Suppose the position of an object moving in a straight line is given by

s left parenthesis t right parenthesis equals 3 t squared plus 5 t plus 2s(t)=3t2+5t+2.

Find the instantaneous velocity when

t equals 6t=6.

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Part 1

The instantaneous velocity at

tequals=66

is

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.

Suppose the position of an object moving in a straight line is given by s(t) = 3t^2 + 5t + 2. Find the instantaneous velocity when t = 6.

Learn how to find the instantaneous velocity of an object moving along a straight path using calculus. This problem demonstrates the differentiation of a quadratic position function s(t) = 3t^2 + 5t + 2 and evaluates the velocity at t = 6. Ideal for students studying AP Calculus or introductory college-level calculus.

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