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