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

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?

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Philomena 4 years 2021-08-22T16:57:17+00:00 1 Answers 15 views 0

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  1. yenoanh
    0
    2021-08-22T16:58:29+00:00
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    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.

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