# An object's two dimensional velocity is given by v(t) = ( t-2 , 5t^2-3t). What is the object's rate and direction of acceleration at t=2 ?

May 26, 2016

$a = 17.03 \text{ } \frac{m}{s} ^ 2$

$\tan \alpha = 17$

#### Explanation:

$v \left(t\right) = \left(t - 2 , 5 {t}^{2} - 3 t\right)$

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

${a}_{x} \left(t\right) = \frac{d}{d t} \left(t - 2\right) = 1 \text{ "a_x=1" constant}$

${a}_{y} \left(t\right) = \frac{d}{d t} \left(5 {t}^{2} - 3 t\right) = 10 t - 3$

${a}_{y} \left(2\right) = 10 \cdot 2 - 3$

${a}_{y} \left(2\right) = 17$

$a = \sqrt{{a}_{x}^{2} + {a}_{y}^{2}}$

$a = \sqrt{{1}^{2} + {17}^{2}}$

$a = 17.03 \text{ } \frac{m}{s} ^ 2$

$\tan \alpha = 17$