# An object's two dimensional velocity is given by v(t) = ( t^2 +2t , cospit - 3t ). What is the object's rate and direction of acceleration at t=1 ?

Dec 25, 2017

$a = 5$
$\theta = - {37}^{o} {\text{ or " 37}}^{o}$ SE

#### Explanation:

$a \left(t\right) = \frac{\mathrm{dv} \left(t\right)}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left({t}^{2} + 2 t , \cos \pi t - 3 t\right) = \left(2 t + 2 , - \pi \sin \pi t - 3\right)$

a(1) = (4, -3)
The magnitude of the acceleration is
$| | a \left(1\right) | | = \sqrt{{4}^{2} + {\left(- 3\right)}^{2}} = 5$

The direction of the acceleration is determined from

$\tan \theta = \frac{{v}_{y}}{{v}_{x}}$
$\theta = {\tan}^{-} 1 \left(\frac{{v}_{y}}{{v}_{x}}\right) = {\tan}^{-} 1 \left(- \frac{3}{4}\right) = - {37}^{o}$
or

$\theta = {37}^{o}$ SE