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

Answer:

Option D is correct.

as the particle always moves to the right

Step-by-step explanation:

Given the function:  $s(t) = \frac{t^3}{3} - \frac{12t^2}{2} + 36t$ , 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, $v(t) =\frac{ds}{dt} > 0$

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

Use derivative formula:

$\frac{dx^n}{dx} = nx^{n-1}$

$\frac{ds}{dt} = \frac{3t^2}{3}-\frac{24t}{2}+36$

Simplify:

$\frac{ds}{dt} = t^2-12t+36$

As $\frac{ds}{dt} > 0$

⇒$t^2-12t+36 > 0$

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

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

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