Calc 1 Riemann Sums w/ velocity and distance

Wm_Davies
Messages
51
Reaction score
0

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.
 

Attachments

  • Graph.jpg
    Graph.jpg
    20.3 KB · Views: 1,896
Last edited:
Physics news on Phys.org
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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Replies
6
Views
2K
Replies
8
Views
2K
Replies
5
Views
2K
Replies
4
Views
2K
Replies
2
Views
3K
Replies
1
Views
7K
Replies
2
Views
3K
Back
Top