vx=dx/dt=4，vy=dy/dt=6t-6

The tangential acceleration vt = √ (VX 2+vy 2) = √ (4 2+(6t-6) 2) = √ (16+36 (t-1) 2).

In order to find the minimum speed, the first derivative of the speed vt is needed.

dvx/dt=36*2(t- 1)/2√( 16+36(t- 1)^2)=36(t- 1)/√( 16+36(t- 1)^2)

Let dvx/dt=0, t- 1=0, t= 1.

Substitute t= 1 into x=4t+2=6 and Y = 3t 2-6t+5 = 2.

The minimum position is (6,2)