Calculating the distance to turn off the highway

[Pushoam]

Homework Statement

Homework Equations

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?

 

[ehild]

Homework Statement

View attachment 207361

View attachment 207362

Homework Equations

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

correct so far

x = l √(n/(1-η) )

Is this correct?

No, this is not correct. What is n?

 

[Pushoam]

What is n?

Sorry for typing mistake,

n is nothing but η i.e. ita.

 

[ehild]

Sorry for typing mistake,

n is nothing but η i.e. ita.

Still not correct.

 

[Pushoam]

-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?

 

[Hiero]

@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.

 

[ehild]

-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?

Yes, good work!

 

[Pushoam]

Yes, good work!

But , irodov gives x = l / √(η² - 1).

So, I think there is something wrong in

Speed in the field = η v, is this what is meant by η times smaller?

 

[Pushoam]

Yes, good work!

But, Irodov says,
x = l / √(η² - 1)

Is there anything wrong in the following step ?

Speed in the field = η v, is this what is meant by η times smaller?

 

[Hiero]

So, I think there is something wrong in

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.

 

[Pushoam]

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]

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.

 

[Hiero]

So, while writing "Speed in the field = η v" I should add η<1 i.e.
Speed in the field = η v, η<1.

Yes that would be a more proper way to do it. You said this is the quoted answer:

But, Irodov says,
x = l / √(η² - 1)

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.