Given the deceleration $a = -2v^3$ m/s², where $v$ is in m/s, we need to determine the velocity and position of the particle at $t = 4$ seconds, given that at $t = 0$, $v = 8$ m/s and $s = 10$ m.

Step 1: Relate acceleration to velocity

The acceleration $a$ is the derivative of velocity with respect to time:

$a = \frac{dv}{dt}$

Given $a = -2v^3$, we have:

$\frac{dv}{dt} = -2v^3$

This is a separable differential equation, so we separate the variables:

$\frac{dv}{v^3} = -2 dt$

Step 2: Integrate both sides

Integrating both sides with respect to their respective variables: