Question

The position of a particle is represented by the following equation.
`x=3t^(3)+7t^(2)+5t+8`
where x is in metres and t in seconds. Find the acceleration of the particle at t = 1 s.
Strategy : `v=(dx)/(dt) and a=(dv)/(dt)`

Answer 1

Given that, `x=3t^(3)+7t^(2)+5t+8`
`v=(dx)/(dt)=9t^(2)+14t+5`
`a=(dv)/(dt)=18t+14`
`a|_(t=1)=18(1)+14=18+14=32m//s^(2)`