Question

From point $A$ located on a highway, one has to get as soon as possible, by a car, to point $B$ located in a field off the highway, at a distance $l$ from point $D$. If the car moves $n$ times slower in the field, at what distance $x$ from $D$ must one turn off the highway?

• A. $\frac{2l}{\sqrt{n^2-1}}$
• B. $\frac{l}{\sqrt{n^2-1}}$
• C. $\frac{l}{\sqrt{n-1}}$
• D. $\frac{l}{\sqrt{n}}$

Answer 1

The correct option is B $\frac{l}{\sqrt{n^2-1}}$

Let the car turn off the highway at point $\text{C}$ and let distance $SD=x$.

If $v$ is the velocity of the car in the field, then $nv$ is the velocity on the highway.


Then, time travelled on highway, $t_1=\frac{AD-x}{nv}$
& time travelled through the field, $t_2=\frac{\sqrt{l^2+x^2}}{v}$
For $t$ to be minimum,
$\frac{d}{dx}(t_1+t_2)=0$
$\Rightarrow \frac{d}{dt}\left[\frac{1}{v}\{\left(\frac{AD-x}{n}\right)+\sqrt{l^2+x^2}\}\right]=-\frac{1}{n}+\frac{x}{\sqrt{l^2+x^2}}=0$
$\Rightarrow l^2+x^2=n^2{x}^2$
or $x=\frac{l}{\sqrt{n^2-1}}$