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We are given a position function \( s(t) = t^4 - 4t^2 + 8t \) that describes the motion of an object. We need to find the velocity and acceleration at \( t = 2 \).

1. **Velocity** is the first derivative of the position function \( s(t) \), which is denoted by \( v(t) = s'(t) \).

2. **Acceleration** is the second derivative of the position function, denoted by \( a(t) = s''(t) \).

### Step 1: Finding the velocity function \( v(t) \)
We differentiate \( s(t) \) with respect to \( t \):
\[
v(t) = s'(t) = \frac{d}{dt}(t^4 - 4t^2 + 8t)
\]
Using power rule:
\[
v(t) = 4t^3 - 8t + 8
\]

Now, evaluate \( v(2) \):
\[
v(2) = 4(2)^3 - 8(2) + 8 = 4(8) - 16 + 8 = 32 - 16 + 8 = 24 \, \text{m/s}
\]

### Step 2: Finding the acceleration function \( a(t) \)
We differentiate the velocity function \( v(t) = 4t^3 - 8t + 8 \) to get the acceleration:
\[
a(t) = s''(t) = \frac{d}{dt}(4t^3 - 8t + 8)
\]
Again, using the power rule:
\[
a(t) = 12t^2 - 8
\]

Now, evaluate \( a(2) \):
\[
a(2) = 12(2)^2 - 8 = 12(4) - 8 = 48 - 8 = 40 \, \text{m/s}^2
\]

### Final Results:
- \( v(2) = 24 \, \text{m/s} \)
- \( a(2) = 40 \, \text{m/s}^2 \)

Would you like more details on the steps involved? Let me know if you have any questions!

Here are five related questions:
1. How do you interpret the velocity function geometrically?
2. What does it mean if the acceleration is negative?
3. How would you find the position of the object at \( t = 2 \)?
4. How does the object’s velocity change between \( t = 1 \) and \( t = 2 \)?
5. What is the significance of the second derivative test in motion problems?

**Tip:** The derivative gives the rate of change at a specific point. For position functions, the first derivative gives velocity, and the second gives acceleration.

Suppose that s(t) = t^4 - 4t^2 + 8t denotes the position in meters of an object moving along a straight line after t seconds. Find the velocity and acceleration of the object when t = 2.

Learn how to calculate the velocity and acceleration of an object given the position function s(t) = t^4 - 4t^2 + 8t. This calculus problem explores key concepts of derivatives, including finding first and second derivatives. Suitable for students in Grades 11-12 or college-level calculus courses.

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