 The velocity of a ball in the horizontal direction is given as the following function:

v(t) = 20 + 3t3,


where t is in seconds and the velocity is in meters per second. Part (a)
Find the change in the position, in meters, between t = 0 and t = 6.6 s. 

To find the change in position of the ball, we need to compute the displacement, which is the integral of the velocity function \( v(t) \) with respect to time over the interval from \( t = 0 \) to \( t = 6.6 \) seconds.

The velocity function is given as:

\[
v(t) = 20 + 3t^3
\]

The change in position (displacement) \( \Delta x \) is the integral of \( v(t) \) from \( t = 0 \) to \( t = 6.6 \):

\[
\Delta x = \int_0^{6.6} v(t) \, dt = \int_0^{6.6} (20 + 3t^3) \, dt
\]

Now, let's compute this integral:

\[
\int (20 + 3t^3) \, dt = 20t + \frac{3t^4}{4}
\]

Now, we'll evaluate this from \( t = 0 \) to \( t = 6.6 \).

\[
\Delta x = \left[ 20t + \frac{3t^4}{4} \right]_0^{6.6}
\]

Substituting the limits \( t = 6.6 \) and \( t = 0 \):

\[
\Delta x = \left( 20(6.6) + \frac{3(6.6)^4}{4} \right) - \left( 20(0) + \frac{3(0)^4}{4} \right)
\]

Let me calculate that for you.The change in the position of the ball between \( t = 0 \) and \( t = 6.6 \) seconds is approximately \( 1555.11 \) meters.

Do you need further details or have any questions?

Here are 5 related questions for further exploration:
1. How would the displacement change if the velocity function were different?
2. What is the physical interpretation of the integral of a velocity function?
3. How would you find the maximum velocity during the interval \( t = 0 \) to \( t = 6.6 \)?
4. Can you compute the acceleration of the ball from the given velocity function?
5. How would you determine the average velocity over the same interval?

**Tip:** The area under a velocity-time graph gives the displacement, and integrating the velocity function serves the same purpose.

The velocity of a ball in the horizontal direction is given as the following function: v(t) = 20 + 3t^3, where t is in seconds and the velocity is in meters per second. Part (a) Find the change in the position, in meters, between t = 0 and t = 6.6 s.

Learn how to calculate the displacement of a ball with a velocity function v(t) = 20 + 3t^3 over the time interval t = 0 to t = 6.6 seconds. This solution uses integral calculus to determine the change in position, yielding a result of approximately 1555.11 meters. Suitable for students in Grades 11-12.

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