The function
s
(
t
)
describes the motion of a particle along a line:
s
(
t
)
=
18
t
−
t
2
.

a) find the velocity function
v
(
t
)
of the particle at any time
t
≥
0.

b) identify the time interval in which the particle is moving in a positive direction.

c) identify the time interval in which the particle is moving in a negative direction.

d) identify the time at which the particle changes direction.

a)

b)

c)

d)

Explanation:

We have the following parts:

(a) find the velocity function of the particle at any time t ≥0 )

For this case w ejust need to take the derivate of s(t) and we got:

(b) identify the time interval(s) in which the particle is moving in a positive direction

The partcile is moving in the positive direction when the velocity is higher than 0. We have a quadratic equation for the velocity so we can solve for the t intercepts  like this:

For this case

And if we replace we got:

And now we can see on which intervals we have the velocity positive or negative:

So the is positive between

(c) identify the time interval(s) in which the particle is moving in a negative direction

From the before part we see that the velocity is negative just on this interval:

(d) identify the time(s) at which the particle changes direction. s(t) = t³ - 5t²+4t

The points where the particle changes direction are given by the critical point because are the points where the derivate is 0 and we have the change of direction and on this case are:

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y