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Using Graph B: http://www.webassign.net/sercp8/2-figure-07.gif Find the average and instantaneous velocity at t = 9.3 s. Got -0.75 m/s for average velocity, not at all positive that's correct. (To be able to view graph, copy and paste the link! Clicking it won't work :( )
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By this graph (b) the region near t=9.3 is linear (i.e. straight line). Thus, the velocity in this region is constant and can be found by calculation the slope of the line BC: \[v_(average) = v(insta.) = (0-10)/(12-8) = -2.5(m/\sec)\]
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