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Physics 21 Online
OpenStudy (anonymous):

help me in this question of vectors

OpenStudy (anonymous):

OpenStudy (amistre64):

hmm, displacement/time = average velocity is what your last question i helped out stated. integration gives us the area under a curve, which is also a definition of displacement

OpenStudy (anonymous):

OpenStudy (amistre64):

\[2\int_{0}^{1}\sqrt{1-x^2}dx\] is my idea

OpenStudy (anonymous):

i got this answer from a site but can not understand it

OpenStudy (amistre64):

ok, thats using the standard definition of displacement

OpenStudy (amistre64):

notice that from start to finish the thing is only displaced by 2 units, the negative is a direction .. downward

OpenStudy (amistre64):

displacement / time = -2/1

OpenStudy (anonymous):

but it is given only 1 m ? how are -2 m there ?

OpenStudy (amistre64):

its a semicircle with a radius of 1m the diameter of the circle measures 2m in a down direction, so -2m

OpenStudy (anonymous):

oh yes thanks thanks thanks thanks thanks a lot :)))))))))))))))))))))))) gr8 help amsitre , i think u deserve the word SIR

OpenStudy (amistre64):

i spose, its on all the letters i get from the collection agencies :)

OpenStudy (anonymous):

nice

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