# 5.4 Integration formulas and the net change theorem  (Page 7/8)

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For a given motor vehicle, the maximum achievable deceleration from braking is approximately 7 m/sec 2 on dry concrete. On wet asphalt, it is approximately 2.5 m/sec 2 . Given that 1 mph corresponds to 0.447 m/sec, find the total distance that a car travels in meters on dry concrete after the brakes are applied until it comes to a complete stop if the initial velocity is 67 mph (30 m/sec) or if the initial braking velocity is 56 mph (25 m/sec). Find the corresponding distances if the surface is slippery wet asphalt.

In dry conditions, with initial velocity ${v}_{0}=30$ m/s, $D=64.3$ and, if ${v}_{0}=25,D=44.64.$ In wet conditions, if ${v}_{0}=30,$ and $D=180$ and if ${v}_{0}=25,D=125.$

John is a 25-year old man who weighs 160 lb. He burns $500-50t$ calories/hr while riding his bike for t hours. If an oatmeal cookie has 55 cal and John eats 4 t cookies during the t th hour, how many net calories has he lost after 3 hours riding his bike?

Sandra is a 25-year old woman who weighs 120 lb. She burns $300-50t$ cal/hr while walking on her treadmill. Her caloric intake from drinking Gatorade is 100 t calories during the t th hour. What is her net decrease in calories after walking for 3 hours?

225 cal

A motor vehicle has a maximum efficiency of 33 mpg at a cruising speed of 40 mph. The efficiency drops at a rate of 0.1 mpg/mph between 40 mph and 50 mph, and at a rate of 0.4 mpg/mph between 50 mph and 80 mph. What is the efficiency in miles per gallon if the car is cruising at 50 mph? What is the efficiency in miles per gallon if the car is cruising at 80 mph? If gasoline costs $3.50/gal, what is the cost of fuel to drive 50 mi at 40 mph, at 50 mph, and at 80 mph? Although some engines are more efficient at given a horsepower than others, on average, fuel efficiency decreases with horsepower at a rate of $1\text{/}25$ mpg/horsepower. If a typical 50-horsepower engine has an average fuel efficiency of 32 mpg, what is the average fuel efficiency of an engine with the following horsepower: 150, 300, 450? $E\left(150\right)=28,E\left(300\right)=22,E\left(450\right)=16$ [T] The following table lists the 2013 schedule of federal income tax versus taxable income. Federal income tax versus taxable income Taxable Income Range The Tax Is … … Of the Amount Over$0–$8925 10%$0
$8925–$36,250 $892.50 + 15%$8925
$36,250–$87,850 $4,991.25 + 25%$36,250
$87,850–$183,250 $17,891.25 + 28%$87,850
$183,250–$398,350 $44,603.25 + 33%$183,250
$398,350–$400,000 $115,586.25 + 35%$398,350
>$400,000$116,163.75 + 39.6% $400,000 Suppose that Steve just received a$10,000 raise. How much of this raise is left after federal taxes if Steve’s salary before receiving the raise was $40,000? If it was$90,000? If it was \$385,000?

[T] The following table provides hypothetical data regarding the level of service for a certain highway.

Highway Speed Range (mph) Vehicles per Hour per Lane Density Range (vehicles/mi)
>60 <600 <10
60–57 600–1000 10–20
57–54 1000–1500 20–30
54–46 1500–1900 30–45
46–30 1900 2100 45–70
<30 Unstable 70–200
1. Plot vehicles per hour per lane on the x -axis and highway speed on the y -axis.
2. Compute the average decrease in speed (in miles per hour) per unit increase in congestion (vehicles per hour per lane) as the latter increases from 600 to 1000, from 1000 to 1500, and from 1500 to 2100. Does the decrease in miles per hour depend linearly on the increase in vehicles per hour per lane?
3. Plot minutes per mile (60 times the reciprocal of miles per hour) as a function of vehicles per hour per lane. Is this function linear?

a.

b. Between 600 and 1000 the average decrease in vehicles per hour per lane is −0.0075. Between 1000 and 1500 it is −0.006 per vehicles per hour per lane, and between 1500 and 2100 it is −0.04 vehicles per hour per lane. c.

The graph is nonlinear, with minutes per mile increasing dramatically as vehicles per hour per lane reach 2000.

For the next two exercises use the data in the following table, which displays bald eagle populations from 1963 to 2000 in the continental United States.

Population of breeding bald eagle pairs
Year Population of Breeding Pairs of Bald Eagles
1963 487
1974 791
1981 1188
1986 1875
1992 3749
1996 5094
2000 6471

[T] The graph below plots the quadratic $p\left(t\right)=6.48{t}^{2}-80.3\phantom{\rule{0.2em}{0ex}}1t+585.69$ against the data in preceding table, normalized so that $t=0$ corresponds to 1963. Estimate the average number of bald eagles per year present for the 37 years by computing the average value of p over $\left[0,37\right].$

[T] The graph below plots the cubic $p\left(t\right)=0.07{t}^{3}+2.42{t}^{2}-25.63t+521.23$ against the data in the preceding table, normalized so that $t=0$ corresponds to 1963. Estimate the average number of bald eagles per year present for the 37 years by computing the average value of p over $\left[0,37\right].$

$\frac{1}{37}{\int }_{0}^{37}p\left(t\right)dt=\frac{0.07{\left(37\right)}^{3}}{4}+\frac{2.42{\left(37\right)}^{2}}{3}-\frac{25.63\left(37\right)}{2}+521.23\approx 2037$

[T] Suppose you go on a road trip and record your speed at every half hour, as compiled in the following table. The best quadratic fit to the data is $q\left(t\right)=5{x}^{2}-11x+49\text{,}$ shown in the accompanying graph. Integrate q to estimate the total distance driven over the 3 hours.

Time (hr) Speed (mph)
0 (start) 50
1 40
2 50
3 60

As a car accelerates, it does not accelerate at a constant rate; rather, the acceleration is variable. For the following exercises, use the following table, which contains the acceleration measured at every second as a driver merges onto a freeway.

Time (sec) Acceleration (mph/sec)
1 11.2
2 10.6
3 8.1
4 5.4
5 0

[T] The accompanying graph plots the best quadratic fit, $a\left(t\right)=-0.70{t}^{2}+1.44t+10.44,$ to the data from the preceding table. Compute the average value of $a\left(t\right)$ to estimate the average acceleration between $t=0$ and $t=5.$

Average acceleration is $A=\frac{1}{5}{\int }_{0}^{5}a\left(t\right)dt=-\frac{0.7\left({5}^{2}\right)}{3}+\frac{1.44\left(5\right)}{2}+10.44\approx 8.2$ mph/s

[T] Using your acceleration equation from the previous exercise, find the corresponding velocity equation. Assuming the final velocity is 0 mph, find the velocity at time $t=0.$

[T] Using your velocity equation from the previous exercise, find the corresponding distance equation, assuming your initial distance is 0 mi. How far did you travel while you accelerated your car? ( Hint: You will need to convert time units.)

$d\left(t\right)={\int }_{0}^{1}|v\left(t\right)|dt={\int }_{0}^{t}\left(\frac{7}{30}{t}^{3}-0.72{t}^{2}-10.44t+41.033\right)dt=\frac{7}{120}{t}^{4}-0.24{t}^{3}-5.22{t}^{3}+41.033t.$ Then, $d\left(5\right)\approx 81.12$ mph $\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{sec}\approx 119$ feet.

[T] The number of hamburgers sold at a restaurant throughout the day is given in the following table, with the accompanying graph plotting the best cubic fit to the data, $b\left(t\right)=0.12{t}^{3}-2.13{t}^{3}+12.13t+3.91,$ with $t=0$ corresponding to 9 a.m. and $t=12$ corresponding to 9 p.m. Compute the average value of $b\left(t\right)$ to estimate the average number of hamburgers sold per hour.

Hours Past Midnight No. of Burgers Sold
9 3
12 28
15 20
18 30
21 45

[T] An athlete runs by a motion detector, which records her speed, as displayed in the following table. The best linear fit to this data, $\ell \left(t\right)=-0.068t+5.14\text{,}$ is shown in the accompanying graph. Use the average value of $\ell \left(t\right)$ between $t=0$ and $t=40$ to estimate the runner’s average speed.

Minutes Speed (m/sec)
0 5
10 4.8
20 3.6
30 3.0
40 2.5

$\frac{1}{40}{\int }_{0}^{40}\left(-0.068t+5.14\right)dt=-\frac{0.068\left(40\right)}{2}+5.14=3.78$

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