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Calculating the distance to turn off the highway

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  1. Jul 18, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-7-18_20-32-37.png

    upload_2017-7-18_20-33-51.png
    2. Relevant equations


    3. The attempt at a solution
    Speed on a highway = v
    Speed in the field = η v, is this what is meant by η times smaller?

    Let's denote CD by x.

    Now, time taken is
    t = (AD - x )/v + {√(x2 + l2)}/ ηv

    For time to be extreme,

    dt/dx = 0
    -1/v + (1/ η v ) [( 1/2 ) * 2x / (√(x2 + l2)) ] = 0

    x = l √(n/(1-η) )

    Is this correct?
     
  2. jcsd
  3. Jul 18, 2017 #2

    ehild

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    correct so far
    No, this is not correct. What is n?
     
  4. Jul 18, 2017 #3
    Sorry for typing mistake,

    n is nothing but η i.e. ita.
     
  5. Jul 18, 2017 #4

    ehild

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    Still not correct.
     
  6. Jul 18, 2017 #5
    -1/v + (1/ η v ) [( 1/2 ) * 2x / (√(x2 + l2)) ] = 0
    η = x / (√(x2 + l2))

    Squaring both sides gives,
    η2( x2 + l2) = x2
    x = η l / √(1-η2)

    Is this correct?
     
  7. Jul 18, 2017 #6
    @Pushoam, yes, that is correct, assuming η<1.
    (It is possible that the author meant for η>1 in which case the speed in the field is actually v/η. If this is the case though, you don't need to resolve the problem; just replace all your η with 1/η.)

    I think it is meant that way, because if "something moves 5 times slower" that means it moves at 1/5 the speed, so if "something moves η times slower" I would think it means it moves at 1/η times the speed.
     
  8. Jul 18, 2017 #7

    ehild

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    Yes, good work!
     
  9. Jul 18, 2017 #8
    But , irodov gives x = l / √(η2 - 1).

    So, I think there is something wrong in
     
  10. Jul 18, 2017 #9
    But, Irodov says,
    x = l / √(η2 - 1)

    Is there anything wrong in the following step ?
     
  11. Jul 18, 2017 #10
    Yes there are two ways to interpret the statement, "moves η times slower." The two interpretations depend on whether η>1 or η<1.

    If you assume η<1 then "η times slower than v" means "ηv"
    If you assume η>1 then "η times slower than v" means "v/η"

    As I said earlier, you don't need to re-solve the problem if you assume the wrong meaning. Just observe that you effectively have the η.in.your.equation=(1/η.in.their.equation). In other words, you can just replace all η with (1/η) in your final equation and it will be as if you started with "field-speed = v/η" instead of "field-speed = ηv"

    Now you know that if this author says "k times smaller than x" he means [x/k with k>1] and not [kx with k<1]

    Which way is meant is just a convention; it won't affect any physical results.
     
  12. Jul 18, 2017 #11
    Thanks for pointing out this.

    So, while writing "Speed in the field = η v" I should add η<1 i.e.
    Speed in the field = η v, η<1.
     
  13. Jul 18, 2017 #12
    Yes that would be a more proper way to do it. You said this is the quoted answer:
    Notice what happens when η<1... the expression is imaginary!
    Since his result is only sensible for η>1, he must have meant η the other way than you assumed.
     
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