# 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=5 ?

Apr 21, 2016

$a \left(5\right) = 12 , 37$
$\theta = 14 , {04}^{o}$

#### Explanation:

$a \left(t\right) = \frac{d}{d t} v \left(t\right)$

${a}_{x} \left(t\right) = \frac{d}{d t} \left({t}^{2} + 2 t\right) = 2 t + 2 \text{ } {a}_{x} \left(5\right) = 2 \cdot 5 + 2 = 12$

${a}_{y} \left(t\right) = \frac{d}{d t} \left(\cos \pi t - 3 t\right) = - \pi \sin \pi t - 3$

${a}_{y} \left(5\right) = - \pi \sin 5 \pi - 3$

$\sin 5 \pi = 0$

${a}_{y} \left(5\right) = - 3$

$a \left(5\right) = \sqrt{{a}_{x}^{2} + {a}_{y}^{2}}$

$a \left(5\right) = \sqrt{{12}^{2} + {\left(- 3\right)}^{2}}$

$a \left(5\right) = \sqrt{144 + 9}$

$a \left(5\right) = \sqrt{153}$

$a \left(5\right) = 12 , 37$

$\tan \theta = - \frac{3}{12}$

$\tan \theta = - 0 , 25$

$\theta = 14 , {04}^{o}$