thus, taking them 4.44s of walking
4.44s * 2 = 8.88m 4.44s of walking @ 2m/s = 8.88m
4.44s * 2.5 = 11.1m 4.44s of walking @ 2m/s = 11.11m

[TheFoxNL]

somehow that made most sence to me then the other replies /images/graemlins/smile.gif
but im a drop out so... figures... lol

[KikoSanchez]

last line should've read:
4.44s * 2.5 = 11.1m 4.44s of walking @ 2.5m/s = 11.11m
obviously.

How the *(#@*)# do you edit in this place?! I don't see the gd button anywhere....Ok I see it for this post, but not for past posts. What gives?

[madnak]

There's a time limit.

[ZeeJustin]

[ QUOTE ]
Ugh. If I was explaining to a remedial middle school student, I wouldn't define a function of time and solve for t.

[/ QUOTE ]

QFT. Leave the equations out.

I would phrase it like this to the student:

The first step is figuring how long they walked for. We know they are going 20 meters, so all we need is speed. Since they are starting and stopping at the same time, we can just combine their speeds and treat them as one person. Their total speed would be 4.5mps. So how long does it take to go 20 meters if you are going 4.5mps?

Once you know how long they are walking for you can treat the second part like a whole new problem.

Using the time, figure out how far mike walked, and that's the distance from his house.