s(t) = (t-1)^3 - 8(t-1). determine the time intervals for when the train is slowing down and speeding up

To determine when the train is speeding up or slowing down, we need to analyze the velocity and acceleration of the train. This involves:

1. Finding the **velocity** \( v(t) \), which is the first derivative of the position function \( s(t) \).
2. Finding the **acceleration** \( a(t) \), which is the derivative of the velocity function (or the second derivative of \( s(t) \)).
3. Determining when the velocity and acceleration have the same sign (speeding up) and when they have opposite signs (slowing down).

### Step 1: Find the velocity \( v(t) \)

The velocity is the first derivative of the position function:

\[
v(t) = \frac{d}{dt} \left[ (t-1)^3 - 8(t-1) \right]
\]

We use the chain rule to differentiate each term:

\[
v(t) = 3(t-1)^2 - 8
\]

### Step 2: Find the acceleration \( a(t) \)

The acceleration is the derivative of the velocity function:

\[
a(t) = \frac{d}{dt} \left[ 3(t-1)^2 - 8 \right]
\]

Differentiating:

\[
a(t) = 6(t-1)
\]

### Step 3: Analyze the signs of \( v(t) \) and \( a(t) \)

- **Speeding up** occurs when \( v(t) \) and \( a(t) \) have the **same sign**.
- **Slowing down** occurs when \( v(t) \) and \( a(t) \) have **opposite signs**.

#### Analyze velocity \( v(t) = 3(t-1)^2 - 8 \)

To find the critical points, set \( v(t) = 0 \):

\[
3(t-1)^2 - 8 = 0
\]

\[
3(t-1)^2 = 8
\]

\[
(t-1)^2 = \frac{8}{3}
\]

\[
t - 1 = \pm \sqrt{\frac{8}{3}}
\]

\[
t = 1 \pm \sqrt{\frac{8}{3}}
\]

So the critical points are:

\[
t \approx 1 \pm 1.632
\]

These correspond to:

\[
t_1 \approx 2.632 \quad \text{and} \quad t_2 \approx -0.632
\]

#### Analyze acceleration \( a(t) = 6(t-1) \)

The acceleration is 0 when \( t = 1 \).

### Step 4: Determine intervals

Now we evaluate the signs of \( v(t) \) and \( a(t) \) over the intervals:

- For \( t < -0.632 \): \( v(t) > 0 \), \( a(t) < 0 \) (slowing down)
- For \( -0.632 < t < 1 \): \( v(t) < 0 \), \( a(t) < 0 \) (speeding up)
- For \( 1 < t < 2.632 \): \( v(t) < 0 \), \( a(t) > 0 \) (slowing down)
- For \( t > 2.632 \): \( v(t) > 0 \), \( a(t) > 0 \) (speeding up)

### Conclusion

- The train is **slowing down** on the intervals \( (-\infty, -0.632) \cup (1, 2.632) \).
- The train is **speeding up** on the intervals \( (-0.632, 1) \cup (2.632, \infty) \).

Would you like more details or have any questions?

Here are some related questions for further practice:

1. How do you determine when an object changes direction?
2. How can you find when the velocity reaches a maximum or minimum?
3. What is the relationship between acceleration and concavity in motion problems?
4. How do you apply these concepts to motion in 2D or 3D?
5. Can you analyze motion without knowing the exact position function?

**Tip**: Always consider the signs of both velocity and acceleration to determine whether an object is speeding up or slowing down.

The query submitted by the user: s(t) = (t-1)^3 - 8(t-1). Determine the time intervals for when the train is slowing down and speeding up.

Learn how to analyze the motion of a train using calculus by determining the time intervals for when the train is slowing down and speeding up. This solution uses derivatives to calculate velocity and acceleration, including a step-by-step breakdown of key intervals. Ideal for college-level calculus students.

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