Respuesta :
Answer:
a) [tex]s'(t) = 3t^2 -10 t +4[/tex]
b) [tex](-\infty <t \leq \frac{5-\sqrt{13}}{3})U(\frac{5+\sqrt{13}}{3}<t < \infty)[/tex]
c) [tex] ( \frac{5-\sqrt{13}}{3} <t < \frac{5+\sqrt{13}}{3})[/tex]
d) [tex] t_1 =\frac{10 -\sqrt{52}}{6}=\frac{5-\sqrt{13}}{3}[/tex]
[tex] t_2 =\frac{10 +\sqrt{52}}{6}=\frac{5+\sqrt{13}}{3}[/tex]
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:
[tex]s'(t) = 3t^2 -10 t +4[/tex]
(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:
[tex]X= \frac{-b \pm \sqrt{b^2 -4ac}}{2a}[/tex]
For this case[tex] a= 3, b = -10, c= 4[/tex]
And if we replace we got:
[tex]t= \frac{10 \pm \sqrt{(-10)^2 -4(3)(4)}}{2(3)}[/tex]
[tex] t_1 =\frac{10 -\sqrt{52}}{6}=\frac{5-\sqrt{13}}{3}[/tex]
[tex] t_2 =\frac{10 +\sqrt{52}}{6}=\frac{5+\sqrt{13}}{3}[/tex]
And now we can see on which intervals we have the velocity positive or negative:
[tex] -\infty <t \leq \frac{5-\sqrt{13}}{3} = +[/tex]
[tex] \frac{5-\sqrt{13}}{3} <t \leq \frac{5+\sqrt{13}}{3}= -[/tex]
[tex] \frac{5+\sqrt{13}}{3}<t <\infty= +[/tex]
So the is positive between [tex]( -\infty <t \leq \frac{5-\sqrt{13}}{3})U(\frac{5+\sqrt{13}}{3}<t<\infty)[/tex]
(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:
[tex](\frac{5-\sqrt{13}}{3} <t < \frac{5+\sqrt{13}}{3})[/tex]
(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:
[tex] t_1 =\frac{10 -\sqrt{52}}{6}=\frac{5-\sqrt{13}}{3}[/tex]
[tex] t_2 =\frac{10 +\sqrt{52}}{6}=\frac{5+\sqrt{13}}{3}[/tex]
