A particle is moving in a straight line is at a distance S from a point O on the line in time t, where `S=t^3-6t^2+8t+5`.Find the velocity when the acceleration is `12 cm//sec^2`.

Explanation

Given, `S=t^3-6t^2+8t+5`
`therefore` velocity,`v=(dS)/dt=3t^2-12t+8`
and acceleration, `a=(dv)/(dt)=6t-12`
But, acceleration, `a=12 cm//sec^2`(given)
`therefore` 6t-12=12 implies t=4`
Required velocity, `v=(3.4)^2-12.4+8=8 cm//sec`