1) Time: 2 s, 6 s
The position of the particle is given by:
[tex]x=t^3 -12t^2 +36t+32[/tex]
where t is the time in seconds and x is the position in feet.
The velocity of the particle can be found by differentiating the position:
[tex]v(t)=x'(t)=3t^2 -24t+36[/tex]
and it is expressed in ft/s.
In order to find the time at which the velocity is v=0 ft/s, we substitute v=0 into the previous equation:
[tex]0=3t^2-24t+36\\0=t^2 -8t+12\\0=(t-2)(t-6)[/tex]
So the two solutions are
t = 2 s
t = 6 s
2) Position: x = 64 ft and x = 32 ft
The position at which the velocity of the particle is v = 0 can be found by susbtituting t = 2 and t = 6 into the equation for the position.
For t = 2 s, we have
[tex]x=(2)^3-12(2)^2 +36(2)+32=64[/tex]
For t = 6 s, we have
[tex]x=(6)^3-12(6)^2 +36(6)+32=32[/tex]
So the two positions are
x = 64 ft
x = 32 ft
3) Acceleration: [tex]-12 ft/s^2[/tex] and [tex]+12 ft/s^2[/tex]
The acceleration of the particle can be found by differentiating the velocity. We find:
[tex]a(t)=v'(t)=6t-24[/tex]
And substituting t = 2 and t = 6, we find the acceleration when the velocity of the particle is zero:
[tex]a(2)=6(2)-24=-12[/tex]
[tex]a(6)=6(6)-24=12[/tex]
So the two accelerations are
[tex]a=-12 ft/s^2[/tex]
[tex]a=12 ft/s^2[/tex]
