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To find the expression for the car's velocity \( v(t) \), we need to take the derivative of the position function \( x(t) \) with respect to time \( t \).

The position function is given as:
\[
x(t) = \frac{2}{3}t^3 - 8t^2 + 30t + 36
\]

Now, let's differentiate each term:

1. The derivative of \( \frac{2}{3}t^3 \) is:
   \[
   \frac{d}{dt} \left(\frac{2}{3}t^3\right) = 2t^2
   \]
2. The derivative of \( -8t^2 \) is:
   \[
   \frac{d}{dt} (-8t^2) = -16t
   \]
3. The derivative of \( 30t \) is:
   \[
   \frac{d}{dt} (30t) = 30
   \]
4. The derivative of the constant \( 36 \) is:
   \[
   \frac{d}{dt} (36) = 0
   \]

Thus, the velocity function \( v(t) \) is:
\[
v(t) = 2t^2 - 16t + 30
\]

Would you like more details on this differentiation process or further clarifications?

### Related questions:
1. What is the car's acceleration \( a(t) \) based on the velocity function?
2. At what time does the car reach its maximum velocity?
3. How can we find the time when the car comes to a stop?
4. How would the behavior of the car change if the \( t^3 \) term was absent in the position function?
5. What is the physical meaning of each term in the velocity function?

**Tip:** Remember that velocity is the first derivative of position, while acceleration is the derivative of velocity.

A car on a track is moving such that its position relative to some fixed point is given by x(t) = (2/3)t^3 - 8t^2 + 30t + 36, where x is measured in centimeters and t is measured in seconds. Write an expression for the car's velocity, v(t).

Learn how to derive the car's velocity from its position function x(t) = (2/3)t^3 - 8t^2 + 30t + 36 by taking the first derivative with respect to time. This problem introduces fundamental calculus concepts, such as derivatives of polynomials, and is suitable for students in Grades 11-12.

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