Problem 116 - College Algebra with Modeling and Visualization

Question

The table lists the velocity and distance raveled by a falling object for various elapsed times. $$ \begin{array}{|rcccccc|} \hline \text { Time (sec) } & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \text { Velocity (fl/sec) } & 0 & 32 & 64 & 96 & 128 & 160 \\ \hline \text { Distance (ft) } & 0 & 16 & 64 & 144 & 256 & 400 \\ \hline \end{array} $$ (a) Make a scatterplot of the ordered pairs determined by (time, velocity) and (time, distance) in the same vicwing rectangle \([-1,6,1]\) by \([-10,450,50]\) (b) Find a function \(v\) that models the velocity. (c) The distance is modeled by \(d(x)=a x^{2} .\) Find \(a\) (d) Find the time when the distance is 200 feet. Find the velocity at this time.

Step-by-Step Solution

Answer

(a) Make a scatterplot from the ordered pairs. (b) Velocity model: v(t) = 32t. (c) Distance model: d(x) = 16x^2. (d) Time at 200 ft: approx. 3.54s, Velocity: approx. 113.28 ft/s.

Step 1: Plotting Points for Scatterplot

To create a scatterplot for both time-velocity and time-distance, list the ordered pairs. For time-velocity: (0,0), (1,32), (2,64), (3,96), (4,128), (5,160). For time-distance: (0,0), (1,16), (2,64), (3,144), (4,256), (5,400). Use the viewing rectangle [-1,6,1] by [-10,450,50] to ensure all points fit within the plot. Plot both sets of points on the same graph, with time on the x-axis, velocity on one y-axis, and distance on another y-axis.

Step 2: Finding Model for Velocity

Notice that the velocity changes at a constant rate. Calculate the rate of change or slope between each consecutive time point. For instance, between time 0 and 1: (32-0)/(1-0)=32. The slope remains constant at 32 ft/s for each interval. This indicates a linear relationship described by the function v(t) = 32t.

Step 3: Modeling Distance Function

The problem provides that the distance is modeled by \( d(x) = ax^2 \). We use one data pair to find \( a \). For example, at \( x = 2, d(2) = 64 \). Substitute into the model: \( 64 = a(2)^2 \), solving for \( a \) gives \( 64 = 4a \) thus \( a = 16 \). The distance function is \( d(x) = 16x^2 \).

Step 4: Solving for Time at Given Distance

We need to find the time \( x \) when distance \( d(x) = 200 \) feet using \( d(x) = 16x^2 \). Set the equation: \( 16x^2 = 200 \). Solving for \( x \), divide both sides by 16, \( x^2 = 12.5 \). Taking the square root: \( x = \sqrt{12.5} \approx 3.54 \) seconds.

Step 5: Calculating Velocity at a Given Time

With the time \( x = 3.54 \), use the velocity function \( v(t) = 32t \). Calculate the velocity: \( v(3.54) = 32 \times 3.54 = 113.28 \) ft/s. Conclude that the velocity at the time when the distance reached 200 feet is approximately 113.28 ft/s.

Key Concepts

Velocity

Velocity refers to the speed of an object in a specific direction. In our exercise, we analyzed how the velocity of a falling object changes over time. Here, the velocity increases at a steady rate, implying a constant acceleration. This behavior is observed in free-falling objects where gravity causes a uniform acceleration. In this case, velocity was measured in feet per second and it shows a linear increase. The linear function for velocity can be represented as \( v(t) = 32t \), where \( v \) is velocity and \( t \) is time. This means for every second that passes, the velocity of the object increases by 32 feet per second. Understanding this concept helps students quantify the motion of objects and predict future velocities given a specific amount of time.

Scatterplot

A scatterplot is a graphical representation that uses dots to plot the values of two different variables. In our original exercise, we created a scatterplot to visualize how time correlates with both velocity and distance for a falling object. Scatterplots are helpful to identify relationships and trends in data.

• Time was plotted on the x-axis
• Velocity and distance were plotted on separate y-axes

By examining the scatterplot, one can easily see the linear relationship between time and velocity, and a nonlinear relationship between time and distance. These plots make complex data easier to understand by providing a visual representation. They are a key tool in data analysis, helping students interpret and present their findings effectively.

Quadratic function

In our exercise, the distance of the falling object is modeled by a quadratic function. A quadratic function is one where the highest exponent of the variable is two, expressed in general form as \( ax^2 + bx + c \). In this case, the function is simplified to \( d(x) = 16x^2 \), where \( d(x) \) represents distance based on time, and \( a = 16 \) is a constant derived from our data. Quadratic functions describe parabolic shapes on a graph. This characteristic is crucial for modeling scenarios like the acceleration of objects under gravity, where the relationship between variables isn't direct or linear. By solving the quadratic equation, students can determine values such as the time it takes to reach a particular distance, which roots the abstract concept in real-world applications.

Linear function

Linear functions are foundational in algebra and describe relationships where the rate of change is constant. This type of function is expressed in the form of \( y = mx + b \), where \( m \) is the slope, and \( b \) is the y-intercept. In this exercise, the velocity of the object over time is modeled by the linear function \( v(t) = 32t \). Here, the slope \( m \) is 32, indicating the object accelerates uniformly at 32 feet per second. The y-intercept is zero, which means when time is zero, velocity is also zero. Linear functions are straightforward and easy to graph, showing a direct proportionality. They're widely used due to their simple nature and serve as a stepping stone to more complex functions, emphasizing the relationship between two variables clearly in a visual and analytical manner.