From point A located on a highway (figure) one has to get by car as soon as possible to point B located in the field at a distance `l` from the highway. It is known that the car moves in the field `eta` times slower than on the highway. At what distance from point D one must turn off the highway?

Answer

`CD=l//sqrt(eta^2-1)`.

Explanation

Let the car turn off the highway at a distance x from the point D.
So, `CD=x`, and if the speed of the car in the field is `nu`, then the time taken by the car to cover the distance `AC=AD-x` on the highway
`t_1=(AD-x)/(etanu)` (1)
and the time taken to travel the distance CB in the field
`t_2=(sqrt(l^2+x^2))/(nu)` (2)
So, the total time elapsed to move the car from point A and B
`t=t_1+t_2=(AD-x)/(etanu)+(sqrt(l^2+x^2))/(nu)`
For t to be minimum
`(dt)/(dx)=0` or `1/nu[-1/eta+(x)/(sqrt(l^2+x^2))]=0`
or `eta^2x^2=l^2+x^2` or `x=(l)/(sqrt(eta^2-1))`