NLM BYU 48

Solution 48

From the given graph in the problem, we identify the two resistance functions:

• Air Resistance ($F_a$): The straight line. It passes through the origin and approximates linear behavior ($F \propto v$).
• Water Resistance ($F_w$): The curve. It approximates quadratic behavior ($F \propto v^2$).

Part (a): Wind speed 5 m/s, Stagnant Water

Let the speed of the boat be $v_b$.

• The wind moves at $5$ m/s in the direction of motion. The relative velocity of the air with respect to the boat is $v_{rel} = 5 – v_b$. This provides the driving force.
• The water is stagnant. The boat moves through it at $v_b$. This provides the resistive force.

At terminal speed, the driving force equals the resistive force:

$$ F_a(5 – v_b) = F_w(v_b) $$

Graphical Solution: We must find the intersection of the “inverted” Air Resistance graph (shifted by 5 units) and the Water Resistance graph. Looking at the graph:

• We need the value of $v$ where the force of air at $(5-v)$ equals the force of water at $v$.
• From the graph, at $v \approx 3.0$ m/s, the water resistance is approximately equal to the air resistance at $2.0$ m/s ($5-3$).

Part (b): Water flow 5 m/s, Stagnant Air

Let the speed of the boat be $v_b$.

• The water flows at $5$ m/s. The relative velocity of the water with respect to the boat is $5 – v_b$. This provides the driving force.
• The air is stagnant. The boat moves through it at $v_b$. This provides the resistive force.

Condition for equilibrium:

$$ F_w(5 – v_b) = F_a(v_b) $$

Graphical Solution: We look for the intersection where the Water Resistance force at $(5-v)$ equals the Air Resistance force at $v$.

• From the graph, at $v \approx 2.0$ m/s, the air resistance is equal to the water resistance at $3.0$ m/s ($5-2$).