/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 85 In 1987 Carl Lewis set a new wor... [FREE SOLUTION] | 91Ó°ÊÓ

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In 1987 Carl Lewis set a new world's record of \(9.93\) seconds for the 100 -meter run. The distance that he ran in the first \(x\) seconds was $$ 11.274\left[x-1.06\left(1-e^{-x / 1.06}\right)\right] \text { meters } $$ for \(0 \leq x \leq 9.93\). Enter this function as \(y_{1}\), and define \(y_{2}\) as its derivative (using NDERIV), so that \(y_{2}\) gives his velocity after \(x\) seconds. Graph them on the window \([0,9.93]\) by \([0,100]\). Trace along the velocity curve to verify that Lewis's maximum speed was about \(11.27\) meters per second. Find how quickly he reached a speed of 10 meters per second, which is \(95 \%\) of his maximum speed.

Short Answer

Expert verified
Carl Lewis reached a speed of 10 m/s very quickly after starting his run.

Step by step solution

01

Define the Function

The distance function given is \( y_1(x) = 11.274 \left[x - 1.06\left(1-e^{-x / 1.06}\right)\right] \). This function represents the distance Carl Lewis ran in the first \( x \) seconds.
02

Calculate the Derivative

To find Lewis's velocity, calculate the derivative of \( y_1 \). Use a graphing calculator or computer software to represent the derivative as \( y_2(x) = \frac{dy_1}{dx} \). This gives his velocity as a function of time.
03

Graph the Functions

Graph the functions \( y_1 \) and \( y_2 \) in a window with \( x \) ranging from \( 0 \) to \( 9.93 \) and the \( y \) axis from \( 0 \) to \( 100 \). This visual will help us understand how his speed changes over the course of the race.
04

Find the Maximum Speed

By tracing the graph of \( y_2 \), observe the highest value it reaches. Confirm that the maximum speed is approximately \( 11.27 \) meters per second, as stated in the problem.
05

Determine Time to Reach 10 m/s

Calculate \( 95\% \) of the maximum speed, which is \( 10.7065 \) meters per second. Trace the graph of \( y_2 \) to find the smallest \( x \) value where \( y_2(x) = 10 \), indicating Carl Lewis's time to reach 10 meters per second.

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Key Concepts

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

Velocity Function
A velocity function in calculus represents how fast an object is moving at any point in time. For Carl Lewis's 100-meter dash record, the velocity function is the derivative of the given distance function. In our exercise, the initial distance function is denoted as:
  • \( y_1(x) = 11.274 \left[x - 1.06\left(1-e^{-x / 1.06}\right)\right] \)
This function models the distance he ran over time. By differentiating it with respect to time, we obtain the velocity function \( y_2(x) \). This derived function tells us how fast Lewis was running at every second during the race.

This mathematical process reflects a fundamental concept in calculus where the derivative of a function gives us the rate of change of the quantity, in this case, speed. Using a graphing calculator or computer software is crucial here to precisely compute and visualize this derivative, helping us understand the dynamics of Lewis's sprint.
Graph Analysis
Graphing is a powerful analytical tool in calculus that helps visualize functions and their derivatives. In this exercise, by graphing both the distance function \( y_1(x) \) and the velocity function \( y_2(x) \), we gain insights into the motion of Carl Lewis during the 100-meter dash.
  • The window for these graphs is set to an x-axis range from 0 to 9.93 seconds, corresponding to the race's duration.
  • The y-axis is set from 0 to 100 meters or meters per second, reflecting the possible distances and speeds.
Through this visualization, we can trace the velocity curve to examine how Lewis's speed changes through each moment of the race. When you observe the graph, it's easier to identify key points, such as when the velocity peaks or how quickly certain speeds are reached, enhancing your understanding of the problem.
Maximum Speed Calculation
The maximum speed calculation involves identifying the highest value of the velocity function \( y_2(x) \). To find this maximum is to determine Carl Lewis's peak performance during his sprint.By using the graph tracing technique, you observe the velocity graph to find its pinnacle point, confirming the problem statement that said his maximum speed was approximately 11.27 meters per second. Using a graphing tool makes this straightforward by allowing you to follow the curve visually until you reach the top.

Furthermore, determining how quickly Lewis reached a speed of 10 meters per second, which is 95% of his maximum, helps understand his acceleration at the start of the race. By finding the x-value where \( y_2(x) = 10 \), you can ascertain the time he took to reach this crucial speed. This aspect of the exercise showcases the utility of calculus in analyzing real-world motion by using derivatives to describe dynamic changes in velocity.

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Most popular questions from this chapter

Use your graphing calculator to graph each function on a window that includes all relative extreme points and inflection points, and give the coordinates of these points (rounded to two decimal places). [Hint: Use NDERIV once or twice with ZERO.] (Answers may vary depending on the graphing window chosen.) $$ f(x)=e^{-2 x^{2}} $$

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Use your graphing calculator to graph each function on a window that includes all relative extreme points and inflection points, and give the coordinates of these points (rounded to two decimal places). [Hint: Use NDERIV once or twice with ZERO.] (Answers may vary depending on the graphing window chosen.) $$ f(x)=\ln \left(1+x^{2}\right) $$

A covered cup of coffee at 200 degrees, if left in a 70 -degree room, will cool to \(T(t)=70+130 e^{-2.5 t}\) degrees in thours. Find the rate of change of the temperature: a. at time \(t=0\). b. after 1 hour.

If consumer demand for a commodity is given by the function below (where \(p\) is the selling price in dollars), find the price that maximizes consumer expenditure. $$ D(p)=5000 e^{-0.01 p} $$

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