The instantaneous velocity of an object is the derivative of its position function $s(t)$ with respect to time $t$.

Given: $s(t) = 3t^2 + 5t + 2$

To find the instantaneous velocity at $t = 6$, we first take the derivative of $s(t)$:

$v(t) = \frac{ds(t)}{dt} = \frac{d}{dt}(3t^2 + 5t + 2)$

Differentiating term by term:

$v(t) = 6t + 5$

Now, substitute $t = 6$ into the velocity equation:

$v(6) = 6(6) + 5 = 36 + 5 = 41$

Thus, the instantaneous velocity at $t = 6$ is:

$v(6) = 41 \, \text{units per time}$

Would you like further details on the differentiation process or have any questions?

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5 Related Questions:

1. How is the derivative of a position function related to velocity?
2. How would you find acceleration from the velocity function?
3. What happens if the position function is more complex, like a cubic function?
4. What is the physical meaning of negative velocity in this context?
5. How would the velocity change if we added more terms to the position function?

Tip: The instantaneous rate of change of any function at a given point is found by taking its derivative.