The function s(t) describes the motion of a particle along a line. s(t) = 14t − t2 (a) Find the velocity function v(t) of the particle at any time t ≥ 0. v(t) = 14−2t (b) Identify the time interval on which the particle is moving in a positive direction. (Enter your answer using interval notation.) (c) Identify the time interval on which the particle is moving in a negative direction. (Enter your answer using interval notation.) (7,[infinity]) (d) Identify the time at which the particle changes direction.

Recall that [tex]s(t) = 14t-t^2[/tex]. We will assume that whenever its velocity is positive, it is moving in a positive direction and that whenever the velocity is negative, it is moving in a negative direction. We know that the particle changes direction when the velocity passes from positive to negative or viceversa, as t goes by.

To calculate the velocity (t), we will derivate s(t) with respect to t. Recall that given a functio of the form [tex] x^n[/tex] its derivative is [tex] nx^{n-1}[/tex]. Thus, using the properties of derivatives we get that

[tex] v(t) = s'(t) = 14-2t[/tex].

We can see that 14-2t>0 if 14>2t which is equivalent that 7>t. So the particle moves in a positive direction whenever [tex]0\leq t<7[/tex]. On the same way, 14-2t<0 when t>7 ([tex](7,\infty)[/tex] and it changes direction at t=7 since it passes from positive to negative.