Physics, 16.07.2019 15:30 astorkid

Aman starts walking north at 2 ft/s from a point p. five minutes later a woman starts walking south at 6 ft/s from a point 500 ft due east of p. at what rate are the people moving apart 15 min after the woman starts walking? (round your answer to two decimal places.)

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7.98 ft/second You can create a right triangle to express the distance between the man and the woman. One leg of the triangle remains constant at 500 ft. (If you think about problem, the east/west separation between the man and the woman will remain constant. And that starts at 500 feet). So the other leg of the triangle needs to be calculated. The man is walking north for 20 minutes (5 minutes before the woman starts and 15 minutes after the woman started), so his distance north is 20 * 60 * 2 = 2400 ft. The woman has been walking south for 15 minutes, so her distance south is 15 * 60 * 6 = 5400 ft. So their exact north south separation is 2400 + 5400 = 7800 feet. And finally, the north south separation between the man and woman continues to increase at the rate of 8 ft per second. Now let's express the distance between the man and woman at time x where x is the number of seconds from where the woman has already walked for 15 minutes. That would be sqrt((7800 + 8x)^2 + 500^2) sqrt((7800 + 8x)^2 + 250000) Since we're looking for rate of change, that screams "first derivative". So let's calculate the first derivative. sqrt((7800 + 8x)^2 + 250000) = 1/2((8x + 7800)^2 + 250000)^(1/2) * d/dx((8x + 7800)^2 + 250000) = (d/dx((8x + 7800)^2 + d/dx(250000))/(2*sqrt((8x+7800)^2 + 250000)) = (2(8x + 7800) * d/dx(8x + 7800))/(2*sqrt((8x+7800)^2 + 250000)) = (8 * d/dx(x) + d/dx(7800))(8x + 7800)/sqrt((8x+7800)^2 + 250000) = 8(8x + 7800)/sqrt((8x+7800)^2 + 250000) = (64x + 62400)/sqrt((8x+7800)^2 + 250000) Now let's calculate the value of the first derivative at x=0, so (64x + 62400)/sqrt((8x+7800)^2 + 250000) =(64*x + 62400)/sqrt((8*0+7800)^2 + 250000) =62400/sqrt(7800^2 + 250000) =62400/sqrt(60840000 + 250000) =62400/sqrt(61090000) =62400/7816.009 =7.983613927 So the rate at which the man and woman are moving away from each other is 7.98 ft/second.

Answer from: Quest

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

$p(t)=0.3t^2+6t+80$

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

$p(0)=0+0+80=80$

according to question, we have said that if the population has twice its initial population, then it becomes

$p(t)=2\times 80=160$

so, time taken to reach this above mentioned population is given by

$160=0.3t^2+6t+80\\\\160-80=0.3t^2+6t\\\\80=0.3t^2+6t\\\\0.3t^2+6t-80=0\\\\\text{using quadratic formula ,we get }\\\\t_{1,\: 2}=\frac{-60\pm \sqrt{60^2-4\cdot \: 3\cdot \: 800}}{2\cdot \: 3}\\\\t=-29.14\ and\ t=9.149$

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

Answer from: Quest

Yup no bull frogs here man

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