Solution 47
Let the drag force exerted by the water be proportional to the relative velocity between the water and the boat. The expression for drag force is $F = k v_{rel}$.
Step 1: Equilibrium Condition (Anchored)
When the boat is anchored, its velocity is $0$. The river flows at velocity $v_0$.
Relative velocity $v_{rel} = v_0 – 0 = v_0$.
The drag force balances the horizontal tension $T_0$:
$$ F_{drag} = T_0 $$ $$ k v_0 = T_0 \implies k = \frac{T_0}{v_0} $$Step 2: Motion After Chain Breaks
Let $v$ be the velocity of the boat at time $t$. The river still flows at $v_0$.
The new relative velocity is $v_{rel}’ = v_0 – v$. The drag force drives the boat downstream.
Using Newton’s Second Law ($F = ma$):
$$ m \frac{dv}{dt} = k (v_0 – v) $$Rearranging the terms to integrate:
$$ \frac{dv}{v_0 – v} = \frac{k}{m} dt $$Integrate from initial velocity $0$ to final velocity $0.5v_0$, and time $0$ to $t$:
$$ \int_{0}^{0.5v_0} \frac{dv}{v_0 – v} = \int_{0}^{t} \frac{k}{m} dt $$ $$ \left[ -\ln(v_0 – v) \right]_{0}^{0.5v_0} = \frac{k}{m} t $$ $$ -\ln(v_0 – 0.5v_0) + \ln(v_0) = \frac{k}{m} t $$ $$ \ln\left( \frac{v_0}{0.5v_0} \right) = \frac{k}{m} t $$ $$ \ln(2) = \frac{k}{m} t $$Substitute $k = \frac{T_0}{v_0}$ from Step 1:
$$ t = \frac{m \ln 2}{k} = \frac{m \ln 2}{(T_0 / v_0)} $$
$$ t = \frac{m v_0 \ln 2}{T_0} $$