How 2 find distance from non uniform velocity time graph where can't use triangle?

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*Double post,from merge of 2 posts,sry.
 
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Hi. *Problem solved.

A trolley of mass 930 g is held on a horizontal surface by means of two springs,one spring on the left and right respectively.The variation with time t of the speed v of the trolley for the first 0.60s of its motion is shown in the fig(It's a v-t graph,max y=8.0cms^-1 ,min x=0.0s & 0.6s) below.

untitled.jpg


Use the fig above to determine the distance moved during the first 0.60s of its motion.

The answer=0.031m +-.001m.I find using a triangle to find half of the distance,then multiply 2 to get the whole distance not satisfying.I used s=ut +at^2/2 also and still didn't get it.Any 1 pls tell of a way to get the ans?


*Edit
 
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Discussion

Im sorry, but I don't understand what your talking about. Its not 'area', it represents something.
 
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Im sorry, but I don't understand what your talking about. Its not 'area', it represents something.
 
It's distance,which is represented under the area under the curve,how to find?
 
You approximate the area using rectangles.

If you know calculus then you can calculate the exact area in many cases by finding an antiderivative.
 
You approximate the area using rectangles.

If you know calculus then you can calculate the exact area in many cases by finding an antiderivative.
 
Crosson said:
You approximate the area using rectangles.

If you know calculus then you can calculate the exact area in many cases by finding an antiderivative.
There's no equation given for the graph,how?
 
Crosson said:
You approximate the area using rectangles.

If you know calculus then you can calculate the exact area in many cases by finding an antiderivative.
There's no equation given for the graph,how?
 
Use the geometry of the curve to estimate the area. Do you have a digital camera to snap a picture of the graph?
 
Use the geometry of the curve to estimate the area. Do you have a digital camera to snap a picture of the graph?
 
A diagram would certinaly be nice!

Can you segment the area under the graph into nice geometric regions? By that I mean triangles and rectangles?
 
I've just added the graph pic on the first post,edited.If u 1 ,refer to that 1.
 
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This thread has been merged with another IDENTICAL question that the OP has cross-posted. So if it appears to make no sense at some spot, it isn't my fault.

Zz.