Turn off at $x=\dfrac{l}{\sqrt{\eta^{2}-1}}$ from point $D$.
(ii) Problem Statement
"From point $A$ located on a highway 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?"
Key idea / reference check: $D$ is the foot of the perpendicular from $B$ to the highway, so $BD=l$. Minimising total time gives $\eta^2x^2=l^2+x^2$, i.e. $x=l/\sqrt{\eta^2-1}$. This matches the reference answer exactly. It needs $\eta\gt1$, and if $AD$ is shorter than this $x$ the car should leave the highway right at $A$.
(iii) The Big Idea in Plain Words
The highway is fast, the field is slow. Going straight to $B$ is the shortest path, but it spends a lot of time in the slow field. Staying on the road too long makes the field leg steep and long. The best turn-off is a balance between the two.
Symbols: $l$ = distance of $B$ from the highway; $x=CD$ = distance from $D$ where the car turns off; $v$ = speed in the field; $\eta v$ = speed on the highway; $AD$ = distance from $A$ to $D$.
Write the total time as a function of $x$, then find the $x$ where it stops decreasing and starts increasing. That is the minimum.
(iv) The Solution Viewed in Each Frame
1. Ground Frame (road map)
Why: the road and field are at rest here, so distances and speeds are direct.
Highway leg $AC=AD-x$ at speed $\eta v$: $t_1=\dfrac{AD-x}{\eta v}$. Field leg $CB=\sqrt{l^2+x^2}$ at speed $v$: $t_2=\dfrac{\sqrt{l^2+x^2}}{v}$. Total $t=t_1+t_2$, minimised at $x=l/\sqrt{\eta^2-1}$.
2. Angle (Snell) View
Why: the condition $dt/dx=0$ is really a statement about the angle $\theta$ of the field path with the perpendicular $CB$ to the road.
Here $\sin\theta=\dfrac{x}{\sqrt{l^2+x^2}}$, and $dt/dx=0$ gives $\sin\theta=\dfrac{1}{\eta}=\dfrac{v}{\eta v}$, the same rule as Snell's law for light.
3. Centre-of-Mass Frame
Does not apply: there is a single moving car and no interacting system.
4. Medium Frame
Does not apply. The road and field do not move, so there is no moving medium.
- Changes: the description: lengths on the map versus the angle $\theta$.
- Same: $x=l/\sqrt{\eta^2-1}$, equivalently $\sin\theta=1/\eta$.
(v) The Solution by Each Method
Method 1: Calculus (the reference solution)
- Why: write time as one function of one unknown.
$t=\dfrac{AD-x}{\eta v}+\dfrac{\sqrt{l^2+x^2}}{v}$. - Why: a minimum needs zero slope.
$\dfrac{dt}{dx}=\dfrac1v\left[-\dfrac1\eta+\dfrac{x}{\sqrt{l^2+x^2}}\right]=0\Rightarrow \eta^2x^2=l^2+x^2$. - Why: solve for $x$.
$x=\dfrac{l}{\sqrt{\eta^2-1}}$. The second derivative $\dfrac{l^2}{v(l^2+x^2)^{3/2}}\gt0$ confirms a minimum.
Method 2: Fermat / Snell Method (fastest)
Least-time paths obey $\sin\theta=v_{\text{field}}/v_{\text{road}}=1/\eta$. Then $\tan\theta=x/l=1/\sqrt{\eta^2-1}$, so $x=l/\sqrt{\eta^2-1}$ in one line.
Method 3: Force Method
Does not suit this problem: no forces are involved, only a kinematic optimisation.
Method 4: Energy Method
Does not suit this problem: speeds are given and constant, so energy plays no role.
Fastest: Snell/Fermat. Most reliable in an exam: calculus. All give the same answer.
(vi) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|
(vii) Common Exam Traps
- Measuring $x$ from $A$ instead of from $D$. The answer is the distance from $D$.
- Swapping the speeds: the highway speed is $\eta v$ (faster), the field speed is $v$. The result needs $\eta\gt1$.
- Forgetting the limit: if $AD\lt l/\sqrt{\eta^2-1}$, the car should turn off at $A$ itself.
Verified: the live simulation, every frame and every method give the same answer, in SI units.