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

Apr 20, 2016

${v}_{x} \left(t\right) = 3 t - \sqrt{t}$

$\implies {a}_{x} \left(t\right) = {\dot{v}}_{x} \left(t\right) = 3 - \frac{1}{2 \sqrt{t}}$

$\therefore {a}_{x} \left(3\right) = 3 - \frac{1}{2 \sqrt{3}}$

${v}_{y} \left(t\right) = {t}^{3} - 3 t$

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

$\therefore {a}_{y} \left(3\right) = 24$

hence, the object's acceleration is given as $\sqrt{{a}_{x}^{2} + {a}_{y}^{2}}$

and its direction is given as $\tan \theta = {v}_{y} / {v}_{x}$