The displacement of a particla moving in straight line is given by `s=t^(4)+2t^(3)+3t^(2)+4`, where s is in meters and t is in seconds. Find the
(a) velocity at `t=1 s`, (b) acceleration at `t=2 s`,
(c ) average velocity during time interval `t=0` to `t=2 s` and
(d) average acceleration during time interval `t=0` to `t=1 s`.

Explanation

`s=t^(4)+2t^(3)+3t^(2)+4`
`v=(ds)/(dt)=4t^(3)+6t^(2)+6t`
`a=(dv)/(dt)=12t^(2)+12t+6`
(a) `At t=1 s, v=4(1)^(3)+6(1)^(2)+6(1)=16 m//s`
(b) `At t=2 s, a =12(2)^(2)+12(2)+6=78 m//s^(2)`
(c ) `At t_(1)=0, s_(1)=4 m`
At `t_(2)=2 s, s_(2)=(2)^(4)+2(2)^(3)+3(2)^(2)+4=48 m`
`bar(v)=(Deltas)/(Deltat)=(s_(2)-s_(1))/(t_(2)-t_(1))=(48-4)/(2-0)=22 m//s`
`At t_(1)=0, v_(1)=0`
At `t_(2)=1 s, v_(2)=4(1)^(3)+6(1)^(2)+6(1)=16 m//s`
`bar(a)=(Deltav)/(Deltat)=(v_(2)-v_(1))/(t_(2)-t_(1))=(16-0)/(1-0)=16 m//s`