Kinematic Problem: Tangential Acceleration and Radius of Curvature at t=1

[Dell]

a body moves according to the following:
a_y=2-4t
a_x=-1
v_y=2t-2t²
v^x=1-t
x₀=2
y₀=3

what is the tangential acceleration when t=1?
what is the radius of curvature of the motion when t=1?

a_T=\vec{a}dot\vec{v}/|v|
at t=1, \vec{v}=0 so there is no tangential acceleration, how can this be true?

the secon part of the question is also not making sence

a_r=v²/r, but here v=0 and ar=3 since there is no a_T and ax=-1, ay=-2?

 

[LowlyPion]

I think what the question is asking is what is the linear acceleration at t=1 given by the expression of the acceleration vector at that time. Isn't this merely a matter of evaluating the a_(t) function at t=1?

For the radius of curvature, I would suggest:
http://www.intmath.com/Applications-differentiation/8_Radius_curvature.php