Let the car turn off the highway at a distance $x$ from the point $D$. So, $C D=x$, and if the speed of the car in the field is $v$, then the time taken by the car to cover the distance $A C=A D-x$ on the highway
$$
t_{1}=\frac{A D-x}{\eta v}
$$
and the time taken to travel the distance $C B$ in the field
$$
t_{2}=\frac{\sqrt{l^{2}+x^{2}}}{v}
$$
So, the total time elapsed to move the car from
$$
t=t_{1}+t_{2}=\frac{A D-x}{\eta v}+\frac{\sqrt{l^{2}+x^{2}}}{v}
$$
For $t$ to be minimum
$$
\frac{d t}{d x}=0 \text { or } \frac{1}{v}\left[-\frac{1}{\eta}+\frac{x}{\sqrt{l^{2}+x^{2}}}\right]=0
$$
or $\quad \eta^{2} x^{2}=l^{2}+x^{2} \quad$ or $\quad x=\frac{l}{\sqrt{\eta^{2}-1}}$