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A person is walking along a straight line. Her velocity function(in meters per second) is given by v(t)=3t-6, 0<=t<=3. Find (a) the displacement, and (b) the distance travelled by this person over the 3 seconds Note: the answers to the two questions are different. Figuring out what they mean is part of the problem.
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integrate v(t) to get \[x(t)=1.5*t^2-6*t\] set v(t) equal to zero and solve to get the farthest distance traveled from the starting point (the max/min). this is x=2. plug in numbers and she walked backwards 6 meters then forward 1.5 meters for a displacement of -4.5 meters and a total distance traveled of 7.5 meters
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