Question

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 1

a) The north boundary of the sidewalk is a horizontal line y = -90.

The south boundary of the sidewalk is a horizontal line y = -110.

b) The water will first strike the sidewalk after 2400 seconds.

c) The water from the sprinkler will fall completely north of the sidewalk after 3600 seconds.

d) The total amount of time water from the sprinkler falls on the sidewalk is 1200 seconds.

e) The area of grass watered after one hour is 400 pi square feet.

(a) The tractor will have x and y coordinates.

Initially, the tractor is 100 feet south of the sidewalk, so its y-coordinate is y = -100.

The tractor is on the hose, which is perpendicular to the sidewalk, which means its x-coordinate is constant.

Let's set the x-coordinate to be x = 0.

The north boundary of the sidewalk is a horizontal line y = -90.

The south boundary of the sidewalk is a horizontal line y = -110.

(b) The water from the sprinkler will first strike the sidewalk when the edge of the sprinkler's circle is on the sidewalk. This happens when the sprinkler is 20 feet north of the sidewalk, which means y = -80.

The time it takes for the sprinkler to move 20 feet north is:

time = distance / rate

time = 20 ft / (1/2 inch/sec) * (12 inches/ft) * (1/60 sec/min)

time = 2400 seconds

So, the water will first strike the sidewalk after 2400 seconds.

(c) The water from the sprinkler will fall completely north of the sidewalk when the center of the sprinkler's circle is north of the sidewalk. This happens when the sprinkler is 30 feet north of the sidewalk, which means y = -70.

The time it takes for the sprinkler to move 30 feet north is:

time = distance / rate

time = 30 ft / (1/2 inch/sec) * (12 inches/ft) * (1/60 sec/min)

time = 3600 seconds

So, the water from the sprinkler will fall completely north of the sidewalk after 3600 seconds.

(d) The total amount of time water from the sprinkler falls on the sidewalk is the difference between the time it takes for the water to first strike the sidewalk and the time it takes for the water to fall completely north of the sidewalk. This is:

total time = 3600 seconds - 2400 seconds

total time = 1200 seconds

So, the total amount of time water from the sprinkler falls on the sidewalk is 1200 seconds.

(e) The area of grass watered after one hour is the area of a circle with radius 20 feet that has been watered for one hour.

The area of a circle is:

area = pi * radius^2

area = pi * (20 ft)^2

area = 400 pi square feet

One hour is 60 minutes, so the area of grass watered after one hour is:

area = 60 minutes * 400 pi square feet / 60 minutes

area = 400 pi square feet

So, the area of grass watered after one hour is 400 pi square feet.

Answer 2

a) see below

b) 32 minutes after turn-on

c) 52 minutes after turn-on

d) 20 minutes

e) 6856.6 ft²

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.

__

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

__

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).

__

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

__

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.