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Chapter 28: Problem 41

Solve the given problems by integration.A marathon runner's speed (in \(\mathrm{km} / \mathrm{h}\) ) is \(v=\frac{12.0}{0.200 t+1} \cdot\) How far does the runner go in \(3.00 \mathrm{h}\) ?

Short Answer

Expert verified
The runner travels approximately 28.2 km in 3 hours.

Step by step solution

01

Set Up the Problem

To find the distance traveled by the marathon runner in 3 hours, we need to integrate the velocity function over the time interval from 0 to 3 hours. The velocity function is given as \( v(t) = \frac{12.0}{0.200t + 1} \).
02

Write the Integral Expression

The distance \( d \) that the runner travels is the integral of the velocity function over the interval from 0 to 3 hours. Therefore, we set up the integral as follows:\[d = \int_0^3 \frac{12.0}{0.200t + 1} \, dt.\]
03

Perform the Integration

To solve the integral \( \int \frac{12.0}{0.200t + 1} \, dt \), identify it as a form that allows a substitution. Let \( u = 0.200t + 1 \), then \( du = 0.200 \, dt \). Therefore, \( dt = \frac{du}{0.200} \).Substitute into the integral:\[d = \int \frac{12.0}{u} \cdot \frac{1}{0.200} \, du = \frac{12.0}{0.200} \int \frac{1}{u} \, du.\]
04

Evaluate the Integral

The integral \( \int \frac{1}{u} \, du \) is the natural logarithm function, \( \ln|u| + C \). Thus:\[d = \frac{12.0}{0.200} \cdot [\ln|u|] = 60 \cdot \ln|0.200t + 1|.\]
05

Evaluate the Integral from 0 to 3

Now, apply the limits of integration, evaluating between 0 and 3:\[d = 60 \cdot [\ln|0.200(3) + 1| - \ln|0.200(0) + 1|].\]Calculate the expressions:\[\ln(0.6 + 1) - \ln(1) = \ln(1.6) - \ln(1) = \ln(1.6).\]
06

Calculate the Distance

Use the value from the previous step to compute the distance:\[d = 60 \cdot \ln(1.6).\]Calculate \( \ln(1.6) \) using a calculator to get approximately 0.470. Then:\[d = 60 \times 0.470 \approx 28.2 \, \mathrm{km}.\]

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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 describes how an object's speed changes over time. In this exercise, the marathon runner's speed is expressed as a function of time: \[ v(t) = \frac{12.0}{0.200t + 1} \]. This tells us that the runner's velocity is not constant but instead depends on the time variable \( t \). Key points to understand here are:
  • The numerator, 12.0, represents the maximum possible speed when time \( t \) approaches zero.
  • The denominator, \( 0.200t + 1 \), indicates how the speed decreases over time, as it increases with \( t \).
By analyzing this function, you can see that as time increases, the effect of the 0.2 multiplied by \( t \) reduces the overall velocity. This means over time, the runner's pace slows slightly. Such functions allow us to model real-world scenarios like the diminishing speed of a runner as fatigue sets in.
Distance Calculation
To find out how far the runner travels in a given time, we need to calculate the distance. The distance an object travels during a time period when its speed varies is found by integrating the velocity function over time.Here's how it's done:
  • You write the integral of the velocity function over your time frame. In this case, it's from 0 to 3 hours: \[d = \int_0^3 \frac{12.0}{0.200t + 1} \, dt.\]
  • This integral sums up all the tiny pieces of distance the runner covers at infinitesimally small time intervals, adding them up for the full 3 hours.
Understanding this process is crucial, as you use integration to transform a function of velocity into a measurable distance. This provides a clear picture of just how dynamics change over time when dealing with variable speeds.
Definite Integral
A definite integral allows us to determine the net quantity鈥攕uch as distance鈥攐ver a specific interval. The definite integral \[d = \int_0^3 \frac{12.0}{0.200t + 1} \, dt\] is evaluated from 0 to 3. This integral essentially computes the total distance by accumulating the runner's speed over three hours.In practice:
  • We first substitute to simplify the integration. Let \( u = 0.200t + 1 \) and \( du = 0.200 \, dt \).
  • Reformulate the integral accordingly: \[d = \int \frac{12.0}{u} \cdot \frac{1}{0.200} \, du.\]
  • The integral of \( \frac{1}{u} \, du \) results in the logarithmic function \( \ln|u| \).
  • By calculating within the limits of 0 to 3, \[d = 60 \cdot [\ln|1.6|] = 28.2 \, \text{km}.\]
Using the definite integral gives precise control over measuring quantities where relationships vary, like non-constant speeds, ensuring an accurate final result.

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Most popular questions from this chapter

Solve the given problems by integration. A ball is rolling such that its velocity \(v\) (in \(\mathrm{cm} / \mathrm{s}\) ) as a function of time \(t\) (in s) is \(v=\frac{0.45}{0.25 t^{2}+1} .\) How far does it move in \(10.0 \mathrm{s} ?\)

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