Question

Given the position function, s of t equals t cubed divided by 3 minus 12 times t squared divided by 2 plus 36 times t , between t = 0 and t = 15, where s is given in feet and t is measured in seconds, find the interval in seconds where the particle is moving to the right. (5 points)

0 < t < 6


5 < t < 15


t > 15


The particle always moves to the right.

Answer 1

Option D is correct.

as the particle always moves to the right

Explanation:

Given the function:
`s(t) = (t^3)/(3) - (12t^2)/(2) + 36t` , 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,
`v(t) =(ds)/(dt) > 0`

Find the derivative of s(t) with respect to t;

Use derivative formula:

`(dx^n)/(dx) = nx^(n-1)`

`(ds)/(dt) = (3t^2)/(3)-(24t)/(2)+36`

Simplify:

`(ds)/(dt) = t^2-12t+36`

As
`(ds)/(dt) > 0`

⇒
`t^2-12t+36 > 0`

`(t-6)^2 >0` [
`(a-b)^2 = a^2-2ab+b^2` ]

⇒ this is always true because square of any number is always positive

Therefore, it means that the particle always moves to the right.