Problem 10 - Thomas Calculus

Question

Lunar projectile motion A rock thrown vertically upward from the surface of the moon at a velocity of 24 \(\mathrm{m} / \mathrm{sec}\) (about 86 \(\mathrm{km} / \mathrm{h}\) ) reaches a height of \(s=24 t-0.8 t^{2}\) meters in \(t\) sec. a. Find the rock's velocity and acceleration at time \(t .\) (The acceleration in this case is the acceleration of gravity on the moon.) b. How long does it take the rock to reach its highest point? c. How high does the rock go? d. How long does it take the rock to reach half its maximum height? e. How long is the rock aloft?

Step-by-Step Solution

Answer

Velocity: \(v(t) = 24 - 1.6t\), Acceleration: \(-1.6\); Max height at \(t = 15\)s, height 180m; Rock aloft for 30s.

Step 1: Find the Velocity Function

The velocity function is the derivative of the position function with respect to time. Given the height function \(s(t) = 24t - 0.8t^2\), we differentiate with respect to \(t\):\[v(t) = \frac{d}{dt}(24t - 0.8t^2) = 24 - 1.6t.\]

Step 2: Find the Acceleration Function

The acceleration function is the derivative of the velocity function with respect to time. Differentiate the velocity function \(v(t) = 24 - 1.6t\) with respect to \(t\):\[a(t) = \frac{d}{dt}(24 - 1.6t) = -1.6.\]This represents the acceleration due to gravity on the moon.

Step 3: Calculate Time to Reach Maximum Height

To find when the rock reaches its maximum height, set the velocity function equal to zero since the velocity is zero at the peak:\[24 - 1.6t = 0\]Solving for \(t\), we get:\[t = \frac{24}{1.6} = 15.\]The rock reaches its maximum height at \(t = 15\) seconds.

Step 4: Determine Maximum Height

Substitute \(t = 15\) seconds back into the height function to find the maximum height:\[s(15) = 24(15) - 0.8(15)^2 = 360 - 180 = 180.\]The rock reaches a maximum height of 180 meters.

Step 5: Calculate Time to Reach Half Maximum Height

Half of the maximum height is \(90\) meters. Set the height function equal to 90 and solve for \(t\):\[24t - 0.8t^2 = 90.\]Rearranging gives:\[0.8t^2 - 24t + 90 = 0.\]Use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 0.8\), \(b = -24\), \(c = 90\):\[t = \frac{24 \pm \sqrt{576 - 288}}{1.6}.\]\[t = \frac{24 \pm \sqrt{288}}{1.6}.\]Solving this, we find two times: \(t \approx 4\) and \(t \approx 11.25\) seconds.

Step 6: Calculate Time Rock is Aloft

Set the height function equal to zero to find when the rock hits the ground:\[24t - 0.8t^2 = 0.\]Factor the expression:\[t(24 - 0.8t) = 0.\]This gives \(t = 0\) for the start, and:\[24 - 0.8t = 0 \Rightarrow t = 30\]The rock is aloft for 30 seconds.

Key Concepts

Velocity Function

In lunar projectile motion, understanding the velocity function is crucial. The velocity function represents how fast and in what direction an object is moving at any given moment. To find it, one must differentiate the position function with respect to time.
Given the position function for a rock launched on the moon, \(s(t) = 24t - 0.8t^2\), the velocity, \(v(t)\), is the first derivative:

• Velocity function: \(v(t) = \frac{d}{dt}(24t - 0.8t^2)\).
• This equals \(v(t) = 24 - 1.6t\).

This function shows that the velocity decreases as a linear function of time due to gravitational pull being exerted on it. Initially, the rock is thrown upwards with a velocity of 24 m/s, but this velocity decreases by 1.6 m/s each second.

Acceleration Function

Acceleration is the rate at which velocity changes over time. On the moon, the constant acceleration due to gravity is much lower than on Earth. Calculating this acceleration is simple once we have the velocity function.
Given our velocity function, \(v(t) = 24 - 1.6t\), the acceleration function, \(a(t)\), is the derivative of velocity:

• Acceleration function: \(a(t) = \frac{d}{dt}(24 - 1.6t)\).
• This equals \(a(t) = -1.6\).

A constant negative acceleration indicates that the rock slows down as it ascends. It is important here since lunar gravity causes all objects to decelerate upwards at a steady rate of 1.6 m/s².

Maximum Height Calculation

When calculating the maximum height of a rock thrown upwards, we need to focus on when its velocity becomes zero, as it reaches the peak of its motion. At this point, upward movement halts and begins its descent.
To find the time it takes to reach this peak:

• Set \(v(t) = 0\), which gives \(24 - 1.6t = 0\).
• Solve for \(t\) to find \(t = 15\) seconds.

Once we know the time, substitute back into the position function \(s(t)\):

• At \(t = 15\), \(s(15) = 24(15) - 0.8(15)^2\).
• This calculates to \(s(15) = 180\) meters.

Thus, the maximum height the rock reaches is 180 meters.

Quadratic Formula in Physics

The quadratic formula is essential in physics for solving motion problems involving time, such as when the rock reaches a certain height. When half of the maximum height, namely 90 meters, is reached, it requires solving a quadratic equation.
Consider the equation formed from the height function:

• Set \(24t - 0.8t^2 = 90\), rearranging to \(0.8t^2 - 24t + 90 = 0\).

Use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

• Here, \(a = 0.8\), \(b = -24\), \(c = 90\).
• Solving, we find \(t = \frac{24 \pm \sqrt{288}}{1.6}\), resulting in two possible \(t\) values: approximately 4 seconds and 11.25 seconds.

These values indicate the times at which the rock is at half its maximum height during its ascent and descent.