Thank you for the clarification. Let's solve the problem with the correct equation provided:

The height $y$ of the rock, in meters, after $t$ seconds is given by:

$y(t) = 16t - 1.86t^2$

Part (a): Find the average velocity over the given time intervals.

The formula for average velocity over the interval $[t_1, t_2]$ is:

$v_{\text{avg}} = \frac{y(t_2) - y(t_1)}{t_2 - t_1}$

1. For the interval $t = 1$ to $t = 2$:

   $v_{\text{avg}} = \frac{y(2) - y(1)}{2 - 1}$

   Calculate $y(2)$ and $y(1)$:

   $y(2) = 16(2) - 1.86(2^2) = 32 - 7.44 = 24.56 \text{ meters}$

   $y(1) = 16(1) - 1.86(1^2) = 16 - 1.86 = 14.14 \text{ meters}$

   Therefore:

   $v_{\text{avg}} = \frac{24.56 - 14.14}{1} = 10.42 \text{ m/s}$

2. For the interval $t = 1$ to $t = 1.5$:

   $v_{\text{avg}} = \frac{y(1.5) - y(1)}{1.5 - 1}$

   Calculate $y(1.5)$:

   $y(1.5) = 16(1.5) - 1.86(1.5^2) = 24 - 4.185 = 19.815 \text{ meters}$

   Therefore:

   $v_{\text{avg}} = \frac{19.815 - 14.14}{0.5} = \frac{5.675}{0.5} = 11.35 \text{ m/s}$

3. For the interval $t = 1$ to $t = 1.1$:

   $v_{\text{avg}} = \frac{y(1.1) - y(1)}{1.1 - 1}$

   Calculate $y(1.1)$:

   $y(1.1) = 16(1.1) - 1.86(1.1^2) = 17.6 - 2.2466 = 15.3534 \text{ meters}$

   Therefore:

   $v_{\text{avg}} = \frac{15.3534 - 14.14}{0.1} = \frac{1.2134}{0.1} = 12.13 \text{ m/s}$

4. For the interval $t = 1$ to $t = 1.01$:

   $v_{\text{avg}} = \frac{y(1.01) - y(1)}{1.01 - 1}$

   Calculate $y(1.01)$:

   $y(1.01) = 16(1.01) - 1.86(1.01^2) = 16.16 - 1.888686 = 14.271314 \text{ meters}$

   Therefore:

   $v_{\text{avg}} = \frac{14.271314 - 14.14}{0.01} = \frac{0.131314}{0.01} = 13.13 \text{ m/s}$

5. For the interval $t = 1$ to $t = 1.001$:

   $v_{\text{avg}} = \frac{y(1.001) - y(1)}{1.001 - 1}$

   Calculate $y(1.001)$:

   $y(1.001) = 16(1.001) - 1.86(1.001^2) = 16.016 - 1.862862 = 14.153138 \text{ meters}$

   Therefore:

   $v_{\text{avg}} = \frac{14.153138 - 14.14}{0.001} = \frac{0.013138}{0.001} = 13.14 \text{ m/s}$

Part (b): Use the answers from Part (a) to estimate the instantaneous velocity at $t = 1$.

To find the instantaneous velocity at $t = 1$, we can observe the average velocities as the time interval gets smaller. The values approach approximately $13.14$ m/s. Thus, the instantaneous velocity at $t = 1$ is approximately $13.14$ m/s.

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Questions for Further Exploration:

1. How can we use derivatives to calculate the exact instantaneous velocity?
2. What is the significance of the coefficient $-1.86$ in the equation?
3. How would the height equation change if the rock were thrown with a different initial velocity?
4. How can we determine the time at which the rock reaches its maximum height?
5. What role does gravity play in the motion of the rock on Mars compared to Earth?

Tip:

When estimating instantaneous velocity from average velocities, observe the trend in the calculated values as the time interval becomes increasingly small. This trend gives insight into the instantaneous velocity.