A particle is moving in a straight line under acceleration `a=54-18 t`. The particle starts from the origin.
(a) Sketch the a-t, v-t and s-t graph.
(b) Find the distance traveled in the first 9 s. (Acceleration is in `m//s^(2)`, time is in second)

Explanation

(a) (i) `a=54-18t`
`t=0, a=54 m//s^(2)`
`a=0implies54-18t=0impliest=3 s`
`t lt 3 s, a= +ve`
`t gt 3 s, a=-ve`
Shape: `y=54-18x`(straight line)

(ii) `a=54-18t`
`(dv)/(dt)=54-18 t`
`int_(0)^(v) dv=int_(0)^(t)(54-18t) dt`
`|v|_(0)^(v)=|54t-18t^(2)/2|_(0)^(t)`
`v=54t-9t^(2)`
`t=0, v=0`
`v=0implies54t-9t^(2)=0`
`9t(6-t)=0impliest=0, t=6 s`
v is maximum if `(dv)/(dt)=a=54-18t=0`
For `t=3 s`
`v=v_(max)=54xx3-9xx3^(2)=162-81=81 m//s`
For t gt 6 s, v=-ve
Shape: `y=54x-9x^(2)` (parabola, open downward)

(iii) s-t graph
`v=54t-9t^(2)`
`(ds)/(dt)=54 t-9 t^(2)`
`int_(0)^(s) ds=int_(0)^(t)(54 t-9 t^(2)) dt`
`s=27 t^(2)-3 t^(3)`
`t=0, s=0`
`s=0implies27 t^(2)-3t^(3)=0implies3t(9-t)=0`
`impliest=0, t=9 s`
s is maximum if `(ds)/(dt)(=v)=0`
`implies t= 6s`
`s_(max)=27(6)^(2)-3(6)^(3)=324 m`
For t gt 9 s, `s=-ve`
For `t=0` to `t=3s`, velocity is increasing, i.e. slope of s-t graph is increasing.
For `t=3` to `6s`, velocity is decreasing and v gt 0, a lt 0, the particle is slowing down, slope of the s-t graph is decreasing.
For t gt 6 s, v lt 0, a 0 lt 0, the particle is speeding up.
For `t=3 s, s=162 m`
For `t=6 s, s=324 m`
For `t=9 s, s=0`

(b) Distance in 9 s
Velocity changes sign at `t=6 s`
Distance`=int_(0)^(6) v dt+|int_(6)^(9)v dt|`
`=324+|-320|=648 m`