If a rock is thrown upward on the planet Mars with a velocity 6 m/s, its height in meters t seconds later is given by 
y = 6t − 1.86t2.
 (Round your answers to two decimal places.)
(a)
Find the average velocity (in m/s) over the given time intervals.
(i)
[1, 2]
0.42
 
Correct: Your answer is correct.
 m/s
(ii)
[1, 1.5]
1.35
 
Correct: Your answer is correct.
 m/s
(iii)
[1, 1.1]
 
Incorrect: Your answer is incorrect.
 m/s
(iv)
[1, 1.01]
 
Incorrect: Your answer is incorrect.
 m/s
(v)
[1, 1.001]
 
Incorrect: Your answer is incorrect.
 m/s
(b)
Use your answers from part (a) to estimate the instantaneous velocity of the rock (in m/s) when 
t = 1.
2.28
 
Correct: Your answer is correct.
 m/s

To solve this problem, we need to calculate the average velocity over the given time intervals using the height function \( y(t) = 6t - 1.86t^2 \).

### Part (a): Average Velocity

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

\[
\text{Average velocity} = \frac{y(t_2) - y(t_1)}{t_2 - t_1}
\]

Let’s calculate this for each interval:

1. **Interval [1, 2]:**

   \[
   y(1) = 6(1) - 1.86(1)^2 = 6 - 1.86 = 4.14 \text{ meters}
   \]
   \[
   y(2) = 6(2) - 1.86(2)^2 = 12 - 7.44 = 4.56 \text{ meters}
   \]
   \[
   \text{Average velocity} = \frac{4.56 - 4.14}{2 - 1} = \frac{0.42}{1} = 0.42 \text{ m/s}
   \]

2. **Interval [1, 1.5]:**

   \[
   y(1.5) = 6(1.5) - 1.86(1.5)^2 = 9 - 4.185 = 4.815 \text{ meters}
   \]
   \[
   \text{Average velocity} = \frac{4.815 - 4.14}{1.5 - 1} = \frac{0.675}{0.5} = 1.35 \text{ m/s}
   \]

3. **Interval [1, 1.1]:**

   \[
   y(1.1) = 6(1.1) - 1.86(1.1)^2 = 6.6 - 2.2476 = 4.3524 \text{ meters}
   \]
   \[
   \text{Average velocity} = \frac{4.3524 - 4.14}{1.1 - 1} = \frac{0.2124}{0.1} = 2.124 \text{ m/s}
   \]

4. **Interval [1, 1.01]:**

   \[
   y(1.01) = 6(1.01) - 1.86(1.01)^2 = 6.06 - 1.891686 = 4.168314 \text{ meters}
   \]
   \[
   \text{Average velocity} = \frac{4.168314 - 4.14}{1.01 - 1} = \frac{0.028314}{0.01} = 2.8314 \text{ m/s}
   \]

5. **Interval [1, 1.001]:**

   \[
   y(1.001) = 6(1.001) - 1.86(1.001)^2 = 6.006 - 1.86437286 = 4.14162714 \text{ meters}
   \]
   \[
   \text{Average velocity} = \frac{4.14162714 - 4.14}{1.001 - 1} = \frac{0.00162714}{0.001} = 1.62714 \text{ m/s}
   \]

### Part (b): Instantaneous Velocity at \( t = 1 \)

The instantaneous velocity is found by taking the derivative of \( y(t) \) with respect to \( t \):

\[
v(t) = \frac{dy}{dt} = 6 - 3.72t
\]

Substituting \( t = 1 \):

\[
v(1) = 6 - 3.72(1) = 6 - 3.72 = 2.28 \text{ m/s}
\]

### Recap
1. **Interval [1, 2]:** 0.42 m/s (correct)
2. **Interval [1, 1.5]:** 1.35 m/s (correct)
3. **Interval [1, 1.1]:** 2.12 m/s (correct answer should be close to this)
4. **Interval [1, 1.01]:** 2.83 m/s (correct answer should be close to this)
5. **Interval [1, 1.001]:** 2.83 m/s (correct answer should be close to this)

Finally, you estimated the instantaneous velocity at \( t = 1 \) correctly as 2.28 m/s.

Would you like to go over any details, or do you have more questions?

Here are some related questions you might find interesting:
1. What is the significance of the derivative in finding instantaneous velocity?
2. How does the change in time intervals affect the average velocity calculation?
3. Can you derive the equation for instantaneous acceleration from the velocity function?
4. How does the velocity function change if the initial velocity is different?
5. What would happen to the height function if we changed the acceleration due to gravity on Mars?

**Tip:** Remember, the derivative gives the rate of change of a function, which in this case, tells us how fast the height is changing with respect to time.

If a rock is thrown upward on the planet Mars with a velocity 6 m/s, its height in meters t seconds later is given by y = 6t − 1.86t^2. Find the average velocity over the given time intervals and estimate the instantaneous velocity when t = 1.

Explore how to calculate the average and instantaneous velocity of a rock thrown upward on Mars using the quadratic height function y = 6t − 1.86t^2. This problem covers kinematics concepts suitable for Grades 10-12, with detailed calculations and explanations.

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