# An object's two dimensional velocity is given by v(t) = ( 3t^2 - 5t , -t^3 +4t ). What is the object's rate and direction of acceleration at t=1 ?

May 4, 2016

$a \left(1\right) = \sqrt{2}$
$\alpha = {45}^{o}$

#### Explanation:

${a}_{x} \left(t\right) = \frac{d}{d t} {v}_{x} \left(t\right) \text{ ; } {a}_{y} \left(t\right) = \frac{d}{d t} {v}_{y} \left(t\right)$

${a}_{x} \left(t\right) = \frac{d}{d t} \left(3 {t}^{2} - 5 t\right) = 6 t - 5$

${a}_{x} \left(1\right) = 6 \cdot 1 - 5 = 1$

a_y(t)=d/(d t) (-t^³+4t)=-3t^2+4

${a}_{y} \left(1\right) = - 3 \cdot {1}^{2} + 4$

${a}_{y} \left(1\right) = - 3 + 4$

${a}_{y} \left(1\right) = 1$

$a \left(1\right) = \sqrt{{a}_{x} {\left(1\right)}^{2} + {a}_{y} {\left(1\right)}^{2}}$

$a \left(1\right) = \sqrt{{1}^{2} + {1}^{2}}$

$a \left(1\right) = \sqrt{2}$

$\tan \alpha = \frac{{a}_{y} \left(1\right)}{{a}_{x} \left(1\right)} = \frac{1}{1} = 1 \text{ ; } \alpha = {45}^{o}$