Solution to Question 16
1. Strategy and Path Analysis
The river is frictionless. Once the hunter steps onto the river, he cannot change his velocity. To cross distance $l$ in minimum time, he must maximize his velocity component perpendicular to the bank. Any velocity parallel to the bank does not help cross the river. Therefore, the optimal path is a straight line perpendicular to the bank.
The Problem: The hunter starts at the river edge (at rest). He cannot accelerate forward because that leads immediately onto the frictionless ice.
The Solution: He must first retreat perpendicular to the bank (move backwards to create a “runway”), stop, and then accelerate forward towards the river to gain the required launch velocity $v$.
2. Time Calculation (On the Ground)
Let the launch velocity be $v$ and the maximum acceleration be $a = \mu g$.
Phase 1: Approach (Accelerating towards the river)
To reach velocity $v$ at the edge from rest, he needs a runway distance $d$:
$$ v^2 – 0 = 2ad \implies d = \frac{v^2}{2a} $$
Time taken for this approach:
$$ t_{approach} = \frac{v}{a} $$
Phase 2: Retreat (Creating the runway)
He must first cover this distance $d$ starting from the edge (at rest) and ending at the turn-around point (at rest). The fastest way to do this is to accelerate for half the time and decelerate for half the time.
Let $t_{retreat}$ be the total retreat time.
$$ d = 2 \times \left( \frac{1}{2} a \left(\frac{t_{retreat}}{2}\right)^2 \right) = \frac{a t_{retreat}^2}{4} $$
Equating the distance $d$:
$$ \frac{a t_{retreat}^2}{4} = \frac{v^2}{2a} \implies t_{retreat}^2 = \frac{2v^2}{a^2} \implies t_{retreat} = \frac{\sqrt{2}v}{a} $$
Total Time on Ground:
$$ t_{ground} = t_{retreat} + t_{approach} = \frac{\sqrt{2}v}{a} + \frac{v}{a} = \frac{(\sqrt{2} + 1)v}{\mu g} $$
3. Optimization
Time to cross the river (coasting at constant $v$): $t_{river} = \frac{l}{v}$.
Total time function $T(v)$:
$$ T(v) = \frac{(\sqrt{2} + 1)v}{\mu g} + \frac{l}{v} $$
To find the minimum time, differentiate with respect to $v$:
$$ \frac{dT}{dv} = \frac{\sqrt{2} + 1}{\mu g} – \frac{l}{v^2} = 0 $$
$$ v^2 = \frac{\mu g l}{\sqrt{2} + 1} $$
4. Minimum Time Expression
Substitute the optimal condition $l/v = v \frac{\sqrt{2}+1}{\mu g}$ back into the time equation:
$$ T_{min} = \frac{(\sqrt{2} + 1)v}{\mu g} + \frac{(\sqrt{2} + 1)v}{\mu g} = 2 \frac{(\sqrt{2} + 1)v}{\mu g} $$
Now substitute $v = \sqrt{\frac{\mu g l}{\sqrt{2} + 1}}$:
$$ T_{min} = \frac{2(\sqrt{2} + 1)}{\mu g} \sqrt{\frac{\mu g l}{\sqrt{2} + 1}} $$
$$ T_{min} = 2 \sqrt{\frac{(\sqrt{2} + 1)^2 \cdot \mu g l}{(\mu g)^2 (\sqrt{2} + 1)}} $$
$$ T_{min} = 2 \sqrt{\frac{(\sqrt{2} + 1)l}{\mu g}} $$
Answer: $$ 2 \sqrt{\frac{(\sqrt{2} + 1)l}{\mu g}} $$
The river is frictionless. Once the hunter steps onto the river, he cannot change his velocity. To cross distance $l$ in minimum time, he must maximize his velocity component perpendicular to the bank. Any velocity parallel to the bank does not help cross the river. Therefore, the optimal path is a straight line perpendicular to the bank.
The Problem: The hunter starts at the river edge (at rest). He cannot accelerate forward because that leads immediately onto the frictionless ice.
The Solution: He must first retreat perpendicular to the bank (move backwards to create a “runway”), stop, and then accelerate forward towards the river to gain the required launch velocity $v$.
Let the launch velocity be $v$ and the maximum acceleration be $a = \mu g$.
Phase 1: Approach (Accelerating towards the river)
To reach velocity $v$ at the edge from rest, he needs a runway distance $d$:
$$ v^2 – 0 = 2ad \implies d = \frac{v^2}{2a} $$
Time taken for this approach:
$$ t_{approach} = \frac{v}{a} $$
Phase 2: Retreat (Creating the runway)
He must first cover this distance $d$ starting from the edge (at rest) and ending at the turn-around point (at rest). The fastest way to do this is to accelerate for half the time and decelerate for half the time.
Let $t_{retreat}$ be the total retreat time.
$$ d = 2 \times \left( \frac{1}{2} a \left(\frac{t_{retreat}}{2}\right)^2 \right) = \frac{a t_{retreat}^2}{4} $$
Equating the distance $d$:
$$ \frac{a t_{retreat}^2}{4} = \frac{v^2}{2a} \implies t_{retreat}^2 = \frac{2v^2}{a^2} \implies t_{retreat} = \frac{\sqrt{2}v}{a} $$
Total Time on Ground: $$ t_{ground} = t_{retreat} + t_{approach} = \frac{\sqrt{2}v}{a} + \frac{v}{a} = \frac{(\sqrt{2} + 1)v}{\mu g} $$
Time to cross the river (coasting at constant $v$): $t_{river} = \frac{l}{v}$.
Total time function $T(v)$:
$$ T(v) = \frac{(\sqrt{2} + 1)v}{\mu g} + \frac{l}{v} $$
To find the minimum time, differentiate with respect to $v$:
$$ \frac{dT}{dv} = \frac{\sqrt{2} + 1}{\mu g} – \frac{l}{v^2} = 0 $$
$$ v^2 = \frac{\mu g l}{\sqrt{2} + 1} $$
Substitute the optimal condition $l/v = v \frac{\sqrt{2}+1}{\mu g}$ back into the time equation: $$ T_{min} = \frac{(\sqrt{2} + 1)v}{\mu g} + \frac{(\sqrt{2} + 1)v}{\mu g} = 2 \frac{(\sqrt{2} + 1)v}{\mu g} $$ Now substitute $v = \sqrt{\frac{\mu g l}{\sqrt{2} + 1}}$: $$ T_{min} = \frac{2(\sqrt{2} + 1)}{\mu g} \sqrt{\frac{\mu g l}{\sqrt{2} + 1}} $$ $$ T_{min} = 2 \sqrt{\frac{(\sqrt{2} + 1)^2 \cdot \mu g l}{(\mu g)^2 (\sqrt{2} + 1)}} $$ $$ T_{min} = 2 \sqrt{\frac{(\sqrt{2} + 1)l}{\mu g}} $$