91Ó°ÊÓ

Careful measurements have been made of Olympic sprinters in the 100 -meter dash. A quite realistic model is that the sprinter's velocity is given by $$v_{x}=a\left(1-e^{-b t}\right)$$ where \(t\) is in \(\mathrm{s}, v_{x}\) is in \(\mathrm{m} / \mathrm{s},\) and the constants \(a\) and \(b\) are characteristic of the sprinter. Sprinter Carl Lewis's run at the 1987 World Championships is modeled with \(a=11.81 \mathrm{m} / \mathrm{s}\) and \(b=0.6887 s^{-1}\) a. What was Lewis's acceleration at \(t=0 \mathrm{s}, 2.00 \mathrm{s},\) and \(4.00 \mathrm{s} ?\) b. Find an expression for the distance traveled at time \(t\) c. Your expression from part \(\mathbf{b}\) is a transcendental equation, meaning that you can't solve it for \(t .\) However, it's not hard to use rial and error to find the time needed to travel a specific distance. To the nearest \(0.01 \mathrm{s}\), find the time Lewis needed to sprint \(100.0 \mathrm{m} .\) His official time was \(0.01 \mathrm{s}\) more than your answer, showing that this model is very good, but not perfect.

Step by step solution

01

Calculating Acceleration at Given Times

The acceleration of Carl Lewis at a particular time t can be obtained by differentiating his velocity function with respect to time. The velocity function is given by \(v_x = a(1 - e^{-bt})\) where \(a = 11.81 m/s\) and \(b = 0.6887 s^{-1}\) Now, differentiating this velocity function with respect to time, we get acceleration function as \(a_x = \frac{dv_x}{dt} = abe^{-bt}\) Now we can substitute the values \(t = 0s\), \(t = 2.00s\), \(t = 4.00s\) into the acceleration function to get the respective accelerations.

02

Deriving the Distance Expression

The distance traveled by Carl Lewis in time t can be obtained by integrating the velocity function with respect to time. Therefore, \(x = \int v_x dt = \int (a - ae^{-bt}) dt = at - a \int e^{-bt} dt\). Applying integral of exponential function, we get \(x = at + \frac{a}{b}e^{-bt}\)

03

Calculating Time Taken for a Specific Distance

Now, using the distance function obtained in the previous step, we can determine the time taken by Carl Lewis to sprint 100.0m. The distance is given by \(x = 100.0m\). So, we need to solve for \(t\) in the equation \(100.0 = 11.81t + \frac{11.81}{0.6887}e^{-0.6887t}\). Since this is a transcendental equation and cannot be solved algebraically for t, we can use numerical methods, for example, a trial and error method or a numerical solver on a calculator or computer software to solve it. The official time was 0.01s more than the calculated time, highlighting that while the model is quite accurate, it is not perfect.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Velocity function

In kinematics, the velocity function describes how the velocity of an object changes with time. For sprinter Carl Lewis's run, his velocity function is given by \(v_{x} = a(1-e^{-bt})\), where \(a\) and \(b\) are constants that depend on the sprinter's characteristics. This equation models his velocity, which starts at zero and increases as time progresses. The term \(e^{-bt}\) represents an exponential decay factor slowing the initial velocity increment. As time \(t\) increases, \(e^{-bt}\) approaches zero, and the velocity stabilizes near the maximum value \(a\). Understanding such velocity functions is crucial for modeling real-world motion accurately.

Acceleration calculation

Acceleration is the rate of change of velocity over time. For Carl Lewis, his acceleration can be obtained by differentiating the velocity function with respect to time. The differentiated function, \(a_x = abe^{-bt}\), shows that the acceleration decreases exponentially over time. Initially, at \(t = 0\), the acceleration is at its highest value \(ab\), equal to \(8.13 \, m/s^2\) given \(a = 11.81 \, m/s\) and \(b = 0.6887 \, s^{-1}\). The exponential term \(e^{-bt}\) causes the acceleration to decrease as \(t\) increases, meaning Lewis's speed increases at a decreasing rate.

Transcendental equation

Transcendental equations, like the one we encountered for Carl Lewis's sprint, have no straightforward algebraic solution. These equations typically include exponential or trigonometric functions that prevent simple solving methods. The distance traveled, modeled by \(100.0 = 11.81t + \frac{11.81}{0.6887}e^{-0.6887t}\), is a classic transcendental equation due to the presence of \(e^{-0.6887t}\). Such equations illustrate the complex nature of real-world phenomena, where the relationship between multiple variables isn't merely linear or polynomial.

Numerical methods

When facing transcendental equations, numerical methods offer practical solutions. These methods include approaches like trial and error, interval bisection, or using numerical solvers available on calculators and computers. For Carl Lewis's sprint, determining the time to cover 100.0 meters involves solving the equation \(100.0 = 11.81t + \frac{11.81}{0.6887}e^{-0.6887t}\) using such methods, as algebraic solutions aren't feasible. Numerical techniques ensure we can approximate solutions to any desired level of precision, allowing athletes and scientists to effectively analyze performance data or physical models. These methods are invaluable tools in both academic and applied physics.