Question

The position function s(t)=t3−8ts(t)=t3−8t gives the position in miles of a freight train where east is the positive direction and tt is measured in hours. determine the direction the train is traveling when s(t)=0.s(t)=0. determine the direction the train is traveling when a(t)=0.a(t)=0. determine the time intervals when the train is slowing down or speeding up.

Answer 1

we are given

$s(t)=t^3-8t$

(a)

we can set s(t)=0

and then we can solve for t

$t^3-8t=0$

$t(t^2-8)=0$

$t(t-\sqrt{8}) (t+\sqrt{8})=0$

so, we get

$t=0,t=\sqrt{8}$............Answer

(b)

Firstly, we will find a(t)

we know that a(t) is second derivative of s(t)

$s'(t)=3t^2-8*1$

now, we can find derivative again

$s''(t)=3*2t^1-8*0$

$a(t)=6t$

now, we can set it to 0

and then we can solve for t

$a(t)=6t=0$

$t=0$............Answer

(c)

Firstly, we will set s'(t)=0 and a(t)=0

and then we solve for t

$s'(t)=3t^2-8*1=0$

$t=1.63299$

$t=0$

now, we can draw number line

and then we can check sign of s'(t) and s''(t) on each intervals

same sign means speeding up

different sign means speeding down

speeding up interval:

$(1.63299,\infty)$

speeding down interval:

$(0, 1.63299)$