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need help on the attachment @Calculus1
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distance traveled is area under the curve. use geometry (count the little squares) to find it
huh?
if you have the velocity you get the distance traveled by integration. in this case you have no function to integrate using calculus, so you just want the area under the curve
if i see the picture correctly the area under the curve from 0 to 3 is 5
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4 1x1 squares plus the 1x2 triangle on the left
how did you know it was 0 to 3
the question says "how far at t = 3?"
so assuming of course that we start counting at t = 0 ( usually correct) then it is from 0 to 3
ok
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4 1x1 squares plus the 1x2 triangle on the left??
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