Question

The displacement x of a particle varies with time t as x=2t3-2t2+2t+8 Find velocity when acceleration is zero

Answer 1

dx/dt=v

differentiating x=2t3-2t2+2t+8 with respect to t we get = 6t^2 - 4t + 2 =v

dv/dt=a

differentiating 6t^2 - 4t + 2 we get =12t-4=a=0(given in qs)

hence t =1/3

substituting into 6t^2 - 4t + 2 we get v= 4.67m/s

i think this is the way it is done please do check if the ans is right