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A motor boat is moving with speed v0 when its engine is switched off. Its deceleration is described by dv/dt = −kv³, where k is a constant. The speed of the boat after time t from the moment of cut-off is

1. v0 / √((2v0²kt)+1)
2. v0 e^(−kt)
3. v0 / 2
4. v0

Correct answer: v0 / √((2v0²kt)+1)

Solution

dv/dt = -k v^3 gives v^-3 dv = -k dt. Integrating from v0 to v: (1/v^2 - 1/v0^2)/2 = k t, so 1/v^2 = 1/v0^2 + 2kt. Hence v = v0/sqrt(1 + 2 v0^2 k t).

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