WPE CYU 7

Solution to Question 7

Problem Analysis:

• Let $v$ be the maximum speed of the boat relative to water with one engine.
• Let $u$ be the velocity of the river flow.
• Resistance is proportional to velocity ($F \propto v$). This implies power $P = Fv \propto v^2$. Since having two engines doubles the power, the new relative velocity $v’$ satisfies $v’^2 = 2v^2$, so $v’ = \sqrt{2}v$.

Step 1: Determining Flow Direction

Given times: $t_1 = 50$ min (1 engine) and $t_2 = 25$ min (2 engines).

Let the distance be $D$. The time taken is $D / V_{net}$.

Equation 1: $50 = \frac{D}{v \pm u}$

Equation 2: $25 = \frac{D}{\sqrt{2}v \pm u}$

Dividing the equations:

$$ \frac{50}{25} = \frac{\sqrt{2}v \pm u}{v \pm u} \implies 2 = \frac{\sqrt{2}v \pm u}{v \pm u} $$

Case A (Downstream, +): $2(v+u) = \sqrt{2}v + u \implies 2v + 2u = \sqrt{2}v + u \implies u = (\sqrt{2}-2)v$. Since $\sqrt{2} < 2$, $u$ would be negative, which contradicts the magnitude definition. Thus, the trip P to Q is Upstream.

Case B (Upstream, -): $2(v-u) = \sqrt{2}v – u \implies 2v – 2u = \sqrt{2}v – u \implies u = (2-\sqrt{2})v$. This yields a positive $u$.

Thus, P to Q is upstream. We need the time for Q to P (downstream).

From the upstream equation: $D = 50(v – u) = 50v [1 – (2-\sqrt{2})] = 50v(\sqrt{2}-1)$.

Part (a): Q to P with One Engine

Speed = $v + u$.

$$ t = \frac{D}{v+u} = \frac{50v(\sqrt{2}-1)}{v + (2-\sqrt{2})v} = \frac{50(\sqrt{2}-1)}{3-\sqrt{2}} $$

Rationalizing the denominator:

$$ t = \frac{50(\sqrt{2}-1)(3+\sqrt{2})}{9-2} = \frac{50(3\sqrt{2} + 2 – 3 – \sqrt{2})}{7} = \frac{50(2\sqrt{2}-1)}{7} $$ $$ t \approx \frac{50(2(1.414)-1)}{7} = \frac{50(1.828)}{7} \approx 13.06 \text{ min} $$

Part (b): Q to P with Both Engines

Speed = $\sqrt{2}v + u$.

$$ t = \frac{D}{\sqrt{2}v+u} = \frac{50v(\sqrt{2}-1)}{\sqrt{2}v + (2-\sqrt{2})v} = \frac{50(\sqrt{2}-1)}{2} $$ $$ t = 25(\sqrt{2}-1) \approx 25(0.414) \approx 10.35 \text{ min} $$