Calculus problem

[paulw]

I remember this problem from school and I have mucked around with solutions.

The problem is: You are crossing a railway line that is x feet across, your speed is y, there is a train approaching at speed z. What is the optimal angle to cross the railway at to maximize the time available to cross?

Any ideas,

P.

[kedaman]

You need to know where both you and the train is initially positioned as well.
As time needs to be maximized, the train will pass by as you are across, that is final position and time are equal
Y = you
T= train
.p intial position vector
.v velocity vector
.t time scalar
the equation is as follows:

Y.p+Y.v*Y.t=T.p+T.v*T.t


substitute your velocity vector with the angle (a) and length (l):

Y.p+(cos(a)*l*j+sin(a)*l*k)*Y.t=T.p+T.v*T.t

j and k are unit vectors. Solve a

[paulw]

Yep - but whats the equation? - I was being lazy and doing it on the train on the way home. I'll bowdlerize yours

Somehow I knew you'd answer this one.

Tx.

P.

[thinktank]

You are really sick.

or what would you call a person who relies on calculus at death's door ?

[kedaman]

|(Y.p+(cos(a)*l*j+sin(a)*l*k)*Y.t)-(T.p+T.v*T.t )|

[RobIII]

    /|
  a/ |b
  /  |
 /   |
/    |
-----
  c

I always learned "b" was the shortest route So fastest....?

Pseudo code:

If Train = NOT NEAR
  Walk
Else
  If Train = NOT Really NEAR
    Run(Fast)
  else
    If You = NOT Stupid
      Wait(TrainPassed)
    else
      Run(Screaming AND Hope_Not_To_Get_Hit)
    end if
End If

[paulw]

Hmmm. I think Kedaman's help is a little more concrete...

Cheers,

P.

[chrisf]

The max. time to cross means that the train will be level with you just as you reach the other side of the tracks.

When you have crossed the tracks, at angle a, then you have also travelled x * y * t * tan a along the track
While the train will have travelled z * (d + y * tan a) * t along the track.

Where d is the starting distance down the track of the train.
Angle a is the angle of crossing from perpendicular to ther tracks.

This gives:

tan a = z * d / (x * y - z * y)

with no calculus.

[kedaman]

Well apparently paulw insisted on having it, so he changed the issue into maximize distance between self and train, I would never spoil the chance to avoid calculus

[chrisf]

I have a degree in avoiding calculus.

[DavidHooper]

*chuckle*

[paulw]

Sadly, I have a degree in Mathematics and I can't remember this

P.

[DavidHooper]

where did you read ure degree??

[chrisf]

Originally posted by DavidHooper
where did you read ure degree??

Mostly from the comfort of my bed, with occasional days out to the Anti-Calculus University of Anti-Hard-Workingshire.

[paulw]

King's College, London.

but it was a long time ago, and I was very young...

Cheers,

P.

[Parke]

I have to question chrisf's solution based on dimensional analysis. Looking at his denominator (x * y - z * y), we have ft * ft/s - ft/s * ft/s. That I have problems with subtracting unlike units. One of the items this problem is missing is how far away is the train at time zero.

_W_|
\ |
\ | X
\ |
\|_________Wo___________

So we have (y*t)^2 = W^2 + X^2
and W + Wo = z*t

This can be viewed as maximizing W, or t or the angle.

I'll leave this to the calculus experts to do the rest.

Enjoy life,

Parke