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".# Show that , after #t# 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_0= #initial velocity while switching the motor to the off position.

1 Answer
Apr 1, 2018

Answer:

See below.

Explanation:

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

#1/2d/(dt) v^2 = -K v^4# and making #u = v^2# we get

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

#(du)/u^2 + 2K dt = 0 rArr u = 1/(2K t +C_0)# and finally

#u = v^2 rArr v = 1/sqrt(2K t + C_0)#

now considering

#v_0 = 1/sqrtC_0# we have

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