The problem involves an object moving in rectilinear motion, and its signed distance $s$ (in meters) from the origin after $t$ seconds is given by the position function:

$s(t) = 3t^2 + 8t$

We are tasked with finding the velocity $v(t)$ at different times, where the velocity is the derivative of the position function $s(t)$.

Step 1: Find the velocity function

The velocity is the derivative of the position function $s(t)$ with respect to time $t$:

$v(t) = \frac{d}{dt}[3t^2 + 8t] = 6t + 8$

Step 2: Evaluate the velocity at specific times

1. At $t_0 = 0$: $v(0) = 6(0) + 8 = 8 \, \text{m/s}$

2. At $t_0 = 2$: $v(2) = 6(2) + 8 = 12 + 8 = 20 \, \text{m/s}$

3. At any time $t_0$: The general expression for velocity is: $v(t_0) = 6t_0 + 8$

Final Answers:

• $v(0) = 8 \, \text{m/s}$
• $v(2) = 20 \, \text{m/s}$
• $v(t_0) = 6t_0 + 8 \, \text{m/s}$

Would you like more details or explanations? Here are five related questions to consider:

1. What is the acceleration function for this object?
2. How does the position function change if the initial velocity changes?
3. What happens to the velocity if we change the function to $s(t) = 3t^2 + 8t + 5$?
4. How can we find the time when the velocity is zero?
5. What is the object's displacement after 4 seconds?

Tip: Remember that velocity is the rate of change of position, and acceleration is the rate of change of velocity.