Question

the position of a particle moving along x-axis given by x=(-2t³+3t²+5). the acceleration of particle at the instant its vilocity becomes zero is??

Answer 1

Hi ...dear...
here is your answer....
X = -2t^3+3t^2+5V
=dx/dt=-6t^2+6tAt t= 1
, v =0 ( -6(1)^2-6(1)=0)
At t=1 velocity =0
then acceleration = dv/dt = -12t+6 at t= 1 , a= -12(1)+6=-6m/s^2.......
.
.
hope it helped you!!!
Regards Brainly Star Community
#shubhendu