Problem 5 - Applied Calculus

Question

A car starts moving at time \(t=0\) and goes faster and faster. Its velocity is shown in the following table. Estimate how far the car travels during the 12 seconds. $$\begin{array}{c|c|c|c|c|c} \hline t \text { (seconds) } & 0 & 3 & 6 & 9 & 12 \\ \hline \text { Velocity (ft/sec) } & 0 & 10 & 25 & 45 & 75 \\ \hline \end{array}$$

Step-by-Step Solution

Answer

The car travels approximately 352.5 feet in 12 seconds.

Step 1: Understanding the Problem

The problem involves estimating the distance traveled by a car over a period of 12 seconds by using its velocity at certain time points. We need to calculate the area under the velocity-time curve, which gives us the distance traveled.

Step 2: Choosing a Method of Estimation

Since we have velocity data at time intervals, we can estimate the distance using the Trapezoidal Rule. This method works well for estimating the area under a curve when we have discrete data points.

Step 3: Applying the Trapezoidal Rule

The Trapezoidal Rule formula for estimating the integral (distance) is given by \( \Delta x \times \frac{f(x_0) + 2f(x_1) + 2f(x_2) + \, ... \, + 2f(x_{n-1}) + f(x_n)}{2} \). Here, \( \Delta x \) is the interval of time between data points. Our data points are: \( (0, 0), (3, 10), (6, 25), (9, 45), (12, 75) \).

Step 4: Calculating \(\Delta x\) for Velocity Data

In this case, \( \Delta x = 3 \) seconds (since each interval spans 3 seconds).

Step 5: Substituting the Values into the Trapezoidal Formula

Now substitute the velocity values into the trapezoidal formula: \[ 3 \times \frac{0 + 2(10) + 2(25) + 2(45) + 75}{2} \] which simplifies to \[ 3 \times \frac{0 + 20 + 50 + 90 + 75}{2} \].

Step 6: Performing the Calculations

Add up the values inside the bracket: \( 0 + 20 + 50 + 90 + 75 = 235 \). Then divide by 2 to get \( \frac{235}{2} = 117.5 \). Finally, multiply by the interval \( 3 \): \[ 3 \times 117.5 = 352.5 \] feet.

Key Concepts

Velocity-Time Graph

A Velocity-Time Graph is a powerful tool to represent how the velocity of an object changes over time. In our exercise, we have velocity data of a car at different time intervals ranging from 0 to 12 seconds. These data points are: 0 seconds with 0 ft/sec, 3 seconds with 10 ft/sec, 6 seconds with 25 ft/sec, 9 seconds with 45 ft/sec, and 12 seconds with 75 ft/sec.

Graphs like these provide a visual insight into the motion of an object. Here's how they help:

• The slope of the graph indicates acceleration. A steeper slope means faster acceleration.
• The area under the curve corresponds to the distance traveled during that time interval.
• They allow us to use numerical methods like the Trapezoidal Rule to calculate distance.

By analyzing the velocity-time graph, we can determine the relationship between velocity and time, which is crucial to solving our distance estimation problem.

Calculating Distance

Calculating Distance from a velocity-time graph involves finding the area under the velocity curve. The area gives the distance traveled during a given time period. Imagine slicing the graph into narrow vertical strips representing short time intervals. The area of each strip adds up to the total distance.

In mathematical terms, if the curve were smooth, we'd use integration to find the area. However, with discrete data points, numerical methods come into play. For our exercise, we apply the Trapezoidal Rule, which treats the area under each segment as a trapezoid:

• Each trapezoid's area is computed and summed up for the intervals.
• The base of each trapezoid is the time interval, here 3 seconds.
• The height is the average velocity at the beginning and end of the interval.

This approach effectively estimates the distance traveled, giving us a practical solution for real-world problems where only sampled data is available.

Numerical Integration

Numerical Integration is the process of calculating the integral of a function using numerical approximation rather than symbolic algebra. This is especially useful for integrals of functions that are difficult or impossible to solve analytically.

In our problem, we use the Trapezoidal Rule, a straightforward method of numerical integration that approximates the area under the curve as a series of trapezoids. Here's why the trapezoidal rule is beneficial:

• It's relatively simple and easy to apply, using available data points.
• It can provide a good approximation when data is discrete or irregular.

We calculate the distance using the formula: \[\Delta x \times \frac{f(x_0) + 2f(x_1) + 2f(x_2) + \ldots + 2f(x_{n-1}) + f(x_n)}{2}\]For our car example, where the time intervals are 3 seconds apart, we substituted the velocities into this formula. The calculations showed a total distance of 352.5 feet traveled during the 12 seconds, demonstrating the power of numerical integration in practical applications.