Problem 7 - Active Calculus

Question

The velocity of a ball that has been tossed vertically in the air is given by \(v(t)=16-32 t\), where \(v\) is measured in feet per second, and \(t\) is measured in seconds. The ball is in the air from \(t=0\) until \(t=2\). a. When is the ball's velocity greatest? b. Determine the value of \(v^{\prime}(1)\). Justify your thinking. c. What are the units on the value of \(v^{\prime}(1) ?\) What does this value and the corresponding units tell you about the behavior of the ball at time \(t=1 ?\) d. What is the physical meaning of the function \(v^{\prime}(t) ?\)

Step-by-Step Solution

Answer

a. Greatest at \(t = 0\) with \(v(0) = 16 \text{ft/s}\). b. \(v^{\prime}(1) = -32 \text{ft/s}^2.\) c. The units are ft/s^2, indicating deceleration at 32 ft/s^2 at t=1. d. \(v^{\prime}(t)\) is the ball's acceleration.

Step 1: Determine the time interval

The ball is in the air from \(t = 0\) to \(t = 2\) seconds. This will help to find the range for velocity calculation.

Step 2: Find when velocity is greatest

The velocity function is v(t) = 16 - 32t. It is a linear function. The velocity is greatest at the smallest value of t. Hence, v(0) is the greatest velocity. Calculate v(0): \[v(0) = 16 - 32 \times 0 = 16 \text{ feet per second} \]

Step 3: Calculate the derivative of v(t)

To determine \(v^{\prime}(t)\), find the derivative of \(v(t) = 16 - 32t\). \[v^{\prime}(t) = -32 \]

Step 4: Determine the value of \(v^{\prime}(1)\)

Substitute \(t = 1\) into \(v^{\prime}(t)\): \[v^{\prime}(1) = -32 \text{ feet per second squared} \]

Step 5: Explain the units of \(v^{\prime}(1)\)

The units of \(v^{\prime}(1)\) are feet per second squared (ft/s^2). This tells us the rate of change of velocity with respect to time.

Step 6: Interpret \(v^{\prime}(1)\)

At \(t = 1\), the velocity of the ball is decreasing by 32 feet per second every second. This means the ball is decelerating at 32 feet per second squared.

Step 7: Determine the physical meaning of \(v^{\prime}(t)\)

\(v^{\prime}(t)\) represents the acceleration of the ball, which is the rate of change of velocity with respect to time.

Key Concepts

Understanding the Derivative in Calculus

The derivative is a fundamental concept in calculus. It represents how a function changes as its input changes. For any function, the derivative tells us the rate at which the function is increasing or decreasing. In this exercise, we looked at the velocity of a ball given by the function \(v(t) = 16 - 32t\).
To find how this velocity changes over time, we took the derivative of the velocity function. Mathematically, we calculated \(v^{\text{'} }(t) = -32\). This gives us the rate of change of velocity with respect to time.
This process of finding the derivative is crucial in physics and many other fields to understand how various quantities change.

What is Acceleration?

Acceleration is a key concept in physics that refers to how quickly an object's velocity changes. It is the derivative of the velocity function. From the exercise, we determined that \(v^{\text{'}}(t) = -32\). This tells us that the acceleration of the ball is constant at \( -32 \text{ ft/s}^2 \).
Negative acceleration indicates that the ball is slowing down; hence, it's often called deceleration.
Understanding acceleration helps us predict how the velocity of the ball changes over time.

The Concept of Rate of Change

The rate of change measures how a quantity changes with respect to another quantity. In calculus, this concept is embodied by the derivative. In the context of the exercise, the rate of change of the velocity function \(v(t) = 16 - 32t\) is represented by the derivative \(v^{\text{'}}(t) = -32\).
This constant rate of change reveals how the velocity of the ball changes every second. Specifically, as each second passes, the velocity decreases by 32 feet per second.
Rates of change are pivotal in understanding how different quantities evolve over time.

Linear Functions and Their Characteristics

A linear function is one in the form \(f(t) = mt + b\), where \(m\) and \(b\) are constants. The velocity function \(v(t) = 16 - 32t\) is a linear function with a slope of -32 and y-intercept of 16.
Linear functions have a constant rate of change, which is reflected in their straight-line graphs. In our case, the slope -32 tells us how steeply the velocity changes.
The y-intercept 16 indicates the initial velocity of the ball when \(t = 0\). Linear functions are straightforward and predictable, making them easy to analyze in physics problems.

Importance of Units in Physics

Units are fundamental in physics because they provide a standard for measurement. In this exercise, velocity is given in feet per second (ft/s), and acceleration in feet per second squared (ft/s²).
Understanding the units helps interpret the physical meaning of calculated values. For instance, \(v^{\text{'}}(1) = -32 \text{ feet per second squared}\) tells us the rate at which the ball's velocity is decreasing each second.
Consistency in units ensures that our calculations are meaningful and comparable. Never underestimate the need for clear units in any physics-related problem.