Algebra Word Problem

• Jul 20th 2013, 10:48 PM
ConfusedStudent123
Algebra Word Problem
Hi guys! I'm kind of struggling to put this into algebraic terms, I know the answer but need to show working and just can not get my head around it, if anyone would be so kind and help me out by giving the working and an explanation as to how you got there it would be greatly appreciated. Here is the question:

Two walkers are 15 km apart and they start walking toward each other. One walks twice as fast as other. If they meet 2 hours later at what average speed are they each walking?
• Jul 20th 2013, 10:49 PM
Prove It
Re: Algebra Word Problem
What have you tried to do so far?
• Jul 20th 2013, 10:55 PM
ConfusedStudent123
Re: Algebra Word Problem
Well I've tried to make equations with what I am given and use the v = d/t formula, where one walker's speed is twice that of the other and tried solving simultaneously but it just doesn't seem to make sense.
• Jul 20th 2013, 11:35 PM
Prove It
Re: Algebra Word Problem
That's a good start.

I would draw a line which can be our 15km distance. Mark in a point (it doesn't have to be accurate) where the two people will meet. The distance from the left end to that point we can call x. What would be the other distance then?
• Jul 20th 2013, 11:45 PM
ConfusedStudent123
Re: Algebra Word Problem
Ahhh yes! I see what you did there. Thank you for the help prove it, I look forward to our next our encounter!
• Jul 21st 2013, 08:46 AM
bjhopper
Re: Algebra Word Problem
Hello ConfusedStudent123,

Rate of closure of two bodies approaching equals closure distance /time
15/2 = 7.5 km/hr
rate of A =x B=2x
3x =7.5 x= 2.5 km/hr 2x = 5.0 km/hr
• Jul 21st 2013, 03:05 PM
Pusekatten
Re: Algebra Word Problem
I would do it very simply.
The slow walker would be X and because the faster walker is twice as fast he would be 2X. Add them together you get 3X. 15/3=5. X=5 2X=10.
Devide those with 2 and you should get the avrage speed in km/h.
• Jul 22nd 2013, 11:47 AM
HallsofIvy
Re: Algebra Word Problem
Quote:

Originally Posted by Pusekatten
I would do it very simply.
The slow walker would be X and because the faster walker is twice as fast he would be 2X. Add them together you get 3X. 15/3=5. X=5 2X=10.

So you set 3X equal to 15? Why? 3X is a speed (the relative speed with which one is approaching the other) and 15 is a distance. They can't be equal!

Quote:

Devide those with 2 and you should get the avrage speed in km/h.
Divide what with 2? And which "average speed"? You understand that they are asking for two average speeds don't you? If you solve "3X= 15, you get X= 5. But you just said "the slow walker would be X" and, in fact, "5 km/hr" is the speed of the faster walker!

Confusedstudent123, Pusekatten ("pussycat"?) does have the basic idea. If you call the speed of the slower walker "X", in km/hr, then the speed of the other, who is walking twice as fast, is 2X. Each is approaching the other at speed 3X. So this is exactly the same as if one person stood still and the other walked toward him at speed 3X. (In very technical terms, this would be "relative" to that person's "frame of reference".)

So this is exactly the same as if one person walked 15 km in 2 hours. That person would be walking at an average of 15/2= 7.5 km/hr. But that is equal to 3X so we have 3X= 7.5, X= 7.5/3= 2.5 km/hr. That is the speed of the slower person. The other person, walking twice as fast, would be walking at 2(2.5)= 5 km/hr.

As a check, the slower walker, at 2.5 km/hr, will have walked, in 2 hours, 2(2.5)= 5 km. The faster walker will have walked 2(5)= 10 km.
Yes, those add to 15 km so they will have met at the end of those 2 hours.