An object's two dimensional velocity is given by v(t) = ( t^2 +2 , cospit - 3t ). What is the object's rate and direction of acceleration at t=2 ?

$a \left(t\right) = \frac{\mathrm{dv}}{\mathrm{dt}} = \left(2 t , - \pi \sin \pi t - 3\right)$ at t=2 one finds
$a \left(2\right) = \left(4 , - 3\right)$ The magnitude f the acceleration vector is
$\sqrt{16 + 9} = 5$
Its direction is given by an angle theta such that $\tan \theta = - \frac{3}{4}$ or $\theta = - 37$ degrees.