s = (1/2) (t^2) − 6t + 23 Find the instantaneous velocity when t = 8

The instantaneous velocity is the derivative of the position function \( s(t) \) with respect to time \( t \). The given function is:

\[
s(t) = \frac{1}{2} t^2 - 6t + 23
\]

To find the instantaneous velocity, we'll differentiate \( s(t) \) to obtain \( v(t) \), the velocity function:

\[
v(t) = \frac{d}{dt} \left( \frac{1}{2} t^2 - 6t + 23 \right)
\]

The derivative of \( s(t) \) is:

\[
v(t) = t - 6
\]

Now, substitute \( t = 8 \) into the velocity function to find the instantaneous velocity at that moment:

\[
v(8) = 8 - 6 = 2
\]

Thus, the instantaneous velocity when \( t = 8 \) is **2 units per time**.

Would you like further details or explanations on this problem?

Here are 5 related questions to explore further:
1. How do you interpret the physical meaning of velocity at a specific time?
2. What is the difference between instantaneous and average velocity?
3. Can you find the acceleration by differentiating the velocity function?
4. What is the general formula for velocity in terms of the position function?
5. How does velocity change with respect to time for quadratic functions?

**Tip:** The instantaneous velocity at any time is the slope of the tangent to the position-time graph at that point.

s = (1/2) (t^2) − 6t + 23. Find the instantaneous velocity when t = 8.

Learn how to find the instantaneous velocity for the quadratic function s(t) = (1/2)t^2 - 6t + 23 by differentiating the position function with respect to time. This calculus-based problem is suitable for students in Grades 11-12 and undergraduate level.

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