The position coordinate of a moving particle is given by x = 6+18t +9`t^(2)` (where x is in metre, t in seconds. What is its velocity and acceleration at t = 2 s.

Explanation

Given position coordinate,
x = 6+18t+9`t^(2)=9t^(2)` +18t+6
Differentiating the above equation with respect to t ,
`(dx)/(dt)` = 18t+18 or, v = 18t +18 `" "cdots(1)`
Given t = 2s
Therefore the velocity at 2 s is
v= 36 +18 = 54 m/s
Again differentiating equation (1) w.r.t. time t,
`(dv)/(dt)= 18`
Therefore the acceleration a= 18 m//`s^(2)`.