Question

A boat is subject to hydrodynamic drag forces of F = 40v 2 [N] where v is the magnitude of velocity in [m/s]. The boat’s mass is 450 kg and it is moving at 10 m/s when the engine is shut down. When the boat’s velocity has decreased to 1 m/s, what distance has it moved from since the engine was shut down?

Answer 1

Answer:

It has moved a distance, S = 25.9 m

Explanation:

F = 40 v²........(1)

$F = mv\frac{dv}{dS}$.........(2)

Equating (1) and (2)

$-mv\frac{dv}{dS} = 40 v^2\\\\v\frac{dv}{dS} = \frac{-40 v^2}{m} \\\\m = 450 kg\\v\frac{dv}{dS} = \frac{-40 v^2}{450}\\\\ \frac{dv}{v} = \frac{40}{450} dS$

Integrate both sides:

$v_1 = 1, v_2 = 10$

$\int\limits^{10}_1 {\frac{1}{v} } \, dv = \frac{-40}{450} \int\limits^S_0 \, dS \\\\ln\frac{1}{10} = \frac{-40}{450} (S-0)\\\\S = \frac{-40}{450} ln(0.1)\\\\S = 25.9 m$.