# Accerlation distance etc

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

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Hootenanny
Staff Emeritus
Gold Member
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?

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

no sorry, that above my level, so are there any other ways then?

thnx for all the help

Hootenanny
Staff Emeritus
Gold Member
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 ???
$$-1.2m/s^2 \neq 1.2m/s^{-2}$$