Accerlation distance etc

  • #1
bascially, not techinally a question, but some basic theory i need to know for the batch of qu's.

basically, i need to know how to find the distance travelled by something using a speed/time graph. i know this is area under it (curvey accerlation then constant speed then curvey retardation). so i what i though i'd do is split the graph into, accel., constant v, and retard. now i know how to find the distance travelled during constant v. but i not sure how to do with accerlation and retardation.

i know the accerlation to be 6.62m/s^2 for 2.9s and the retardation to be -1.2m/s^2 for 1s, or i 'think' it can be written as 1.2m/s^-2 ???

anyways, can someone please explain how i would find the distance travelled please :D and say if im doint he wrong way too. thnx
 

Answers and Replies

  • #2
Hootenanny
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Is it a requirement of the problem that you are to solely use the s/t graph? Perhaps a kinematic equation would be more appropriate here. Have you met such equations before?
 
  • #3
dont think i have, i'm only 16 and are doing the exam all other 16 year olds are doing.

well providing i could use this 'kinematic equation' i dont see why i shouldn't

i'm not exactly sure how i would find the distance travelled solely using the s/t time graph :S

thnx
 
  • #5
no sorry, that above my level, so are there any other ways then?

thnx for all the help
 
  • #6
Hootenanny
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no sorry, that above my level, so are there any other ways then?

thnx for all the help
I suppose you could approximate the graph as a trapezium or various other shapes and calculate an approximate area; but I'm not sure. Perhaps someone with more knowlage of GCSE level questions could chip in here...

Just one more point;
i know the accerlation to be 6.62m/s^2 for 2.9s and the retardation to be -1.2m/s^2 for 1s, or i 'think' it can be written as 1.2m/s^-2 ???
[tex]-1.2m/s^2 \neq 1.2m/s^{-2}[/tex]
 

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