# 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.

Apr 1, 2018

See below.

#### Explanation:

If $a = \dot{v} = - K {v}^{3}$ then after multiplication by $v$ in both sides we have

$\frac{1}{2} \frac{d}{\mathrm{dt}} {v}^{2} = - K {v}^{4}$ and making $u = {v}^{2}$ we get

$\frac{1}{2} \frac{d}{\mathrm{dt}} u = - K {u}^{2}$ This is a separable differential equation with solution

$\frac{\mathrm{du}}{u} ^ 2 + 2 K \mathrm{dt} = 0 \Rightarrow u = \frac{1}{2 K t + {C}_{0}}$ and finally

$u = {v}^{2} \Rightarrow v = \frac{1}{\sqrt{2 K t + {C}_{0}}}$

now considering

${v}_{0} = \frac{1}{\sqrt{C}} _ 0$ we have

$v = {v}_{0} / \sqrt{1 + 2 K {v}_{0}^{2} t}$