# An object's velocity is given by v(t) = (t^2 -t +1 , t^3- 3t ). What is the object's rate and direction of acceleration at t=6 ?

May 6, 2016

$a \left(6\right) = 105.57$
$\theta = {84.02}^{o}$

#### Explanation:

${a}_{x} \left(t\right) = \frac{d}{d t} = \left({t}^{2} - t + 1\right)$

${a}_{x} \left(t\right) = 2 t - 1 \text{ } {a}_{x} \left(6\right) = 2 \cdot 6 - 1 = 11$

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

${a}_{y} \left(t\right) = 3 {t}^{2} - 3 \text{ } {a}_{y} \left(6\right) = 3 \cdot {6}^{2} - 3 = 108 - 3 = 105$

a(6)=sqrt(a_x(6)^2+a_y(6)^2

$a \left(6\right) = \sqrt{{11}^{2} + {105}^{2}}$

$a \left(6\right) = \sqrt{121 + 11025}$

$a \left(6\right) = 105.57$

$\tan \theta = \frac{{a}_{y} \left(6\right)}{{a}_{x} \left(6\right)} = \frac{105}{11}$

$\theta = {84.02}^{o}$