Answer:
Option D is correct.
as the particle always moves to the right
Step-by-step explanation:
Given the function: [tex]s(t) = \frac{t^3}{3} - \frac{12t^2}{2} + 36t[/tex] , 0<t<15;
where
s(t) represent the distance in feet.
t represents the time in second.
Since, when the particle is moving to the right,
⇒ s(t) is increasing.
so, [tex]v(t) =\frac{ds}{dt} > 0[/tex]
Find the derivative of s(t) with respect to t;
Use derivative formula:
[tex]\frac{dx^n}{dx} = nx^{n-1}[/tex]
[tex]\frac{ds}{dt} = \frac{3t^2}{3}-\frac{24t}{2}+36[/tex]
Simplify:
[tex]\frac{ds}{dt} = t^2-12t+36[/tex]
As [tex]\frac{ds}{dt} > 0[/tex]
⇒[tex] t^2-12t+36 > 0[/tex]
[tex](t-6)^2 >0[/tex] [[tex](a-b)^2 = a^2-2ab+b^2[/tex] ]
⇒ this is always true because square of any number is always positive
Therefore, it means that the particle always moves to the right.
