A particle moving along a straight line is subjected to a deceleration a = (-2v3) m/s2, where v is in m/s. If it has a velocity v = 8m/s and a position s = 10m when t = 0, determine its velocity and position when t = 4 s.

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:



Learn how to determine the velocity and position of a particle moving along a straight line with a given deceleration. This problem involves solving a separable differential equation and understanding the relationship between velocity, acceleration, and position. Suitable for college-level physics and mathematics students.

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