A particle is moving with a velocity of `v=(3+ 6t +9t^2) m/s.` Find out
(a) the acceleration of the particle at `t=3 s.`
(b) the displacement of the particle in the interval `t=5s` to `t=8 s.`

Explanation

(a) Velocity of the moving particle `v = (3+6t+9t^(2))`
Acceleration of particle, `a=(dv)/(dt)=(6+18 t)cm^(-2)`
At t = 3 s, `a=(6+18xx3)cm//s^(2)=60 cms^(-2)`
(b) Given, velocity `v =(3+6t+9t^(2))cm//s`
or `(ds)/(dt)=(3+6t+9t^(2))` or `ds = (3+6t+9t^(2))dt`
`thereofre` By integration `int_(0)^(s)ds=int_(5)^(8)(3+6t+9t^(2))dt`
The displacement of the particle in the interval t = 5 s to t = 8 s
`s=[3t+3t^(2)+3t^(3)]_(5)^(8)` or `s = 1287 cm`