The displacement x of a particle varies with time 't' as, `x=3t^(2)-4t+30.` Find the position, velocity and acceleration of the particle at t = 0.

Explanation

Position of a particle `x=3t^(2)-4t+30.`
velocity `v=(dx)/(dt)=(d)/(dt)(3t^(2)-4t+30)`
`=1+6t-4" "[because x^(n)=nx^(n-1)]`
`"Acceleration "a=(dv)/(dt)=(d)/(dt)(6t-4)`
`a=6`
At time t = 0, we have,
`"Position x"=3t^(2)-4t+30`
`=3(0)^(2)-4(0)+30`
x = 30 m.
`"Velocity v"=6t-4=6(0)-4`
`=-4m//s=-4ms^(-1)`
`"Acceleration a"=6ms^(-2)`
`because" position x"="30 m, velocity v"=-4ms^(-1)`,
`"acceleration "=6ms^(-2)`