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

It has moved a distance, S = 25.9 m

Step-by-step explanation:

F = 40 v²........(1)

`F = mv(dv)/(dS)`.........(2)

Equating (1) and (2)

`-mv(dv)/(dS) = 40 v^2\\\\v(dv)/(dS) = (-40 v^2)/(m) \\\\m = 450 kg\\v(dv)/(dS) = (-40 v^2)/(450)\\\\ (dv)/(v) = (40)/(450) dS\\`

Integrate both sides:

`v_1 = 1, v_2 = 10`

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