The deceleration experienced by a moving motor boat, after its engine is cut off is given by $\dfrac{{dv}}{{dt}} = - k{v^3}$ where $k$ is a constant. If ${v_0}$ is the magnitude of the velocity at the cut off, the magnitude of the velocity at a time $t$ after the cut-off is:
A) $\dfrac{{{v_0}}}{2}$
B) ${v_0}$
C) ${v_0}{e^{ - \dfrac{k}{2}}}$
D) $\dfrac{{{v_0}}}{{\sqrt {(2{v_0}^2kt + 1)} }}$

Hint: Recall what is meant by deceleration. To calculate the magnitude of velocity $v$ at time $t$ after the cut-off we need to integrate the given equation of deceleration and find the equation of $v$ and verify that equation with the given options.

Complete step by step solution:
Deceleration is nothing but negative acceleration or decrease in speed, which in simple words means slowing down.
In this question, the deceleration experienced by the moving motor boat is given by $\dfrac{{dv}}{{dt}} = - k{v^3}$
In this equation, this negative sign indicates deceleration or negative acceleration.
Now, to calculate the equation of velocity at time $t$ after cut-off, we integrate the deceleration equation.
$\dfrac{{dv}}{{dt}} = - k{v^3}$ (here $k$ is a constant)
Rearranging the terms to integrate the equation,
$ \Rightarrow \dfrac{{dv}}{{{v^3}}} = - kdt$
$ \Rightarrow \int {\dfrac{{dv}}{{{v^3}}} = \int { - kdt} } $
$ \Rightarrow \dfrac{{ - 1}}{{2{v^2}}} = - kt + c$ ---( $1$ )
To find the value of $c$ in this equation, we find the value of the equation at $t = 0$ and $v = {v_0}$
Therefore, substituting these values we get,
$ \Rightarrow \dfrac{{ - 1}}{{2{v_0}^2}} = c$
Substituting this value of $c$ in the equation ( $1$ ) , we get
$ \Rightarrow \dfrac{{ - 1}}{{2{v^2}}} = - kt + \dfrac{{ - 1}}{{2{v_0}^2}}$
$ \Rightarrow \dfrac{{ - 1}}{{2{v^2}}} = - kt - \dfrac{1}{{2{v_0}^2}}$
Cancelling out the negative sign from both the sides of the equation, we get
$ \Rightarrow \dfrac{1}{{2{v^2}}} = kt + \dfrac{1}{{2{v_0}^2}}$
$ \Rightarrow \dfrac{1}{{2{v^2}}} = \dfrac{{2{v_0}^2kt + 1}}{{2{v_0}^2}}$
Simplifying the equation and putting all terms other than $v$ on one side, we get
$ \Rightarrow \dfrac{{2{v_0}^2}}{{2{v^2}}} = 2{v_0}^2kt + 1$
$ \Rightarrow \dfrac{{{v_0}^2}}{{{v^2}}} = 2{v_0}^2kt + 1$
$ \Rightarrow \dfrac{{{v_0}^2}}{{2{v_0}^2kt + 1}} = {v^2}$
$ \Rightarrow {v^2} = \dfrac{{{v_0}^2}}{{2{v_0}^2kt + 1}}$
$ \Rightarrow v = \dfrac{{{v_0}}}{{\sqrt {2{v_0}^2kt + 1} }}$

Therefore, option (D), $v = \dfrac{{{v_0}}}{{\sqrt {2{v_0}^2kt + 1} }}$ is the correct option.

Note: Integrating the acceleration gives us velocity and integrating the velocity gives us the distance. On the contrary differentiating the distance with respect to time gives us velocity and differentiating the velocity with respect to time gives us acceleration.