Answer to Question #2723 in Integral Calculus for samankhan

Question

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.

Answer 1

We have that a(t) = v'(t) = 8-t. Since v(0)=0, we have that
v(t) =∫₀^t a(s)ds = ∫₀^t(8-s)ds = 8t - t²/2
Hence the distance is given by the formula:
S(t)=∫₀^t v(s)ds = ∫₀^t (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