# 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 ?

Jan 30, 2017

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 $5 \frac{m}{s} ^ 2$
in a direction of 36.9° south of east.

#### Explanation:

Since the acceleration is the first derivative with respect to time, the acceleration must be

a(t) = (6t-2, -3)

and at time $t = 1$ this becomes

$a = \left(4 , - 3\right)$

To change to standard form, we use the following operations:

$| a | = \sqrt{{a}_{x}^{2} + {a}_{y}^{2}} = \sqrt{{4}^{2} + {\left(- 3\right)}^{3}} = 5 \frac{m}{s} ^ 2$

The direction is given by tan^-1 (a_y/a_x) = tan^-1(-3/4)=-36.9°#

So, the acceleration is $5 \frac{m}{s} ^ 2$ in a direction of 36.9° south of east.