Answer:
E
Step-by-step explanation:
We are given that a particle's position along the x-axis at time t is modeled by:
[tex]x(t)=2t^3-21t^2+72t-53[/tex]
And we want to determine at which time(s) t is the particle at rest.
If the particle is at rest, this suggests that its velocity at that time is 0.
Since are we given the position function, we can differentiate it to find the velocity function.
So, by differentiating both sides with respect to t, we acquire:
[tex]\displaystyle x^\prime(t)=v(t)=\frac{d}{dt}\big[2t^3-21t^2+72t-53\big][/tex]
Differentiate. So, our velocity function is:
[tex]v(t)=6t^2-42t+72[/tex]
So, we will set the velocity to 0 and solve for t. Hence:
[tex]0=6t^2-42t+72[/tex]
We can divide both sides by 6:
[tex]0=t^2-7t+12[/tex]
Factoring yields:
[tex](t-3)(t-4)=0[/tex]
By the Zero Product Property:
[tex]t-3\text{ and } t-4=0[/tex]
Hence:
[tex]t=3\text{ and } t=4[/tex]
Therefore, at the 3rd and 4th seconds, the velocity of the particle is 0, impling that the particle is at rest.
Our answer is E.
