If a rock is thrown upward on the planet Mars with a velocity of 
17 m/s,
 its height above the ground (in meters) after t seconds is given by 
H = 17t − 1.86t2.
(a)
Find the velocity (in m/s) of the rock after 1 second.

To find the velocity of the rock after 1 second, we need to calculate the derivative of the height function \( H(t) = 17t - 1.86t^2 \) with respect to time \( t \). The derivative of the height function represents the velocity.

The velocity function \( v(t) \) is given by:

\[
v(t) = \frac{dH}{dt} = \frac{d}{dt} \left( 17t - 1.86t^2 \right)
\]

Differentiating:

\[
v(t) = 17 - 2(1.86)t = 17 - 3.72t
\]

Now, substitute \( t = 1 \) second to find the velocity after 1 second:

\[
v(1) = 17 - 3.72(1) = 17 - 3.72 = 13.28 \, \text{m/s}
\]

Thus, the velocity of the rock after 1 second is **13.28 m/s**.

Would you like further details or clarification? Here are 5 related questions to extend your understanding:

1. What is the velocity at the instant the rock is released (at \( t = 0 \))?
2. How would you find the time at which the rock reaches its maximum height?
3. What is the significance of the negative coefficient in the \( t^2 \) term for the height function?
4. Can you calculate the velocity at \( t = 2 \) seconds?
5. How would you determine the total time the rock spends in the air before hitting the ground?

**Tip:** The derivative of a position function with respect to time always gives the velocity. Keep this in mind for motion problems!

If a rock is thrown upward on the planet Mars with a velocity of 17 m/s, its height above the ground (in meters) after t seconds is given by H = 17t − 1.86t^2. (a) Find the velocity (in m/s) of the rock after 1 second.

Learn how to find the velocity of a rock thrown upward on Mars with an initial velocity of 17 m/s. This problem involves differentiation of a quadratic function representing height. Explore the concepts of velocity and kinematics with detailed steps. Suitable for students in Grades 10-12.

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