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

Hence 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}/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-t^{2}/2 = 0, whence either t=0 or t=4.

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

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

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