Problem 249 - Calculus Volume 1

Question

ITI The accompanying graph plots the best quadratic fit, \(a(t)=-0.70 t^{2}+1.44 t+10.44,\) to the data from the preceding table. Compute the average value of \(a(t)\) to estimate the average acceleration between \(t=0\) and \(t=5\)

Step-by-Step Solution

Answer

The average acceleration is approximately 8.21.

Step 1: Setup Integration Bounds

We need to calculate the average value of a function over an interval. For our quadratic function \( a(t) = -0.70 t^2 + 1.44 t + 10.44 \), the interval is from \( t = 0 \) to \( t = 5 \). The formula for an average value of a function \( f(t) \) over the interval \([a, b]\) is given by: \[ \text{Average value} = \frac{1}{b-a} \int_a^b f(t)\, dt \] In our case, \( a = 0 \) and \( b = 5 \).

Step 2: Integrate the Function

Next, we need to integrate the given function \( a(t) = -0.70 t^2 + 1.44 t + 10.44 \). The integral of this function from \( t = 0 \) to \( t = 5 \) can be calculated as follows:\[ \int_0^5 (-0.70 t^2 + 1.44 t + 10.44) \, dt = \left[ -0.70 \cdot \frac{t^3}{3} + 1.44 \cdot \frac{t^2}{2} + 10.44 \cdot t \right]_0^5 \]

Step 3: Evaluate the Definite Integral

Now, evaluate the definite integral from \( t = 0 \) to \( t = 5 \):\[ \left[ -0.70 \cdot \frac{5^3}{3} + 1.44 \cdot \frac{5^2}{2} + 10.44 \cdot 5 \right] - \left[ -0.70 \cdot \frac{0^3}{3} + 1.44 \cdot \frac{0^2}{2} + 10.44 \cdot 0 \right] \]

Step 4: Simplify and Calculate the Integral Result

Compute the expression further:\[ = \left[ -0.70 \cdot \frac{125}{3} + 1.44 \cdot 12.5 + 52.2 \right] \]Simplifying each term:\[ = \left[ -29.1667 + 18.00 + 52.2 \right] \]\[ = 41.0333 \]

Step 5: Calculate the Average Value

We now use the formula for the average value of a function over an interval:\[ \text{Average value} = \frac{1}{5-0} \cdot 41.0333 \]\[ = \frac{1}{5} \cdot 41.0333 \]\[ = 8.2066 \]

Key Concepts

Quadratic Functions

A quadratic function is a type of polynomial function characterized by its highest degree term being squared. Often expressed in the form, \( f(x) = ax^2 + bx + c \), it can be visualized as a parabola on a graph. Quadratic functions are significant in mathematics due to their simplicity yet versatile structure, allowing them to model a wide array of real-world phenomena.
In the context of the exercise, the quadratic function, \( a(t) = -0.70 t^2 + 1.44 t + 10.44 \), represents a curve fit to a data set. This specific form indicates:

• The coefficient \( -0.70 \) suggests the parabola opens downwards, due to its negative value.
• The linear coefficient \( 1.44 \) influences the slope or tilt of the parabola.
• The constant \( 10.44 \) shifts the entire parabola vertically on the graph.

Understanding how these coefficients affect the parabolic shape is essential for interpreting the physical meaning within data, such as estimating average acceleration over a given time period.

Definite Integrals

Definite integrals are a fundamental concept in calculus, providing a way to calculate the area under a curve, which is crucial for understanding changes and accumulations in various contexts. For a function \( f(x) \), the definite integral over an interval \([a, b]\) is represented as \( \int_a^b f(x) \, dx \).
This tool helps us summarize the behavior of functions over specific intervals, such as calculating distances or average values, like acceleration.
In the given exercise, we use definite integrals to compute the average value of the quadratic function \( a(t) = -0.70 t^2 + 1.44 t + 10.44 \) from \( t = 0 \) to \( t = 5 \).

• First, interpreting each term of the integral is crucial for summarizing the solution.
• Using the Fundamental Theorem of Calculus, we connected antiderivatives and definite integrals to produce the average acceleration value.

Through this process, integrating a polynomial like our quadratic function involves determining antiderivatives and plugging upper and lower bounds before simplifying.

Average Acceleration

Average acceleration in physics represents the rate of change of velocity over time and is a key concept in motion analysis. It is calculated by determining the change in velocity over a particular time interval. In a mathematical sense, if velocity can be represented as a function of time, the average acceleration is obtained by finding the average rate of change of this function.
In our exercise, we calculate the average acceleration from \( t = 0 \) to \( t = 5 \), using the average value of the quadratic function \( a(t) \). This entails computing:

• The definite integral of the acceleration function, \( a(t) \), over the defined time interval.
• Dividing this integral by the interval length \( (5 - 0) \).

This computation reflects how the velocity changes across the interval, giving insight into whether the object involved is speeding up, slowing down, or maintaining a constant rate of velocity change. Understanding this concept in both algebraic and graphical forms aids in grasping real-world motions and their implications.