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.