Question

Help me to solve the interesting fact ?

After stopping the engine of a moving motor boat , the acceleration of the boat follows the equation `color(red)(a=-Kv^3` where `K` is a constant. What is the velocity of the boat `(v)`after `color(red)(t` time of stopping the engine? `v_0` is the velocity at the instance of stopping the engine.

Answer 1

`v = v_0/sqrt[1 + 2 K t v_0^2]`

Explanation:

The differential equation

`a = (dv)/(dt)= -K v^3` can be solved grouping variables

`-(dv)/v^3= K dt` Now integrating both sides

`1/2 v^-2 = K t + C_0`. The initial conditions imposes

`1/2 v_0^-2=C_0` then

`1/2 v^-2= K t + 1/2 v_0^-2` or

`v = v_0/sqrt[1 + 2 K t v_0^2]`