How Do You Solve for Distance When Two Waves Travel at Different Speeds?

  • Thread starter vorcil
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    Algebra
Now, divide both sides by 3500: [itex]d= 120*4500*8000/3500[/itex] If you don't like that "8000", multiply both sides by [itex]8000/8000[/itex]. [tex]d= 120*4500*8000/3500= 120*4500*(8000/3500)= 120*4500*8/7= 1234285.714285714285714.
  • #1
vorcil
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(please show how you rearrange the formula in blue i just need help showing me how you re arranged the question)

two waves traveling reach the same destination
wave 1 traveling at 8000m/s
wave 2 traveling at 4500m/s
wave 2 arrives 2 minutes after the first wave
how far did they travel?


v1=d/t1
v2=d/(t1+120)

t1 = d/8000

4500 = d / ((d/8000)+120)

every time i try solve for the awnser manually it's 274m (which is wrong)
and i get the correct awnser from putting it in my graphics calculator and it's 1234285 meters (1.2*10^6 m)

i have no idea how to re arrange it!

4500 * ( (d/8000) + 120 ) = d
4500 * (d/8000) = d-120
4500 * d = (d-120) / (1/8000)
i don't know
 
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  • #2
Try distributing the 4500 first and then clearing the denominator :smile:
 
  • #3
On the 2nd to last step, you didn't distribute. Your 120 should be 120/4500 when it goes to the right hand side.
 
  • #4
Pengwuino said:
On the 2nd to last step, you didn't distribute. Your 120 should be 120/4500 when it goes to the right hand side.

Yeah guys i still don't understand, can you please give me the rule for dealing with
4500((d/8000)+120) = d

this is how i would solve it using what you've told me

4500 ( d + 120 ) = d(8000)
4500d + 4500(120) = d(8000)
4500(120) = d(8000) - 4500(d)
540,000 = 3500d
540,000/3500 = d

:\

can you please show your step by step up to the awnser? (how I've shown it)
 
  • #5
You're not distributing before dividing by 8000.

[tex]\begin{array}{l}
4500((\frac{d}{{8000}}) + 120) = d \\
\frac{{4500*d}}{{8000}} + 4500*120 = d \\
\end{array}[/tex]

You can work from there.
 
  • #6
Pengwuino said:
You're not distributing before dividing by 8000.

[tex]\begin{array}{l}
4500((\frac{d}{{8000}}) + 120) = d \\
\frac{{4500*d}}{{8000}} + 4500*120 = d \\
\end{array}[/tex]

You can work from there.

how do i deal with the ds'?
everytime i do something it gets canceled or 8000d-4500d
 
  • #7
8000d - 4500d is simply 3500d
 
  • #8
Pengwuino said:
8000d - 4500d is simply 3500d

yup, i know, it's just that 3500d = 4500*120
540000/3500 = 154 which is wrong
 
  • #9
You are still not "distributing" the 8000!

From [tex] 4500((\frac{d}{{8000}}) + 120) = d [/tex]
multiplying the equation by 8000, you get 4500d+ 4500*8000*120= 8000d or
3500d= 4500*8000*120.
 
  • #10
HallsofIvy said:
You are still not "distributing" the 8000!

From [tex] 4500((\frac{d}{{8000}}) + 120) = d [/tex]
multiplying the equation by 8000, you get 4500d+ 4500*8000*120= 8000d or
3500d= 4500*8000*120.

what would this be called?
distributing algebra? if you could tell me, ill google it and practise
sorry for not really getting it I've never really dealed with this in physics or maths before
 
  • #11
It's just the "distributive law"- surely you have done that before!

Multiplying out the left side,
[tex]\frac{4500d}{8000}+ 120*4500)= d[/itex]

Now, multiply both sides of the equation by 8000:
[tex]4500d+ 120*4500*8000= 8000d[tex]
and, subtracting 4500d from both sides
[tex]3500d= 120*4500*8000[/tex]
It's that "8000" on the right side you are missing.
 

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