Physics Help Forum Help with a graph problem
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 Nov 16th 2017, 11:07 AM #1 Junior Member   Join Date: Nov 2017 Posts: 3 Help with a graph problem I have a physics exam coming up and in my practice problems there is a problem that I just can't solve sine I was absent from the class it was discussed at. I know it's simple, but I just can't find a way to solve it, I've looked everywhere online! I've inserted a picture of the said problem. It shows a speed/time graph and the text of the problem says: "What is the value of the shift of position from second 2 to second 8?" Below it, it says to solve by calculating the surface of the squares below the curvature of the graph. And I just can't figure it out! Any help will be greatly appreciated!
Nov 16th 2017, 07:48 PM   #2
Senior Member

Join Date: Aug 2008
Posts: 113

 "What is the value of the shift of position from second 2 to second 8?"
each small square represents 1/4 m displacement.

if I counted correctly ...

$\displaystyle \Delta x = \int_2^8 v(t) \,dt = \int_2^5 v(t) \,dt + \int_5^8 v(t) \,dt = 6-8.75 = -2.75 \, m$

 Nov 17th 2017, 02:59 AM #3 Senior Member     Join Date: Jun 2016 Location: England Posts: 741 Note that Skeeter has introduced the integration symbol in his reply, You should recognise that the integral of speed with respect to time will give displacement (shift of position). Summing the squares under the graph is a basic numerical integration method, It will give an estimate of the integral of the graph, and thus an estimate of the change in position. topsquark likes this. __________________ ~\o/~
 Nov 17th 2017, 05:56 AM #4 Junior Member   Join Date: Nov 2017 Posts: 3 After thinking the solution over for a while, I have to say I still do not understand how this is calculated, specifically, how did we end up with two integrals and how do you get them? Could you explain step by step? Also, is there a way to solve this without using integration? Your help is greatly appreciated!
 Nov 17th 2017, 06:33 AM #5 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 995 skeeter didn't actually do the integrals he counted the grid squares hatched in green. (hope he didn't have too much mesquite juice first) Note he broke the integrals into two because you require the area under the graph to obtain the distance moved, and part is positive and part negative. One part is positive because the fly is moving forwards in the first part above the x axis. That is further and further from the original position. The second part is negative which means that the fly is moving backwards. Because the fly is moving backwards it is getting closer and closer to the original position. So the final change in position is the difference of the forwards distance moved and the backwards distance moved. Attached Thumbnails   Last edited by studiot; Nov 17th 2017 at 06:44 AM.
 Nov 17th 2017, 07:10 AM #6 Senior Member   Join Date: Aug 2010 Posts: 388 The "integral" or, in this case, the distance moved given a velocity vs time graph is the area between the curve and the x-axis. That can be very difficult to calculate for many curves and required the invention of "integral caculus". But here, the graph is given as straight lines, horizontal or from one vertex of a square to the diagonally opposite vertex. The "area under the graph" is always made of whole squares, with area 1, or half squares, with area 1/2. From 2 to 2.5 seconds, I see a stack of 7 whole squares and one half square: that area is 7.5. From 2.5 to 3.0, I see a stack of 6 whole squares and one half square: area 6.5. From 3.0 to 3.5, I see a stack of 4 whole squares and 2 partial squares. Neither of the partial squares is exactly a half but since the graph goes from the upper left vertex of the upper square to the lower right vertex of the lower square it cuts those two squares in half- (1/2)(2)= 1. The area is 4+ 1= 5. From 3.5 to 4, I see 2 whole squares and then two partials: 2+ 1= 3. From 4 to 4.5 I see one whole and one half square: 1.5. From 4.5 to 5, I see a half square: .5. So far that is an "area under the curve" or 7.5+ 6.5+ 5+ 3+ 1.5+ .5= 23.0. During that time, from 2 to 5 seconds, the object has moved, to the right, 23.0 meters. After 5 seconds, the velocity becomes negative which means it has started moving back to the left. From 5 to 5.5 seconds, I see 2 partial squares: (1/2)(2)= 1 but since this is below the t-axis, that is -1. From 5.5 to 6 seconds, I see a total of two full squares and four partial squares but because the graph is from one vertex to another, that is a total area of 2+ (1/2)(4)= 4. Again, because it is below the t-axis, that is -4. From 6 to 6.5 seconds, I see 6 whole squares and half of one square for a total of 6.5 squares: -6.5. From 6.5 to 7 seconds, I see 7 whole squares and half of one square for a total of 7.5 squares. Since it is below the t-axis, that is -7.5. Finally, from 7 to 7.5 seconds and from 7.5 to 8 seconds, I see 8 whole squares for a total of 16 between 7 and 8 seconds. Again, that is -16 since it is below the t-axis. From t= 5 to t= 8, that is a total of -1- 4- 6.5- 7.5- 16= -35. Between t= 5 and t= 8 seconds the object has moved 35 meters back to the left. The "displacement", the actual signed distance from the starting point to the end point, is 23- 35= -12. The object ends up 12 meters to the left of its starting point.
 Nov 18th 2017, 07:22 AM #7 Junior Member   Join Date: Nov 2017 Posts: 3 Okay, I finally get it! Thank you studiot for highlighting the area I need to focus on, skeeter for calculating me and giving me the means to check my result and HalssofIvy for showing me the way to do it! Although, one small thing, HallsofIvy you forgot to multiply your result with the surface of the tiny squares, which is 0.25. What you calculated was the number of the surfaces of all the tiny squares that are under the curve. But nontheless, thank you again!

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