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A feather dropped on the moon On the moon, a feather will fall to the ground at the same rate as a heavy stone. Suppose a feather is dropped from a height of \(40 \mathrm{m}\) above the surface of the moon. Its height (in meters) above the ground after \(t\) seconds is \(s=40-0.8 t^{2} .\) Determine the velocity and acceleration of the feather the moment it strikes the surface of the moon.

Short Answer

Expert verified
Answer: The velocity of the feather when it strikes the surface of the moon is approximately -11.31 m/s, and the acceleration is -1.6 m/s².

Step by step solution

01

Determine the time it takes for the feather to hit the ground

To determine the time when the feather hits the ground, we need to set the height function to zero and solve for time t: \(s(t) = 40 - 0.8t^2 = 0\) We'll isolate \(t^2\) and then find \(t\): \(t^2 = \frac{40}{0.8} = 50\) This gives us: \(t = \sqrt{50} \approx 7.07\) So the time it takes for the feather to hit the ground is about 7.07 seconds.
02

Calculate the first and second derivative of the height function

First, let's find the first derivative of the height function, which gives us the velocity function: \(v(t) = \frac{d}{dt}s(t) = \frac{d}{dt}(40 - 0.8t^2) = -1.6t\) Now, let's find the second derivative of the height function, which gives us the acceleration function: \(a(t) = \frac{d}{dt}v(t) = \frac{d}{dt}(-1.6t) = -1.6\)
03

Find the velocity and acceleration when the feather hits the ground

We can now plug in the time it takes for the feather to reach the ground into the velocity and acceleration functions: \(v(7.07) = -1.6(7.07) \approx -11.31\) \(a(7.07) = -1.6\) So, the velocity of the feather when it strikes the surface of the moon is approximately -11.31 m/s, and the acceleration is -1.6 m/s².

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Velocity and Acceleration
When studying motion in physics, velocity and acceleration are fundamental concepts that must be understood. Velocity describes how fast an object is moving and in what direction, while acceleration represents the rate of change of velocity.

Let's apply these concepts to an example from the moon, where gravity is weaker than on Earth. Remember the presented problem: a feather falling on the moon with the position function being defined as \( s = 40 - 0.8 t^2 \). The initial velocity of the feather is zero since it's dropped, not thrown.

What is Velocity?

Using calculus, the velocity, denoted as \( v(t) \), can be found by taking the first derivative of the position function with respect to time, \( t \). So, if we have the position function \( s(t) \), the velocity function \( v(t) = \frac{d}{dt}s(t) \). In our feather scenario, this yields \( v(t) = -1.6t \), which indicates that the feather's velocity increases linearly over time due to the moon's gravity.

What is Acceleration?

Acceleration, denoted as \( a(t) \), is the second derivative of the position function. It tells us how velocity changes with time. In our example, the acceleration of the feather is a constant \( -1.6 \, \text{m/s}^2 \), as revealed by the second derivative of the position function. This value corresponds to the moon's gravitational acceleration affecting the feather.

Understanding both velocity and acceleration requires comprehension of their relationship through derivatives, and knowing how to interpret these derivatives gives insight into an object's motion.
Derivative of Position Function
In calculus, derivatives illustrate how a function changes at any given point. For motion, the derivative of the position function with respect to time provides us with the object's velocity.

Consider our lunar feather once again, with its position function \( s(t) = 40 - 0.8t^2 \). To find its state of motion at any time \( t \), we differentiate this function. The first derivative, representing the feather's velocity, is \( v(t) = -1.6t \). This result tells us that the velocity is a function of time, specifically indicating a linear relationship under constant acceleration.

Interpreting the Derivative

The negative sign in the velocity function reveals the direction of motion—downwards, towards the moon's surface. Also, we understand that the velocity increases in magnitude over time due to constant acceleration (gravity).

Meanwhile, the second derivative of the position function gives us acceleration. In free fall, under gravity only, this value should be constant. This perfectly constant acceleration is what we see on the moon with \( a(t) = -1.6 \, \text{m/s}^2 \), reflecting the gravitational acceleration's consistency.

By studying derivatives, students can predict and analyze motion, such as a feather's trajectory on the moon, providing valuable insights into the fundamentals of kinematics in physics.
Free Fall Acceleration
Free fall acceleration is a type of uniformly accelerated motion under the sole influence of gravity. On Earth, this value is approximately \( 9.81 \, \text{m/s}^2 \), but the moon's weaker gravity results in a different acceleration for falling objects.

As shown in the textbook problem, the free fall acceleration on the moon can be worked out from the second derivative of the position function of a falling object when only the force of gravity is acting on it. Through our analysis, we've discovered that the gravitational acceleration affecting the feather is \( -1.6 \, \text{m/s}^2 \).

Importance of Direction

The negative sign in the free fall acceleration value indicates the direction of the acceleration is downwards (toward the surface of the moon), which is why the velocity of the feather increases in magnitude but remains negative.

Understanding free fall on different celestial bodies, like the moon, not only aids in comprehending universal principles of gravity but also emphasizes how varying environments affect motion. This example also dispels the misconception that lighter objects fall slower under gravity, as without an atmosphere, a feather and a stone fall at the same rate, debunking the common myth through the lens of physics and calculus.

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