How Can the Area Under a Velocity-Time Graph Determine Displacement?

[telko]

OK I have a v-t graph and I wish to find the change in displacement/distance/position w/e.

REFER TO THE PICTURE FOR THE ACTUAL QUESTION THANKS! (see the pic attached)

My teacher taught me a method of seperating the LINE (ref.picture) into segments then calculating the area of each one then adding then up to get the final displacement but I am wondering if my method will work. (not that I don't like her method but i don't find it to be of a sufficient accuracy due to human error.)

Def of ellipse: An ellipse is a set of points in the plane, whose sum of distances from two fixed points is constant. I cannot draw a perfect ellipse i don't have graphing calculator so i can't derive the question of my answer from trying to get the sum of two points over and over again.

Anybody?

 

[marlon]

I don't get this...

What is it you want to do ?

marlon

PS : the system your teachet taught you is just the discrete version of what we call an integral. This is the approach to construct the integral formalism using the concepts of the Darboux Sums.

 

[telko]

lol the longer this question remains here the longer i begin to suspect its retardness. Anyway,

On this site if you scroll to the bottom and read the last paragraph, more specifically the information regarding "(d)"

QUOTE""
Even though wehave no formula for the area of a funny shape like (d), we can approximateits area by dividing it up into smaller areas like rectangles, whose area iseasier to calculate. If someone hands you a graph like (d) and asks you tofind the area under it, the simplest approach is just to count up the littlerectangles on the underlying graph paper, making rough estimates offractional rectangles as you go along.
/QUOTE""

Here it says there is no formula but I just made one in the first post and am wondering if it is correct.

Also to re-iterate my first post for clarity: If the Line on a V-T graph is one continuous line across the whole graph.(as i have drawn in the picture in post 1)
Then, can I not take that line and create 4 identical ones to create an Ellipse?
And from there on use basic calculations to subtract a portion (1/4) of it from a invented square resulting in the area being that of whatever is below the curve?

err.. thanks.

 

[andrevdh]

Your ellipse approach only helps to calculate the total distance covered. To draw a position time graph you have to evaluate the area under the curve at several time intervals, so I am afraid you will have to draw the graph on graph paper and count squares or stick a photocopy of such over the original graph.