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.

To solve this problem, we need to determine the particle's velocity \( v \) and position \( s \) at \( t = 4 \) seconds, given that its deceleration is \( a = -2v^3 \) m/s\(^2\).

### Step 1: Write down the given information.
- Deceleration: \( a = -2v^3 \) m/s\(^2\)
- Initial velocity: \( v_0 = 8 \) m/s at \( t = 0 \)
- Initial position: \( s_0 = 10 \) m at \( t = 0 \)

### Step 2: Use the relationship between acceleration and velocity.
Since acceleration is the derivative of velocity with respect to time, we have:
\[
a = \frac{dv}{dt} = -2v^3
\]
This can be rearranged to separate variables:
\[
\frac{dv}{v^3} = -2 \, dt
\]

### Step 3: Integrate both sides.
Integrate the left side with respect to \( v \) and the right side with respect to \( t \):
\[
\int \frac{dv}{v^3} = -2 \int dt
\]

The integral of \( v^{-3} \) with respect to \( v \) is:
\[
\int v^{-3} \, dv = -\frac{1}{2v^2}
\]

So, the equation becomes:
\[
-\frac{1}{2v^2} = -2t + C
\]

### Step 4: Solve for the constant \( C \).
We use the initial condition \( v = 8 \) m/s when \( t = 0 \):
\[
-\frac{1}{2(8)^2} = C
\]
\[
C = -\frac{1}{128}
\]

Thus, the equation is:
\[
-\frac{1}{2v^2} = -2t - \frac{1}{128}
\]

### Step 5: Solve for velocity \( v \) at \( t = 4 \) seconds.
Substitute \( t = 4 \) seconds into the equation:
\[
-\frac{1}{2v^2} = -2(4) - \frac{1}{128}
\]
\[
-\frac{1}{2v^2} = -8 - \frac{1}{128}
\]
\[
-\frac{1}{2v^2} = -\frac{1025}{128}
\]

Taking the reciprocal and solving for \( v \):
\[
\frac{1}{2v^2} = \frac{1025}{128}
\]
\[
2v^2 = \frac{128}{1025}
\]
\[
v^2 = \frac{64}{1025}
\]
\[
v = \sqrt{\frac{64}{1025}} = \frac{8}{\sqrt{1025}} \approx 0.25 \text{ m/s}
\]

### Step 6: Find the position \( s \) at \( t = 4 \) seconds.
The velocity \( v = \frac{ds}{dt} \). To find the position, integrate the velocity function:
\[
v(t) = \frac{8}{\sqrt{1025}}
\]
Given that \( \frac{ds}{dt} = v(t) \), integrate to find \( s(t) \):
\[
s(t) = \int v(t) \, dt = \int \frac{8}{\sqrt{1025}} \, dt
\]
\[
s(t) = \frac{8}{\sqrt{1025}} \cdot t + s_0
\]
Substitute \( t = 4 \) seconds and \( s_0 = 10 \) m:
\[
s(4) = \frac{8 \cdot 4}{\sqrt{1025}} + 10 \approx 10.996 \text{ m}
\]

### Final Answer:
- The velocity of the particle at \( t = 4 \) seconds is approximately \( 0.25 \) m/s.
- The position of the particle at \( t = 4 \) seconds is approximately \( 10.996 \) m.

Do you want any details on a specific part of this solution or have any questions?

Here are 5 related questions:
1. How does the deceleration affect the velocity over time?
2. What would happen if the initial velocity were different?
3. How would the position change if the initial position was at the origin?
4. What is the significance of the deceleration being proportional to \( v^3 \)?
5. Can this method be used to solve problems with different forms of acceleration?

**Tip:** Understanding the relationship between acceleration and velocity through integration is crucial when dealing with non-constant acceleration.

Learn how to determine the velocity and position of a particle moving with non-constant deceleration using calculus. This problem involves integrating acceleration to find velocity and position functions over time. Suitable for advanced high school students.

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