A particle moves along the x-axis according to `s=t^(3)-15/2t^(2)+12t+5`, where the symbols have their usual meaning.
(a) Determine the time when speed is increasing//decreasing.
(b) At what time, the velocity changes its direction?
(c ) Find the distance traveled in the first six seconds.

Explanation

`s=t^(3)-15/2t^(2)+12t+5`
`v=(ds)/(dt)=3t^(2)-15t+12=3(t^(2)-5t+4)=3(t-1)(t-4)`
`a=(dv)/(dt)=6t-15=3(2t-5)`
When the velocity and acceleration have the same sign, the speed is increasing.
When the velocity and acceleration have opposite signs, the speed is decreasing.
When `t lt 2.5 s, a lt 0`
When `t gt 2.5 s, a gt 0`
(a) When `t lt 1 s, v gt 0, a lt 0`, the speed is decreasing
When `1 s lt t lt 2.5 s, v lt 0, a lt 0`, the speed is increasing
When `2.5 s lt t lt 4 s, v lt 0, a gt 0`, the speed is decreasing
When `t gt 4 s, v gt 0, a gt 0`, the speed is increasing
(b) `v=0=3(t-1)(t-4)impliest=1 s, 4s`
At `t=1 s` and `t=4 s`, the velocity changes direction
(c ) At `t=0, s=5m`
At `t=1s, s=10.5 m`
At `t=4s, s=-3m`
At `t=6s, s=23m`