# An object's two dimensional velocity is given by v(t) = ( sqrt(t^2-1)-2t , t^2-5). What is the object's rate and direction of acceleration at t=6 ?

Jun 30, 2016

$a \left(6\right) = 11.96$

$\theta = - {85.33}^{o}$

#### Explanation:

$\text{Horizontal component of acceleration :}$
${a}_{x} \left(t\right) = \frac{d}{d t} \left(\sqrt{{t}^{2} - 1} - 2 t\right)$

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

${a}_{x} \left(6\right) = \frac{2 \cdot 6}{2 \cdot \sqrt{{6}^{2} - 1}} - 2$

${a}_{x} \left(6\right) = \frac{6}{\sqrt{35}} - 2$

${a}_{x} \left(6\right) = \frac{6 - 2 \cdot \sqrt{35}}{\sqrt{35}} = \frac{- 5.83}{5.92} = - 0.98$

$\text{Vertical component of acceleration :}$

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

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

${a}_{y} \left(6\right) = 2.6 = 12$

$a \left(6\right) = \sqrt{{a}_{x} {\left(6\right)}^{2} + {a}_{y} {\left(6\right)}^{2}}$

$a \left(6\right) \sqrt{{\left(- 0.98\right)}^{2} + {12}^{2}}$

$a \left(6\right) = \sqrt{0.96 + 144}$

$a \left(6\right) = 11.96$

$\tan \theta = \frac{12}{- 0.98}$

$T a n \theta = - 12.24$

$\theta = - {85.33}^{o}$