Can you help with velocity of a boat please.?

After switching off the engine of a speedy motorboat, the acceleration of the boat follows the equation, color(red)(a=-Kv^3"; "K="constant".a=Kv3; K=constant. Show that , after tt time from switching the engine off, the velocity of the boat is color(green)(v=v_0/(sqrt(1+2K cdot t cdot v_0 ^2)),v=v01+2Ktv20, v_0= v0=initial velocity while switching the motor to the off position.

1 Answer
Apr 1, 2018

See below.

Explanation:

If a = dot v = -K v^3a=.v=Kv3 then after multiplication by vv in both sides we have

1/2d/(dt) v^2 = -K v^412ddtv2=Kv4 and making u = v^2u=v2 we get

1/2 d/(dt)u = -K u^212ddtu=Ku2 This is a separable differential equation with solution

(du)/u^2 + 2K dt = 0 rArr u = 1/(2K t +C_0)duu2+2Kdt=0u=12Kt+C0 and finally

u = v^2 rArr v = 1/sqrt(2K t + C_0)u=v2v=12Kt+C0

now considering

v_0 = 1/sqrtC_0v0=1C0 we have

v = v_0/sqrt(1+2K v_0^2 t)v=v01+2Kv20t