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A crawling tractor sprinkler is located as pictured below, 100 feet South of a sidewalk. Once the water is turned on, the sprinkler waters a circular disc of radius 20 feet and moves North along the hose at the rate of 1/2 inch/second. The hose is perpendicular to the 10 ft. wide sidewalk. Assume there is grass on both sides of the sidewalk.
a) Impose a coordinate system. Describe the initial coordinates of the sprinkler and find equations of the lines forming and find equations of the lines forming the north and south boundaries of the sidewalk.

b) When will the water first strike the sidewalk?

c) When will the water from the sprinkler fall completely north of the sidewalk?

d) Find the total amount of time water from the sprinkler falls on the sidewalk.

e) Find the area of grass watered after one hour.

Answer :

sqdancefan

Answer:

  a) see below

  b) 32 minutes after turn-on

  c) 52 minutes after turn-on

  d) 20 minutes

  e) 6856.6 ft²

Step-by-step explanation:

a) We have elected to put the origin at the point where the hose crosses the south edge of the sidewalk. Units are feet. Then the sprinkler starts at (0, -100). After 1 hour, 3600 seconds, the sprinkler is 1800 inches, or 150 ft north of where it started, so stops at (0, 50).

The lines forming the sidewalk boundaries are y=0 and y=10.

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b) Water will first strike the sidewalk when the sprinkler is 20 feet south of it, or 80 feet north of where it started. The sprinkler travels that distance in ...

  (80 ft)(12 in/ft)/(1/2 in/s)(1 min/(60 s)) = 32 min . . . time to start sprinkling sidewalk

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c) The sprinkler has to travel to a point 130 ft north of its starting position for the water to fall north of the sidewalk. That distance is traveled in ...

  (130 ft)(2/5 min/ft) = 52 min . . . time until end of sprinkling sidewalk

Note that we have combined the scale factors in the expression of part b into one scale factor of (2/5 min/ft).

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d) The difference of times in parts b and c is the time water falls on the sidewalk: 20 minutes.

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e) In one hour, the sprinkler travels a distance of ...

  (60 min)(5/2 ft/min) = 150 ft

Of that distance, 10 feet is sidewalk. So, the sprinkler covers an area of grass that is a 140 ft by 40 ft rectangle and a circle of 20 ft radius. The total area of that is ...

  A = LW + πr² = (140 ft)(40 ft) +π(20 ft)² = (14+π)(400) ft² ≈ 6856.6 ft²

The area of grass watered in 1 hour is about 6856.6 ft².

${teks-lihat-gambar} sqdancefan

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