91Ó°ÊÓ

Derivatives are a fundamental concept in calculus, often used to express how a quantity changes with respect to another. In physics, derivatives are crucial for determining the rate of change of physical quantities. In this exercise, we are interested in calculating the velocity of a rock thrown upwards on Mars. The height of the rock as a function of time is given by \[ h(t) = 15t - 1.86t^2. \]
To find the velocity, we must determine how the height changes with time by taking its derivative. This process involves calculating the instantaneous rate of change of height, which is the velocity:

• Derive the height function \(h(t)\) with respect to time \(t\).
• Use the formula \( v(t) = \frac{dh}{dt} \).

The derivative of our function gives:\[ v(t) = 15 - 3.72t. \]This function now defines the rock's velocity at any given time \(t\). By substituting specific time values into this derived formula, we can determine the rock's velocity at those instances.

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \). A critical point of this exercise involves solving for \(t\) using the equation of height \(15t - 1.86t^2 = 25\). Setting it to zero helps us form:\[ 1.86t^2 - 15t + 25 = 0. \]
This is a quadratic equation, and we solve it using the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]where:

• \(a = 1.86\),
• \(b = -15\),
• \(c = 25\).

After plugging in these values and calculating the discriminant \((b^2 - 4ac)\), we find the roots giving the times when the rock reaches a height of 25 meters. These roots represent two moments: once on its way up and once on its way down.

Velocity is the rate of change of an object's position with respect to time. It is a vector quantity, meaning it has both magnitude and direction. In this problem, we compute velocity to understand the kinematics of a rock thrown vertically on Mars. Given the derivative: \[ v(t) = 15 - 3.72t, \]
we find:

• The initial velocity is \( v(0) = 15 \text{ m/s} \).
• At \( t = 2 \text{ seconds} \), the velocity is \( 7.56 \text{ m/s} \).

Velocity indicates how quickly and in what direction the rock's position changes. Positive values show upward movement, while negative values denote downward movement. Calculating velocities after solving the quadratic equation roots tells us how fast the rock travels at 25 meters.

Mars presents a unique environment for physics problems due to its distinct gravity, which is about \(0.38\) that of Earth's. Calculating the motion of objects like a rock thrown upward involves considering this lower gravitational pull. In the given problem, the formula for height derived from Mars's gravity is:\[ h(t) = 15t - 1.86t^2. \]
The parameter \(1.86\) reflects the acceleration due to Mars's gravity, approximately \( 3.72 \text{ m/s}^2 \), which influences how quickly the rock decelerates upward and accelerates downward. Such exercises help us understand the dynamics of motion in different planetary environments, revealing how gravity impacts object movement and velocity on Mars compared to Earth. This opens perspectives on planning real-world space missions or understanding mars' geophysical phenomena.