Question

x-coordinate of a particle moving along this axis is `x = (2+t^2 + 2t^3).` Here, x is in meres and t in seconds. Find (a) position of particle from where it started its journey, (b) initial velocity of particle and (c) acceleration of particle at `t=2s.`

Answer 1

Correct Answer - A::B::C
(a) At `t=0, x=2.0 m`
(b) `v=(dx)/(dt)=2t+6t^2`
At `t=0,v=0`
(c) `a=(dv)/(dt)=2+12t`
At `t=2s`
`a=26 m//s^2`