Question

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.

Answer 1

Answer:

a) v(t) = 14-2t.

b) The particle is moving in a positive direction in the interval [0,7) and it's moving a negative direction in the interval $(7,\infty)$

c) the particles changes direction at t=7.

Step-by-step explanation:

Recall that $s(t) = 14t-t^2$. 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 $x^n$ its derivative is $nx^{n-1}$. Thus, using the properties of derivatives we get that

$v(t) = s'(t) = 14-2t$.

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 $0\leq t$. On the same way, 14-2t<0 when t>7 ($(7,\infty)$ and it changes direction at t=7 since it passes from positive to negative.