Question

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.

Answer 1

a)
`s'(t) = 3t^2 -10 t +4`

b)
`(-\infty <t \leq (5-√(13))/(3))U((5+√(13))/(3)<t < \infty)`

c)
`( (5-√(13))/(3) <t < (5+√(13))/(3))`

d)
`t_1 =(10 -√(52))/(6)=(5-√(13))/(3)`

`t_2 =(10 +√(52))/(6)=(5+√(13))/(3)`

Step-by-step 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:

`s'(t) = 3t^2 -10 t +4`

(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:

`X= (-b \pm √(b^2 -4ac))/(2a)`

For this case
`a= 3, b = -10, c= 4`

And if we replace we got:

`t= (10 \pm √((-10)^2 -4(3)(4)))/(2(3))`

`t_1 =(10 -√(52))/(6)=(5-√(13))/(3)`

`t_2 =(10 +√(52))/(6)=(5+√(13))/(3)`

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

`-\infty <t \leq (5-√(13))/(3) = +`

`(5-√(13))/(3) <t \leq (5+√(13))/(3)= -`

`(5+√(13))/(3)<t <\infty= +`

So the is positive between
`( -\infty <t \leq (5-√(13))/(3))U((5+√(13))/(3)<t<\infty)`

(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:

`((5-√(13))/(3) <t < (5+√(13))/(3))`

(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:

`t_1 =(10 -√(52))/(6)=(5-√(13))/(3)`

`t_2 =(10 +√(52))/(6)=(5+√(13))/(3)`