- #1
inv
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*Double post,from merge of 2 posts,sry.
Last edited:
How 2 find distance from non uniform velocity time graph where can't use triangle?
- Thread starter inv
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Homework Help Overview
The discussion revolves around determining the distance traveled by a trolley based on a non-uniform velocity-time graph. The original poster describes a scenario involving a trolley held by springs, with a specific velocity-time graph provided for the first 0.60 seconds of motion.
Discussion Character
- Exploratory, Conceptual clarification, Problem interpretation
Approaches and Questions Raised
- Participants discuss the concept of area under the curve as it relates to distance, with some suggesting the use of rectangles for approximation. Others express confusion about the terminology and the representation of the area.
Discussion Status
There is ongoing exploration of how to calculate the area under the curve without a clear consensus. Some participants have suggested using geometric approximations, while others have questioned the lack of a given equation for the graph.
Contextual Notes
The original poster has indicated difficulties with traditional methods such as using triangles and specific equations, leading to a search for alternative approaches. A graph has been added for reference, but there are still questions regarding its interpretation and the best method for calculating the distance.
- #2
inv
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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.
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
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.
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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- #3
Cyrus
- 3,246
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What is the area under the curve?
- #4
Cyrus
- 3,246
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What is the area under the curve?
- #5
Cyrus
- 3,246
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Im sorry, but I don't understand what your talking about. Its not 'area', it represents something.
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- #6
Cyrus
- 3,246
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Im sorry, but I don't understand what your talking about. Its not 'area', it represents something.
- #7
inv
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It's distance,which is represented under the area under the curve,how to find?
- #8
Crosson
- 1,256
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You approximate the area using rectangles.
If you know calculus then you can calculate the exact area in many cases by finding an antiderivative.
If you know calculus then you can calculate the exact area in many cases by finding an antiderivative.
- #9
Crosson
- 1,256
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You approximate the area using rectangles.
If you know calculus then you can calculate the exact area in many cases by finding an antiderivative.
If you know calculus then you can calculate the exact area in many cases by finding an antiderivative.
- #10
inv
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There's no equation given for the graph,how?Crosson said:
- #11
inv
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There's no equation given for the graph,how?Crosson said:
- #12
Cyrus
- 3,246
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Use the geometry of the curve to estimate the area. Do you have a digital camera to snap a picture of the graph?
- #13
Cyrus
- 3,246
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Use the geometry of the curve to estimate the area. Do you have a digital camera to snap a picture of the graph?
- #14
Integral
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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?
Can you segment the area under the graph into nice geometric regions? By that I mean triangles and rectangles?
- #15
inv
- 46
- 0
I've just added the graph pic on the first post,edited.If u 1 ,refer to that 1.
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- #16
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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.
Zz.
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