SOLUTION: A flag is raised while an onlooker watches from a distance of 12 feet away from the base of the flag pole (see the figure below). The flag rises vertically at a rate of 7 inches pe
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Question 9592: A flag is raised while an onlooker watches from a distance of 12 feet away from the base of the flag pole (see the figure below). The flag rises vertically at a rate of 7 inches per second. Let t denote the time (in seconds) after the flag begins to rise (For simplicity, assume that when the flag begins to rise it is 0 inches above the ground). Express the distance d (in feet) between the flag and the onlooker as a function of t. d (t) = ?
Answer by prince_abubu(198) (Show Source): You can put this solution on YOUR website!
For another added simplicity, let's say that the person who is 12 feet away has his eyes on ground level. This way, we can express the distance from the ground 12 feet away to the flag as it rises.
At t=0, the onlooker's eye is 12 feet away from the flag.
At t=1 second, the flag is 7 inches off the ground, so that must have increased the (diagonal) distance from the observer's eye to the ground, but by how much?
Suppose that the flag was raised at that constant rate for t seconds. This means that the distance up the flagpole would be 7*t feet. We're going to use the pythagorean theorem here. We'll let a be the distance up the flagpole after time t. We'll let b be the ground distance, which is the 12 feet, and d be the hypotenuse, which is the distance from the eye/ground DIRECTLY.
<----- Pythagorean theorem
<----- Substitute according to what we said on the paragraph above.
<----- To solve for d, we must square root both sides.
<---- Now you have the distance d from the eye/ground to the flag. If you think about it, d depends on t. In that case, d is a function of t. We can then rewrite the equation this way:
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