Let's solve this problem step by step.

Given:

The height $y(t)$ of the rock at time $t$ seconds is given by the equation:

$y(t) = 6t - 1.86t^2$

Part (a): Average Velocity

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

$\text{Average Velocity} = \frac{y(t_2) - y(t_1)}{t_2 - t_1}$

We need to calculate this for the different time intervals.

(i) $[1, 2]$

1. Calculate $y(1)$:

$y(1) = 6(1) - 1.86(1)^2 = 6 - 1.86 = 4.14 \text{ meters}$

2. Calculate $y(2)$:

$y(2) = 6(2) - 1.86(2)^2 = 12 - 7.44 = 4.56 \text{ meters}$

3. Calculate the average velocity:

$\text{Average Velocity} = \frac{4.56 - 4.14}{2 - 1} = \frac{0.42}{1} = 0.42 \text{ m/s}$

(ii) $[1, 1.5]$

1. Calculate $y(1.5)$:

$y(1.5) = 6(1.5) - 1.86(1.5)^2 = 9 - 4.185 = 4.815 \text{ meters}$

2. Calculate the average velocity:

$\text{Average Velocity} = \frac{4.815 - 4.14}{1.5 - 1} = \frac{0.675}{0.5} = 1.35 \text{ m/s}$

(iii) $[1, 1.1]$

1. Calculate $y(1.1)$:

$y(1.1) = 6(1.1) - 1.86(1.1)^2 = 6.6 - 2.24826 = 4.35174 \text{ meters}$

2. Calculate the average velocity:

$\text{Average Velocity} = \frac{4.35174 - 4.14}{1.1 - 1} = \frac{0.21174}{0.1} = 2.12 \text{ m/s}$

(iv) $[1, 1.01]$

1. Calculate $y(1.01)$:

$y(1.01) = 6(1.01) - 1.86(1.01)^2 = 6.06 - 1.882986 = 4.177014 \text{ meters}$

2. Calculate the average velocity:

$\text{Average Velocity} = \frac{4.177014 - 4.14}{1.01 - 1} = \frac{0.037014}{0.01} = 3.70 \text{ m/s}$

(v) $[1, 1.001]$

1. Calculate $y(1.001)$:

$y(1.001) = 6(1.001) - 1.86(1.001)^2 = 6.006 - 1.86000372 = 4.14599628 \text{ meters}$

2. Calculate the average velocity:

$\text{Average Velocity} = \frac{4.14599628 - 4.14}{1.001 - 1} = \frac{0.00599628}{0.001} = 5.99628 \text{ m/s}$

Part (b): Estimate the Instantaneous Velocity at $t = 1$

The instantaneous velocity at $t = 1$ can be estimated by observing the trend in the average velocities as the time interval decreases.

From the calculations above, the average velocities were:

1. (i) $0.42$ m/s
2. (ii) $1.35$ m/s
3. (iii) $2.12$ m/s
4. (iv) $3.70$ m/s
5. (v) $5.996$ m/s

As the time interval gets smaller, the average velocity approaches the instantaneous velocity. Based on the trend, the instantaneous velocity at $t = 1$ can be estimated as approximately:

$\text{Instantaneous Velocity at } t = 1 \text{ is approximately } 6.00 \text{ m/s}$

Final Answers:

• (a)
  • (i) The average velocity over [1, 2] is $0.42$ m/s.
  • (ii) The average velocity over [1, 1.5] is $1.35$ m/s.
  • (iii) The average velocity over [1, 1.1] is $2.12$ m/s.
  • (iv) The average velocity over [1, 1.01] is $3.70$ m/s.
  • (v) The average velocity over [1, 1.001] is $5.996$ m/s.
• (b) The estimated instantaneous velocity at $t = 1$ is approximately $6.00$ m/s.

Would you like more details or have any questions?

Follow-up Questions:

1. What is the physical meaning of the instantaneous velocity in this context?
2. How would the velocity change if the initial speed of the rock were higher?
3. What would the graph of $y(t)$ look like, and how does it relate to the velocity?
4. How does the value of instantaneous velocity compare to the initial velocity of the rock?
5. Can you calculate the maximum height the rock reaches on Mars?

Tip:

Instantaneous velocity can be found by taking the derivative of the position function with respect to time. The derivative gives the exact rate of change at any given moment.