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We are given a velocity function of time for a bicyclist:

\[
v(t) = -0.023t^2 + 1.2t + 1.3
\]

and need to calculate two things:

1. The **bicyclist's displacement** between \( t = 2 \) seconds and \( t = 11 \) seconds.
2. The **bicyclist's average velocity** between \( t = 2 \) seconds and \( t = 11 \) seconds.

### 1. Bicyclist's Displacement

The displacement can be found by integrating the velocity function over the interval from 2 to 11 seconds:

\[
\text{Displacement} = \int_2^{11} v(t) \, dt
\]

Let me compute the integral first to find the displacement.The bicyclist's displacement between \( t = 2 \) seconds and \( t = 11 \) seconds is approximately **71.76 meters**, which matches the provided value of 71.8 meters.

### 2. Bicyclist's Average Velocity

The average velocity over a time interval \([t_1, t_2]\) is given by:

\[
\text{Average velocity} = \frac{\text{Displacement}}{t_2 - t_1}
\]

Substituting \( t_1 = 2 \) and \( t_2 = 11 \), and using the displacement we just calculated, we can compute the average velocity.The bicyclist's average velocity between \( t = 2 \) seconds and \( t = 11 \) seconds is approximately **7.97 m/s**.

Would you like further details on any of the steps or have any other questions?

Here are some related questions you might find interesting:
1. How would you compute the bicyclist's velocity at \( t = 5 \) seconds?
2. Can the velocity function predict the exact moment the bicyclist comes to a stop?
3. How would you determine the total distance traveled between \( t = 0 \) and \( t = 14 \) seconds?
4. What does the negative coefficient of \( t^2 \) in the velocity function indicate about the bicyclist's motion?
5. How would you calculate the instantaneous acceleration at \( t = 10 \) seconds?

**Tip:** The area under a velocity-time graph represents the displacement, which is why integrating velocity over time gives the displacement.

A bicyclist begins at a position of 0m at a time of 0s and then increases in speed for 14s. The bicyclist's velocity (in m/s) is described by the following function of time: v(t) = -0.023t^2 + 1.2t + 1.3. What is the bicyclist's displacement between 2s and 11s, and what is the bicyclist's average velocity between 2s and 11s?

Learn how to calculate the bicyclist's displacement and average velocity using the velocity function v(t) = -0.023t^2 + 1.2t + 1.3 between 2s and 11s. This problem involves calculus and integration, suitable for students in Grades 11-12.

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