# 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=2 ?

May 3, 2016

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

${a}_{x} \left(t\right) = {\dot{v}}_{x} \left(t\right) = 2 t - 1$

$\therefore {a}_{x} \left(2\right) = 3$

${v}_{y} \left(t\right) = {t}^{3} - 3 t$
${a}_{y} \left(t\right) = {\dot{v}}_{y} \left(t\right) = 3 {t}^{2} - 3$
$\therefore {a}_{y} \left(2\right) = 9$

Hence, $| a | = \sqrt{{3}^{2} + {9}^{2}} = \sqrt{90} = 3 \sqrt{10}$

And direction is given as:
$\tan \theta = \frac{9}{2}$