Problem 69 - Calculus and its applications

Question

On Earth, all free-fall distance functions are of the form \(s(t)=4.905 t^{2},\) where \(t\) is in seconds and \(s(t)\) is in meters. The second derivative always has the same value. What does that value represent?

Step-by-Step Solution

Answer

The second derivative value, 9.81, represents Earth's gravitational acceleration.

Step 1: Understand the Free-Fall Distance Function

The given function for the free-fall distance is \( s(t) = 4.905 t^2 \). This equation represents the distance \( s(t) \) an object falls in free fall as a function of time \( t \) in seconds on Earth.

Step 2: Identify the First Derivative

To find the second derivative, we first need the first derivative of the function. The first derivative \( s'(t) \) represents the velocity of the falling object. For \( s(t) = 4.905 t^2 \), the first derivative is:\[ s'(t) = \frac{d}{dt}(4.905 t^2) = 9.81 t \]

Step 3: Calculate the Second Derivative

The second derivative \( s''(t) \) represents the acceleration. It is the derivative of the velocity function \( s'(t) = 9.81 t \). Calculating this gives:\[ s''(t) = \frac{d}{dt}(9.81 t) = 9.81 \]

Step 4: Interpret the Second Derivative

The second derivative \( s''(t) = 9.81 \) is a constant value. This represents the constant acceleration due to gravity on Earth's surface, measured in meters per second squared \( \text{m/s}^2 \).

Key Concepts

Free-Fall Distance

In physics, free-fall refers to the motion of an object falling under the influence of gravity alone, with no other forces acting on it. On Earth, the mathematical representation of the free-fall distance is given by the function \( s(t) = 4.905 t^2 \). This equation shows how far an object falls within a given time \( t \), measured in seconds.
To understand this function, consider that it predicts the distance \( s(t) \) in meters that an object will have fallen after \( t \) seconds. Assumptions made here include no air resistance and that the only force acting on the object is Earth's gravitational pull.
This simple quadratic function reveals that the distance fallen is directly proportional to the square of the time in free fall. In simple terms, the longer the fall, the further (and faster) the object will travel.

Derivatives

In calculus, derivatives are tools that help us understand the rates of change of functions. They are fundamental in finding how a function's output changes with respect to changes in input. For the free-fall distance function \( s(t) = 4.905 t^2 \), derivatives can tell us about velocity and acceleration.

• First Derivative: The first derivative of \( s(t) \), denoted as \( s'(t) \), represents the instantaneous rate of change of distance, which is velocity in the context of motion. For our function, \( s'(t) = 9.81t \). This means that the velocity of a free-falling object increases linearly with time.


• Second Derivative: The second derivative, \( s''(t) \), gives us the acceleration. In this case, \( s''(t) = 9.81 \), a constant value indicating uniform acceleration.

Understanding derivatives is crucial because they allow us to transcend static scenarios and analyze dynamic systems, such as an object falling under gravity.

Acceleration due to Gravity

Acceleration due to gravity, often denoted as \( g \), is the acceleration of an object as it is pulled towards the center of the Earth. In the context of free fall, this concept is central, as it explains the uniform acceleration observed.
For objects falling near Earth's surface, the standard gravity value is approximately \( 9.81 \text{ m/s}^2 \). In our free-fall distance equation \( s(t) = 4.905 t^2 \), when you take the second derivative, you find that this value appears as a constant, reinforcing its role as the gravitational acceleration.
Here, \( g \):

• Is a consistent value irrespective of the object’s mass, meaning all objects fall at the same rate when only gravity acts on them, assuming no air resistance.
• Plays a key role in other formulas relating to motion, such as kinetic energy and potential energy.

Grasping the concept of gravitational acceleration presents a foundational step in understanding the broader principles of physics and calculus combined.