1. Many people know that the weight of an object varies on different planets, but did you know that the weight of an object on earth also varies according to the elevation of the object? In particular, the weight of an object follows this equation: w=Cr -2 where C is a constant and r is the distance that the object is from the center of the earth.
a. Solve the equation w=Cr -2 for r.
b. Suppose that an object is 150 pounds when it is at sea level. Find the value of C that makes the equation true. (Sea level is 3,963 miles from the center of the earth.)
c. Use the value of C you found in the previous question to determine how much the object would weigh in
i. Death Valley (282 feet below sea level)
ii. The top of Mt McKinley (20,430 feet above sea level)
2. After an accident, how do the police determine the speed at which the car had be travelling prior to the accident?
The formula s = 2√(5L) can be used to approximate the speed s, in miles per hour of a car that has left skid marks of length L, in feet.
a. How far will a car skid at 55 mph?
b. How far will a car skid at 100 mph?
3. The equation D= 1.2√h gives the distance, D, in miles that a person can see to the horizon from a height, h, in feet.
a. How far can you see from an airplane window at 32,000 feet?
b. If you can see 13 miles to the horizon from the top of an observation
tower, what is the height of the tower?
do ur own work
1a. W = Cr-2,
Cr = w + 2,
r = (W+2) / C.
b. C = (w+2) / r,
C = (150+2) / 3963 =0.03835.
2a. S = 2sqrt(5L) = 55mi/h,
sqrt(5L) = 27.5.
Square both sides:
5L = 756.25,
L = 151.3 Ft.
b. Same procedure as a.
3a. Solve the given Eq for D.
b. D = 1.2sqrt(h) = 13mi,
sqrt(h) = 10.83,
h = 117.4 Ft.
1. To solve the equation w = Cr - 2 for r, we need to isolate r on one side of the equation.
a. Start with the equation: w = Cr - 2
Add 2 to both sides: w + 2 = Cr
Divide both sides by C: (w + 2)/C = r
So the equation w = Cr - 2 can be solved for r as r = (w + 2)/C.
b. To find the value of C that makes the equation true, we substitute the given values into the equation and solve for C.
Given: w = 150 pounds, r = 3,963 miles
Substituting these values in the equation: 150 = C * 3,963 - 2
To find C, we need to isolate it on one side of the equation:
150 + 2 = C * 3,963
152 = C * 3,963
C = 152 / 3,963
So the value of C is 0.03837 (approximately).
c. i. To determine the weight of the object in Death Valley (282 feet below sea level), we need to calculate the distance from the center of the earth for this location.
Given: r = 3,963 miles - (282 feet / 5280 feet/mile)
r = 3,963 miles - 0.05341 miles
r = 3,962.946 miles
Using the value of C (0.03837) from the previous question:
w = C * r - 2
w = 0.03837 * 3,962.946 - 2
w = 152.276 - 2
w ≈ 150.276 pounds
So the object would weigh approximately 150.276 pounds in Death Valley.
c. ii. To determine the weight of the object at the top of Mt McKinley (20,430 feet above sea level), we need to calculate the distance from the center of the earth for this location.
Given: r = 3,963 miles + (20,430 feet / 5280 feet/mile)
r = 3,963 miles + 3.86742 miles
r = 3,966.87742 miles
Using the value of C (0.03837) from the previous question:
w = C * r - 2
w = 0.03837 * 3,966.87742 - 2
w = 152.420 - 2
w ≈ 150.420 pounds
So the object would weigh approximately 150.420 pounds at the top of Mt McKinley.
2. After an accident, the police can approximate the speed at which the car was traveling prior to the accident using the skid marks left by the car. The formula s = 2√(5L) can be used to estimate the speed s, in miles per hour, of the car based on the length of the skid marks L, in feet.
a. To determine how far a car will skid at 55 mph, we need to substitute the given speed into the formula and solve for L.
Given: s = 55 mph
Substituting this value in the formula: 55 = 2√(5L)
To solve for L, we need to isolate it on one side of the equation:
55/2 = √(5L)
27.5 = √(5L)
Squaring both sides of the equation to eliminate the square root:
(27.5)^2 = (√(5L))^2
756.25 = 5L
L = 756.25/5
So the car will skid approximately 151.25 feet at 55 mph.
b. To determine how far a car will skid at 100 mph, we follow the same process as in part a.
Given: s = 100 mph
Substituting this value in the formula: 100 = 2√(5L)
To solve for L, we need to isolate it on one side of the equation:
100/2 = √(5L)
50 = √(5L)
Squaring both sides of the equation:
(50)^2 = (√(5L))^2
2500 = 5L
L = 2500/5
So the car will skid approximately 500 feet at 100 mph.
3. The equation D = 1.2√h gives the distance, D, in miles, that a person can see to the horizon from a height, h, in feet.
a. To determine how far you can see from an airplane window at 32,000 feet, we need to substitute the given height into the formula and solve for D.
Given: h = 32,000 feet
Substituting this value in the formula: D = 1.2√(32,000)
Evaluating the square root: D = 1.2 * √(32,000)
D = 1.2 * 178.8854
D ≈ 214.6625 miles
So you can see approximately 214.6625 miles from an airplane window at 32,000 feet.
b. To determine the height of an observation tower if you can see 13 miles to the horizon, we need to substitute the given distance into the formula and solve for h.
Given: D = 13 miles
Substituting this value in the formula: 13 = 1.2√h
To solve for h, we need to isolate it on one side of the equation:
13/1.2 = √h
10.8333 = √h
Squaring both sides of the equation:
(10.8333)^2 = (√h)^2
117.4656 = h
So the height of the observation tower is approximately 117.4656 feet.