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.