given the equation for velocity: v(t) = 6t^2 - 30t -36 and acceleration: a(t) = 12t -30, find where the object is slowing down and speeding up

To determine where the object is speeding up or slowing down, we need to analyze the velocity \( v(t) \) and the acceleration \( a(t) \). Here's the step-by-step method:

1. **Speeding Up**: This happens when the velocity and acceleration have the same sign, meaning both are either positive or negative at the same time.
   
2. **Slowing Down**: This occurs when the velocity and acceleration have opposite signs, i.e., one is positive and the other is negative.

### Step 1: Find the critical points of velocity

We are given the velocity function:

\[
v(t) = 6t^2 - 30t - 36
\]

To find critical points, we set \( v(t) = 0 \) and solve for \( t \):

\[
6t^2 - 30t - 36 = 0
\]

Dividing the entire equation by 6:

\[
t^2 - 5t - 6 = 0
\]

Factoring the quadratic:

\[
(t - 6)(t + 1) = 0
\]

So the solutions are:

\[
t = 6 \quad \text{or} \quad t = -1
\]

These are the critical points where the velocity changes sign.

### Step 2: Analyze the velocity intervals

Now, let's check the sign of \( v(t) \) in the intervals divided by \( t = -1 \) and \( t = 6 \).

- For \( t < -1 \) (e.g., \( t = -2 \)): 
  \[
  v(-2) = 6(-2)^2 - 30(-2) - 36 = 24 + 60 - 36 = 48 \quad \text{(positive)}
  \]
- For \( -1 < t < 6 \) (e.g., \( t = 0 \)): 
  \[
  v(0) = 6(0)^2 - 30(0) - 36 = -36 \quad \text{(negative)}
  \]
- For \( t > 6 \) (e.g., \( t = 7 \)): 
  \[
  v(7) = 6(7)^2 - 30(7) - 36 = 294 - 210 - 36 = 48 \quad \text{(positive)}
  \]

Thus, the velocity is:
- Positive for \( t < -1 \) and \( t > 6 \)
- Negative for \( -1 < t < 6 \)

### Step 3: Analyze the acceleration

We are given the acceleration function:

\[
a(t) = 12t - 30
\]

Set \( a(t) = 0 \) to find when the acceleration changes sign:

\[
12t - 30 = 0 \implies t = 2.5
\]

Now, check the sign of \( a(t) \) in the intervals:

- For \( t < 2.5 \) (e.g., \( t = 0 \)): 
  \[
  a(0) = 12(0) - 30 = -30 \quad \text{(negative)}
  \]
- For \( t > 2.5 \) (e.g., \( t = 3 \)): 
  \[
  a(3) = 12(3) - 30 = 36 - 30 = 6 \quad \text{(positive)}
  \]

Thus, the acceleration is:
- Negative for \( t < 2.5 \)
- Positive for \( t > 2.5 \)

### Step 4: Determine intervals of speeding up and slowing down

- **Slowing down** occurs when the velocity and acceleration have opposite signs.
- **Speeding up** occurs when the velocity and acceleration have the same sign.

#### Slowing Down:
- For \( -1 < t < 2.5 \), velocity is negative and acceleration is negative → speeding up.
- For \( 2.5 < t < 6 \), velocity is negative and acceleration is positive → slowing down.

#### Speeding Up:
- For \( t < -1 \), both velocity and acceleration are positive → speeding up.
- For \( t > 6 \), both velocity and acceleration are positive → speeding up.

### Conclusion:

- **Speeding up**: \( t < -1 \) and \( t > 6 \)
- **Slowing down**: \( 2.5 < t < 6 \)

---

Would you like further details or explanations on this? Here are some related questions to explore:

1. How do velocity and acceleration determine motion behavior?
2. What happens at the points where velocity or acceleration is zero?
3. Can we visualize these intervals on a graph?
4. How do changes in velocity affect total displacement?
5. How can we extend this analysis to more complex functions?

**Tip**: Always check critical points and sign changes for velocity and acceleration to analyze motion effectively.

Given the equation for velocity: v(t) = 6t^2 - 30t -36 and acceleration: a(t) = 12t -30, find where the object is slowing down and speeding up.

Learn how to determine where an object is speeding up or slowing down using the equations for velocity v(t) = 6t^2 - 30t - 36 and acceleration a(t) = 12t - 30. This detailed step-by-step solution explains key calculus concepts such as critical points, sign changes, and motion analysis. Suitable for high school calculus students.

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