Problem 17 - Applied Calculus

Question

Your velocity is given by \(v(t)=t^{2}+1\) in \(\mathrm{m} / \mathrm{sec}\), with \(t\) in seconds. Estimate the distance, \(s\), traveled between \(t=0\) and \(t=5 .\) Explain how you arrived at your estimate.

Step-by-Step Solution

Answer

The distance traveled is approximately 46.67 meters.

Step 1: Understand the problem

We want to estimate the distance traveled by calculating the integral of the velocity function, which will provide the displacement over a given time interval. Specifically, we need to find the area under the curve of the velocity function from time \(t = 0\) to \(t = 5\).

Step 2: Set up the integral

The distance traveled, \(s\), can be found using the definite integral of the velocity function \(v(t) = t^2 + 1\) over the interval \([0, 5]\). Therefore, we need to evaluate the integral: \[s = \int_{0}^{5} (t^2 + 1) \, dt\]

Step 3: Integrate the velocity function

To integrate \(v(t) = t^2 + 1\), we find the antiderivative. For \(t^2\), the antiderivative is \(\frac{t^3}{3}\), and for \(1\), it is \(t\). Thus, the antiderivative of the velocity function is \(\frac{t^3}{3} + t\).

Step 4: Evaluate the definite integral

Substitute the upper and lower limits into the antiderivative: \[s = \left[ \frac{t^3}{3} + t \right]_{0}^{5} = \left( \frac{5^3}{3} + 5 \right) - \left( \frac{0^3}{3} + 0 \right)\].

Step 5: Calculate the result

Calculate \(\frac{5^3}{3} = \frac{125}{3}\) and add \(5\) to get \(\frac{125}{3} + 5 = \frac{125}{3} + \frac{15}{3} = \frac{140}{3}\). So, the distance traveled is approximately \(\frac{140}{3} \approx 46.67\) meters.

Key Concepts

Velocity Function

When we're asked to determine how far something has traveled over time, we often start with its velocity function. In this exercise, the velocity function is defined as \(v(t) = t^2 + 1\), where \(v(t)\) is in meters per second and \(t\) is time in seconds. A velocity function provides a way to describe how fast an object is moving and in what direction at any given moment.

Let's break it down:

• Velocity can be thought of as the speed with which an object moves in a particular direction.
• The given function \(v(t) = t^2 + 1\) represents the velocity as a function of time \(t\).
• As time increases, the velocity changes according to the velocity function.

By knowing the velocity function, we can proceed to find the distance traveled over a specific period using calculus methods, such as integration.

Antiderivative

To determine the approximate distance traveled, we must first identify the antiderivative of the velocity function. The concept of an antiderivative is crucial in calculus as it helps in finding the integral of a function.

An antiderivative is basically the reverse of a derivative, and here's how you can understand it:

• To find the antiderivative of \(v(t) = t^2 + 1\), we consider each term separately.
• The antiderivative of \(t^2\) is \(\frac{t^3}{3}\), while the antiderivative for the constant \(1\) is simply \(t\).
• Thus, the antiderivative of \(v(t) = t^2 + 1\) is \(\frac{t^3}{3} + t\).

This antiderivative describes the overall position of the object over time, given by integrating the velocity function. Once we have the antiderivative, we can compute a definite integral to find the distance traveled.

Distance Estimation

Distance estimation using integration involves evaluating the definite integral of the velocity function over a specified time interval. This concept helps in estimating how far an object moves within that period.

Here's a concise break-down of the process:

• Set up the definite integral of the velocity function \(v(t)\) over the interval \([0, 5]\), which translates into the mathematical expression \(s = \int_{0}^{5} (t^2 + 1) \, dt\).
• Use the antiderivative \(\frac{t^3}{3} + t\), previously calculated, and substitute the upper and lower bounds (\(t = 5\) and \(t = 0\)).
• Evaluate the antiderivative at each bound and subtract: \[s = \left[ \frac{t^3}{3} + t \right]_{0}^{5} = \left( \frac{5^3}{3} + 5 \right) - \left( \frac{0^3}{3} + 0 \right)\]
• Carry out the calculations, \(\frac{5^3}{3} = \frac{125}{3}\) and add the extra \(5\) to get the final estimated distance as approximately \(46.67\) meters.

The entire integration process gives us a robust method to approximate the distance traveled over the specified timing, based on the velocity function.