Answer to Question #2723 in Integral Calculus for samankhan
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.
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
Dread it, run from it, the end of summer arrives all the same. And now it's here. It’s finally August.…
APPROVED BY CLIENTS
Thank you so much for your hard work. very impressive indeed ! I really appreciate your effort, however, the alias-listing is missing half the names. I got a list with 9 names only in it after I compiled the code, but very good job over all nonetheless. thank you all again and special thanks to the one worked on this assignment, and I'll def contact you in the future should I ever feel the need of an expert!!