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Answer to Question #2723 in Integral Calculus for samankhan

Question #2723
A body moves in a straight line with acceleration of magnitude a m/s2 given by a = 8 – t at time t seconds. If initially the speed is zero, find
(a) the distance travelled in the first three seconds
(b) the maximum distance from the initial point.
Expert's answer
We have that a(t) = v'(t) = 8-t. Since v(0)=0, we have that
v(t) =∫0t a(s)ds = ∫0t(8-s)ds = 8t - t2/2
Hence the distance is given by the formula:
S(t)=∫0t v(s)ds = ∫0t (8t-t2/2)ds = 8t2-t3/6

a) the distance travelled in the first three seconds uis equal to
S(3) = 8*9-27/6 = 67.5

b) the maximum distance from the initial point is achieved at point, where S'(t) = v(t) = 0:
8t-t2/2 = 0, whence either t=0 or t=4.
Then S(4) = 8*16-64/6=117.33

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