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Chapter 5: Problem 5

When a car skids to a stop, the length \(L\), in feet, of the skid marks is related to the speed \(S\), in miles per hour, of the car by the power function \(L=\frac{1}{30 h} S^{2}\). Here the constant \(h\) is the friction coefficient, which depends on the road surface. \({ }^{11}\) For dry concrete pavement, the value of \(h\) is about \(0.85\). a. If a driver going 55 miles per hour on dry concrete jams on the brakes and skids to a stop, how long will the skid marks be? b. A policeman investigating an accident on dry concrete pavement finds skid marks 230 feet long. The speed limit in the area is 60 miles per hour. Is the driver in danger of getting a speeding ticket? c. This part of the problem applies to any road surface, so the value of \(h\) is not known. Suppose you are driving at 60 miles per hour but, because of approaching darkness, you wish to slow to a speed that will cut your emergency stopping distance in half. What should your new speed be? (Hint: You should use the homogeneity property of power functions here. By what factor should you change your speed to ensure that \(L\) changes by a factor of \(0.5\) ?)

Short Answer

Expert verified
a) 118.63 ft; b) Yes, about 76.6 mph; c) Reduce speed to 42.43 mph.

Step by step solution

01

Equation for Skid Marks

The equation to determine the skid mark length, given the speed and friction coefficient, is \(L = \frac{1}{30h}S^2\), and for dry concrete, \(h = 0.85\).
02

Solve for Part (a)

Substitute \(S = 55\) mph and \(h = 0.85\) into the equation: \(L = \frac{1}{30 \times 0.85} \times 55^2\). Calculate \(L = \frac{1}{25.5} \times 3025 = 118.63\) feet.
03

Solve for Part (b)

Use the equation \(230 = \frac{1}{30 \times 0.85} \times S^2\). Solve for \(S\): \(230 \times 25.5 = S^2\), giving \(S = \sqrt{5865} \approx 76.6\) mph. The driver was above the speed limit.
04

Solve for Part (c)

You need the skid distance to halve, so \(L' = 0.5L\). Since \(L \propto S^2\), the relationship is \(0.5 = \left(\frac{S'}{S}\right)^2\). Solving this gives \(\frac{S'}{S} = \sqrt{0.5} \approx 0.7071\), meaning the new speed \(S' \approx 0.7071 \times 60 = 42.43\) mph.

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Key Concepts

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

Power Functions and Their Properties
A power function is a mathematical expression of the form \( f(x) = k x^n \), where \( k \) is a constant, \( x \) is the variable, and \( n \) is a real number that represents the power to which \( x \) is raised. In the equation for skid marks, \( L = \frac{1}{30h}S^2 \), the skid mark length \( L \) depends on speed \( S \) raised to the power of 2, making \( n = 2 \) and \( k = \frac{1}{30h} \). This type of function is related to quadratic functions due to the squared term, which tells us that the length of skid marks increases quadratically as speed increases.
Understanding power functions is crucial because they model many real-world scenarios where one quantity depends on another raised to a power. In this exercise, the homogeneity property helps us determine how changes in speed affect the stopping distance: halving the stopping distance involves reducing the speed by a factor equivalent to the square root of 0.5, indicating the consistent relationship between \( S \) and its square.
The characteristic feature of power functions like \( L=S^2 \) is their predictable behavior, which allows us to solve problems involving proportional scaling by altering speed or adjusting other parameters like friction coefficients.
Friction Coefficient and Its Role in Motion
The friction coefficient \( h \) is a critical factor in determining how a vehicle comes to a stop. It represents the amount of friction between the road surface and the tires of a vehicle, thereby influencing how quickly a vehicle can come to a halt once the brakes are applied. In the provided exercise, the friction coefficient for dry concrete is approximately \( 0.85 \).
Friction is the resistance that one surface or object encounters when moving over another. A higher friction coefficient indicates more grip and shorter stopping distances, while a lower coefficient implies less traction and potentially longer skid marks.
  • Concrete surfaces generally provide a good amount of friction, which is why they allow for shorter stopping distances compared to surfaces like ice or wet pavement.
  • Friction is affected by multiple factors, including surface conditions (dry or wet), the type of tires, and speed.
  • Understanding the friction coefficient helps drivers better anticipate how their vehicle will perform under different conditions, making it essential for safety in driving.
By incorporating the friction coefficient in calculations for skid marks, we can accurately predict stopping distances and create safer driving conditions by adjusting speed accordingly.
Effective Problem Solving Strategies in Algebra
Problem solving in algebra involves understanding the relationships between variables and applying mathematical principles to find unknown values. In this exercise, we use a step-by-step approach to calculate the lengths of skid marks and investigate speed, employing essential algebraic skills.
First, identifying the correct formula is crucial: \( L=\frac{1}{30h}S^2 \). This guides the entire problem-solving process by highlighting the relationship between speed \( S \), the skid marks \( L \), and friction coefficient \( h \).
To solve Part (a), substituting given values for \( S \) and \( h \) into the equation allows us to find the skid marks' length when travelling at 55 mph.
  • For Part (b), reverse calculation by using the skid marks' length to determine speed establishes whether the driver exceeded the speed limit.
  • Part (c) requires manipulating the power function's homogenous property to find a new speed where stopping distance is halved, demonstrating practical application of scaled relationships.
Good problem-solving in algebra involves clear understanding and execution of mathematical operations, anticipating outcomes based on variable changes, and knowing how to reverse calculations when necessary.

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Most popular questions from this chapter

In the study of population genetics, an important measure of inbreeding is the proportion of homozygous genotypes-that is, instances in which the two alleles carried at a particular site on an individual's chromosomes are both the same. For populations in which bloodrelated individuals mate, there is a higher than expected frequency of homozygous individuals. Examples of such populations include endangered or rare species, selectively bred breeds, and isolated populations. In general, the frequency of homozygous children from matings of blood-related parents is greater than that for children from unrelated parents. \(^{67}\) Measured over a large number of generations, the proportion of heterozygous genotypes-that is, nonhomozygous genotypes-changes by a constant factor \(\lambda_{1}\) from generation to generation. The factor \(\lambda_{1}\) is a number between 0 and 1 . If \(\lambda_{1}=0.75\), for example, then the proportion of heterozygous individuals in the population decreases by \(25 \%\) in each generation. In this case, after 10 generations the proportion of heterozygous individuals in the population decreases by \(94.37 \%\), since \(0.75^{10}=\) \(0.0563\), or \(5.63 \%\). In other words, \(94.37 \%\) of the population is homozygous. For specific types of matings, the proportion of heterozygous genotypes can be related to that of previous generations and is found from an equation. For matings between siblings, \(\lambda_{1}\) can be determined as the largest value of \(\lambda\) for which $$ \lambda^{2}=\frac{1}{2} \lambda+\frac{1}{4}. $$ This equation comes from carefully accounting for the genotypes for the present generation (the \(\lambda^{2}\) term) in terms of those of the previous two generations (represented by \(\lambda\) for the parents' generation and by the constant term for the grandparents' generation). a. Find both solutions to the quadratic equation above and identify which is \(\lambda_{1}\). (Use a horizontal span from \(-1\) to 1 in this exercise and the following two exercises.) b. After 5 generations what proportion of the population will be homozygous? c. After 20 generations what proportion of the population will be homozygous?

The speed at which certain animals run is a power function of their stride length, and the power is \(k=1.7\). (See Figure 5.39.) If one animal has a stride length three times as long as another, how much faster does it run?

Show that the following data can be modeled by a quadratic function, and find a formula for a quadratic model. $$ \begin{array}{|l|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 \\ \hline Q(x) & 5 & 6 & 13 & 26 & 45 \\ \hline \end{array} $$

There are 3600 commercial bee hives in a region threatened by African bees. Today African bees have taken over 50 hives. Experience in other areas shows that, in the absence of limiting factors, the African bees will increase the number of hives they take over by \(30 \%\) each year. Make a logistic model that shows the number of hives taken over by African bees after \(t\) years, and determine how long it will be before 1800 hives are affected.

When a force is applied to muscle tissue, the muscle contracts. Hill's law is an equation that relates speed of muscle contraction with force applied to the muscle. \({ }^{73}\) The equation is given by the rational function $$ S=\frac{\left(F_{\ell}-F\right) b}{F+a}, $$ where \(S\) is the speed at which the muscle contracts, \(F_{\ell}\) is the maximum force of the muscle at the given length \(\ell, F\) is the force against which the muscle is contracting, and \(a\) and \(b\) are constants that depend on the muscle tissue itself. This is valid for non-negative \(F\) no larger than \(F_{\ell}\). For a fast-twitch vertebrate muscle - for example, the leg muscle of a sprinter-we may take \(F_{\ell}=300 \mathrm{kPa}, a=81\), and \(b=6.75\). These are the values we use in this exercise. a. Write the equation for Hill's law using the numbers above for fast-twitch vertebrate muscles. b. Graph \(S\) versus \(F\) for forces up to \(300 \mathrm{kPa}\). c. Describe how the muscle's contraction speed changes as the force applied increases. d. When is \(S\) equal to zero? What does this mean in terms of the muscle? e. Does the rational function for \(S\) have a horizontal asymptote? What meaning, if any, does the asymptote have in terms of the muscle? f. Does the rational function for \(S\) have a vertical asymptote? What meaning, if any, does the asymptote have in terms of the muscle?
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