An object's two dimensional velocity is given by #v(t) = ( 3t^2 - 2t , 1- 3t )#. What is the object's rate and direction of acceleration at #t=1 #?
We take the first derivative with respect to time to get the component (rectangular) form of the acceleration, then change this to standard (polar) form, to get the acceleration as
in a direction of 36.9° south of east.
Since the acceleration is the first derivative with respect to time, the acceleration must be
#a(t) = (6t-2, -3)
and at time
To change to standard form, we use the following operations:
The direction is given by
So, the acceleration is