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Integral question: velocity function of a car is v(t) = 2t^2 + t - 6 find the distance covered by the car during t=0 & t=3
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integrate using limits.
\[\int\limits_{0}^{3} 2t^2 + t -6\]
part a) find displacement of the car between t=0 & t=3 i figured this out and got 4.5 feet (v(t) is expressed in feet/sec btw) part b) find distance covered by the car duing this period of time part b i'm having trouble with, don't know what the hell it's asking
part b sounds exactly like part a
the answer is supposed to be 15.75ft if that helps
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disp is vector while dist is scalar, you know about this right?
4.5 is the disp i think. which is above the x-axis.
if you graph the curve. you will get something like, |dw:1335237647776:dw|
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