Question

A model for the startup of a runner in a short race results in the velocity function \(v(t)=a\left(1-e^{-t / c}\right),\) where \(a\) and \(c\) are positive constants and \(v\) has units of \(\mathrm{m} / \mathrm{s}\). (Source: A Theory of Competitive Running, Joe Keller, Physics Today 26 (September 1973)) a. Graph the velocity function for \(a=12\) and \(c=2 .\) What is the runner's maximum velocity? b. Using the velocity in part (a) and assuming \(s(0)=0\), find the position function \(s(t),\) for \(t \geq 0\) c. Graph the position function and estimate the time required to run \(100 \mathrm{m} .\)

Short answer

Based on the model for a runner's velocity as a function of time given by \(v(t)=a\left(1-e^{-t / c}\right)\), with \(a=12\) and \(c=2\), the maximum velocity is 12 m/s. The corresponding position function for the runner is \(s(t) = 12t - 24e^{-t / 2} + 24\). By using the graph of the position function or numerical methods, we can estimate the time required for the runner to run 100 meters.

Step-by-step solution

Graph the velocity function

Given \(a=12\) and \(c=2\), our velocity function becomes \(v(t)=12\left(1-e^{-t / 2}\right)\). To graph this function, generate a set of \((t, v(t))\) points for various values of \(t\).

Find the maximum velocity

To find the maximum velocity, observe that as \(t\) approaches infinity, the term \(e^{-t / 2}\) approaches zero. Thus, the maximum velocity is \(v_{max}=a\left(1-0\right)=a\), which for our case is \(v_{max}=12\ \mathrm{m} / \mathrm{s}\).

Solve for the position function

We know that the position function, \(s(t)\), is the integral of the velocity function, \(v(t)\). So, we need to compute the integral of \(v(t)=12\left(1-e^{-t / 2}\right)\). To do this, we will integrate term by term: \(\int 12 \ dt = 12t + C_1\) \(\int 12e^{-t / 2} \ dt = -24e^{-t / 2} + C_2\) Taking both constants together in a single constant term, \(C\), we have: \(s(t) = \int v(t) \ dt = 12t - 24e^{-t / 2} + C\) As \(s(0) = 0\), we can find the value of \(C\): \(s(0) = 0 = 12 \cdot 0 - 24 \cdot 1 + C \Rightarrow C = 24\) Thus, our position function is: \(s(t) = 12t - 24e^{-t / 2} + 24\)

Graph the position function

To graph the position function, generate a set of \((t, s(t))\) points for various values of \(t\), using the equation \(s(t) = 12t - 24e^{-t / 2} + 24\).

Estimate the time required to run 100 meters

To estimate the time required to run 100 meters, we need to find the value of \(t\) when \(s(t) = 100\). That is, solve the equation \(12t - 24e^{-t / 2} + 24 = 100\) for \(t\). This can be difficult to solve analytically, but we can use numerical methods, such as approximating the value by looking at the graph of \(s(t)\), using a calculator or software.

Key concepts

Velocity Function

A velocity function describes how an object's velocity changes over time. It is a crucial concept in calculus, particularly when analyzing motion. In the exercise, we have the velocity function given by \(v(t) = a(1 - e^{-t/c})\). This formula models how the velocity of a runner changes as time progresses.
- The constant \(a\) represents the potential maximum velocity the runner can achieve.
- The constant \(c\) controls how quickly the runner approaches this maximum velocity.
As \( t \) increases, the term \( e^{-t/c} \) approaches zero, and hence the velocity approaches \( a \). Therefore, the function shows an exponential increase in velocity that slows as time continues, effectively plateauing at the maximum velocity \(v_{max} = a\).
Graphing this function helps visualize how quickly the runner reaches near-maximum velocity.

Position Function

The position function, \(s(t)\), gives the location of a moving object at any given time. To find the position function, we integrate the velocity function.
In the exercise, the integration of the velocity function \(v(t) = 12(1 - e^{-t/2})\) results in the position function \(s(t) = 12t - 24e^{-t/2} + C\), where \(C\) is a constant.
- Since \(s(0) = 0\), we solve for \(C\) by substituting into the equation: \(s(0) = 0 = 12 \cdot 0 - 24 \, \cdot 1 + C \). This gives us \(C = 24\).
Thus, the complete position function is: \(s(t) = 12t - 24e^{-t/2} + 24\).
The position function indicates the cumulative distance traveled over time, factoring in the initial velocity buildup.

Integration

Integration is the calculus process we use to find a function given its derivative. When moving from a velocity function to a position function, we integrate the velocity function. This means that we are looking for the area under the curve of the velocity graph as a function of time. In simple terms, integration accumulates how much the position changes over time.
For the velocity function \(v(t) = 12(1 - e^{-t/2})\), it's broken into two integrals:

• \(\int 12 \, dt = 12t + C_1\)
• \(\int -12e^{-t/2} \, dt = -24e^{-t/2} + C_2\)

These results combine to form the position function. Understanding integration is key to interpreting changes in physical systems, like the growth of a runner's position over time.

Graphing Functions

Graphing functions allows us to visually analyze and interpret mathematical models. For this exercise, we graph both the velocity and position functions. Graphing the velocity function \(v(t)\) shows how quickly the velocity approaches its maximum value.
Similarly, graphing the position function \(s(t) = 12t - 24e^{-t/2} + 24\) illustrates the total distance covered by the runner over time, effectively showing a smoother curve as it reflects accumulated displacement.
By evaluating these graphs, students can:

• Determine rates of change visually.
• Estimate key values, such as the time to reach specific distances.
• Gain deeper insights into the behavioral aspects of the motion.

Graphing is not just an exercise in plotting points but a tool for exploratory learning and discovery in calculus.