Differentiate both sides of that equation with respect to time to find the velocity:
The derivative of x^2 with respect to t is 2x dx/dt and the derivative of 2+ t is 1. 2x dx/dt= 2xv= 1.
Differentiate again to find the acceleration.
Using the product rule, the derivative of 2x v is 2 v^2+ 2x dv/dt= 2v^2+ 2xa. Of course the derivative of 1 is 0.
So we have 2v^2+ 2xa= 0. From 2xv= 1 , v= 1/(2x) so we can write 2(1/(4x^2))+ 2x a= 1/(2x^2)+ 2x a= 0. The acceleration is 1/(4x^3).
