Problem 41 - Calculus and its applications

Question

In \(t\) seconds, an object dropped from a certain height will fall \(s(t)\) feet, where $$ s(t)=16 t^{2} $$ a) Find \(s(5)-s(3)\). b) What is the average rate of change of distance with respect to time during the period from 3 to 5 sec? This is known as average velocity, or speed.

Step-by-Step Solution

Answer

a) 256 feet. b) 128 feet per second.

Step 1: Understanding the Function

The function given is \( s(t) = 16t^2 \), which represents the distance in feet that an object falls in \( t \) seconds.

Step 2: Substitute Values for s(5) and s(3)

To find \( s(5) - s(3) \), first compute \( s(5) \) and \( s(3) \) using the equation \( s(t) = 16t^2 \). Substitute \( t = 5 \) and \( t = 3 \) into the equation. Calculate \( s(5) = 16 \times 5^2 = 16 \times 25 = 400 \) and \( s(3) = 16 \times 3^2 = 16 \times 9 = 144 \).

Step 3: Calculate s(5) - s(3)

Subtract the two values obtained from substituting into the function: \( s(5) - s(3) = 400 - 144 = 256 \).

Step 4: Calculate Average Velocity

The average rate of change, or average velocity, is calculated as the change in distance divided by the change in time. We have \( \frac{s(5) - s(3)}{5 - 3} = \frac{256}{2} = 128 \) feet per second.

Key Concepts

Average Velocity

Average velocity is a crucial concept in calculus, particularly when dealing with functions that describe the motion of objects. Simply put, average velocity tells us how fast something is moving over a certain period of time.

To calculate average velocity, you need two things:

• The total distance the object has travelled.
• The amount of time it took to travel that distance.

In mathematical notation, average velocity is expressed as the change in distance divided by the change in time:
\[\text{Average Velocity} = \frac{\Delta s}{\Delta t} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}\]Here, \(\Delta s\) represents the change in distance and \(\Delta t\) represents the change in time.

In the problem provided, we computed the average velocity over a period from 3 to 5 seconds by subtracting the distances at these respective times and then dividing by the total time (2 seconds). This allows us to determine that the average velocity is 128 feet per second.

Rate of Change

In calculus, the rate of change is an important concept as it tells us how one quantity changes in relation to another. It is especially useful in understanding real-world behaviors and phenomena.

For any function, the average rate of change is determined by the change in the value of the function divided by the change in the independent variable. It’s akin to finding the slope of the line that connects two points on a graph.

• If you have a function \( f(x) \), the average rate of change between two points \( x_1 \) and \( x_2 \) is:\[ \frac{f(x_2) - f(x_1)}{x_2 - x_1} \]

The exercise demonstrates by calculating the average rate of change (average velocity) between \( t = 3 \) and \( t = 5 \) seconds. This rate of change equates to the average speed the object travels within those seconds.

Quadratic Functions

Quadratic functions are polynomials of degree two and have the general form:
\[ f(x) = ax^2 + bx + c \]They graph as parabolas, either opening up or downwards depending on the sign of the parameter \(a\).

In this exercise, the function \( s(t) = 16t^2 \) is a quadratic function. It specifically describes how far an object will fall under gravity over time when starting from rest and neglecting air resistance. The function lacks both linear and constant terms, making it a simplified quadratic known as a "pure" quadratic function.

• The term \(16t^2\) tells us that distance increases with the square of the time elapsed.
• This means that as time increases, the rate at which distance increases also grows. This is what makes quadratic functions especially interesting and complex when compared to linear functions.

Understanding quadratic functions is necessary when analyzing motion, as they reveal complex relationships between time and distance in various physical phenomena.