Question

Given the position of the particle and using interval notation, when is the particle moving left and when is it moving right

Answer 1

we have the expression

`(2)/(3)t^3-(9)/(2)t^2-18t`

Factor (2/3)t

`(2)/(3)t(t^2-(27)/(4)t^{}-27)`

Solve the quadratic equation

using the formula

a=1

b=-27/4

c=-27

`t=\frac{-(-(27)/(4))\pm\sqrt[]{(-(27)/(4))^2-4(1)(-27)}}{2(1)}`
`t=\frac{(27)/(4)\pm\sqrt[]{(2457)/(16)}}{2}`
`t=(27)/(8)\pm\frac{\sqrt[]{2457}}{8}`

The values of t are

t=-2.82 and t=9.57

therefore

the intervals are

(-infinite, -2.82)