Calc 1 Riemann Sums w/ velocity and distance

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The discussion focuses on estimating the distance an object traveled using Riemann sums based on given velocity data over a time interval. The user successfully calculated the displacement as -11 feet using a left endpoint Riemann sum but struggled to determine the total distance traveled. It was clarified that distance must always be positive, requiring the calculation of the area under the velocity curve without considering the sign. The user ultimately resolved the confusion by summing the areas as positive values, leading to the correct total distance traveled. Understanding the distinction between displacement and distance was key to solving the problem.
Wm_Davies
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Homework Statement



This is somewhat a repost... except I have figured out some of it and I have cleaned up the question.

Your task is to estimate how far an object traveled during the time interval 0<= t >= 8 , but you only have the following data about the velocity of the object.

\frac{time (sec)}{velocity (feet/sec)}\frac{0}{4}\frac{1}{1}\frac{2}{-2}\frac{3}{-3}\frac{4}{-4}\frac{5}{-3}\frac{6}{-1}\frac{7}{-3}\frac{8}{-1}

"See the attached graph."

(PART 'A') Using the left endpoint Riemann sum, find approximately how far the object traveled. Your answers must include the correct units.

Total displacement = "I have 11ft which is the right answer."

Total distance traveled = "I cannot figure this out"




Homework Equations



Distance = time * velocity
Displacement = time * velocity


The Attempt at a Solution


So I went ahead and got the Riemann sum of the left endpoint on the graph below.

\DeltaX = 1

So I just added the y values.

The sum added up to -11 which was the answer for the displacement. I do not know why this is not the answer for the total distance but maybe I am missing something elementary.
 

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To find the distance traveled find the area of the shaded region. Area is always positive BTW :D
 
computerex said:
To find the distance traveled find the area of the shaded region. Area is always positive BTW :D

I tried to compute the area, but I am not getting it. Also area if area is always positive then why would the area of a curve under the x-axis be negative?
 
Wm_Davies said:
I tried to compute the area, but I am not getting it. Also area if area is always positive then why would the area of a curve under the x-axis be negative?

Distance can never be negative.
 
computerex said:
Distance can never be negative.

O.k. that actually makes tons of sense (I figured I was making some elementary mistake). So, I added up the areas as positive numbers and everything was correct. Thanks for the help.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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