Problem 80 - Precalculus with Limits

Question

FREE-FALLING OBJECT In Exercises 79 and 80, use the position function \(s(t) = -16t^2 + 256\) which gives the height (in feet) of a free-falling object. The velocity at time \(t = a\) seconds is given by \(\lim_{t \to a} [s(a) - s(t)]/(a-t)\). Find the velocity when \(t=2\) seconds.

Step-by-Step Solution

Answer

The velocity of the free-falling object when \(t=2\) seconds is 64 ft/sec.

Step 1: Understand the position function

The function \(s(t) = -16t^2 + 256\) is given as the position function of the object, where \(t\) is the time in seconds, and \(s(t)\) is the height in feet. This function describes how the height of the object changes over time.

Step 2: Identify the formula for the velocity

The formula for finding the velocity of the object at some time \(t = a\) is given by \(\lim_{t \to a} [s(a) - s(t)]/(a-t)\). This formula calculates the rate at which the height is changing at a particular moment, which is the velocity of the object.

Step 3: Substituting \(t=2\) in position function

To get the value of the function \(s(t)\) at time \(t=2\) seconds, substitute \(t=2\) into the position function: \(s(2) = -16(2)^2 + 256 = -64 + 256 = 192\) feet.

Step 4: Apply the limit

Now, to find the velocity, we substitute \(a=2\) and \(s(2)\) into the limit formula for the velocity: \(\lim_{t \to 2} [(192) - s(t)]/(2-t)\). However, this limit cannot be directly computed as it leads to a 0/0 undefined situation. Hence, apply L'Hopital's rule, which states that when the limit leads to an undefined form, you can find the derivative of the upper and lower parts separately and find the limit again: \(\lim_{t \to 2} [(192) - s(t)]/(2-t) = \lim_{t \to 2} {-32t}/{-1} = \lim_{t \to 2} {32t} = 32*2 = 64\) ft/sec.

Key Concepts

Position Function

In free-fall motion, the position function is essential for understanding how the height of an object changes over time as it falls freely under the influence of gravity. Here, the position function is given by:

• \( s(t) = -16t^2 + 256 \)

This formula allows us to calculate the height \( s(t) \) of the object at any given time \( t \). The term \(-16t^2\) accounts for the acceleration due to gravity, which is approximately \(-32\text{ ft/sec}^2\), but it halves here because traditionally, this function assumes an initial velocity of zero from twice the gravitational acceleration.
The \(+256\) term represents the initial height from which the object is dropped. By substituting different values of \( t \) into this function, you can see the trajectory of the object over time. This helps visualize how fast and how far the object falls.

Velocity Calculation

Velocity is a measure of how quickly something is moving in a specific direction. In this context, for free-fall, it's about how fast and in what direction the height of the object is decreasing. To calculate velocity at a specific moment, we apply:

• \( \lim_{t \to a} \frac{s(a) - s(t)}{a-t} \)

This formula gives us the instantaneous rate of change in the object's height, which is essentially its velocity. \( s(a) \) and \( s(t) \) denote the positions at times \( t = a \) and a moment before, \( t \), reflecting the change in the position over time. This approach comes from the definition of the derivative in calculus, representing how the position changes per unit time.

L'Hôpital's Rule

L'Hôpital's Rule is a tool we use in calculus to resolve indeterminate forms, such as \( 0/0 \) or \( \infty/\infty \). When trying to find the velocity using limits, we end up with an indeterminate form:

• \( \lim_{t \to 2} \frac{192 - s(t)}{2-t} \)

At first glance, substituting \( t = 2 \) leads to \( 0/0 \). This is where L'Hôpital's Rule becomes handy. It states that we can find the derivatives of the numerator and the denominator separately and then re-evaluate the limit.
Using this rule here enables us to solve the limit by determining:

• First Derivative of Numerator: \(-32t\)
• First Derivative of Denominator: \(-1\)

The limit calculation simplifies to \( 32 \times 2 = 64 \text{ ft/sec} \), which is the velocity at \( t = 2 \).

Instantaneous Rate of Change

The concept of instantaneous rate of change is pivotal in calculus and free-fall motion, offering a snapshot of how a quantity changes at a specific point in time. When we talk about the instantaneous rate of change of the position of a free-falling object, we're effectively discussing its velocity.
In mathematical terms, we employ a limit to calculate this rate of change using the formula:

• \( \lim_{t \to a} \frac{s(a) - s(t)}{a-t} \)

This definition mirrors the derivative of the position function \( s(t) \) with respect to time. By computing this for \( t = 2 \) seconds, we capture how the object's speed varies precisely at that moment.
This principle is not only limited to physics but extends across diverse fields, wherever predicting sudden changes in trend or behavior is crucial.