# 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.

(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) =∫

Hence the distance is given by the formula:

S(t)=∫

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-t

Then S(4) = 8*16-64/6=117.33

v(t) =∫

_{0}^{t}a(s)ds = ∫_{0}^{t}(8-s)ds = 8t - t^{2}/2Hence the distance is given by the formula:

S(t)=∫

_{0}^{t}v(s)ds = ∫_{0}^{t}(8t-t^{2}/2)ds = 8t^{2}-t^{3}/6a) 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-t

^{2}/2 = 0, whence either t=0 or t=4.Then S(4) = 8*16-64/6=117.33

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