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

Apr 23, 2016

$\vec{a} = - 0 , 85 \hat{i} + 4 \hat{j}$
$\theta = - {78}^{o}$

#### Explanation:

$\vec{a} \left(t\right) = \vec{a} \hat{i} + \vec{a} \hat{j}$

$\vec{a} \left(t\right) = \frac{d}{d t} \left(\sqrt{{t}^{2} - 1} - 2 t\right) \hat{i} + \frac{d}{d t} \left({t}^{2}\right) \hat{j}$

$\vec{a} \left(t\right) = \left(\frac{2 t}{2 \cdot \sqrt{{t}^{2} - 1}} - 2\right) \hat{i} + \left(2 t\right) \hat{j}$

$\vec{a} \left(2\right) = \left(\frac{2 \cdot 2}{2 \cdot \sqrt{{2}^{2} - 1}} - 2\right) \hat{i} + \left(2 \cdot 2\right) \hat{j}$

$\vec{a} \left(2\right) = \left(\frac{2}{\sqrt{3}} - 2\right) \hat{i} + 4 \hat{j}$

$\vec{a} \left(2\right) = \left(\frac{2 - 2 \sqrt{3}}{\sqrt{3}}\right) \hat{i} + 4 \hat{j}$

$\vec{a} = \left(\frac{2 \sqrt{3} - 6}{3}\right) \hat{i} + 4 \hat{j}$

$\vec{a} = - 0 , 85 \hat{i} + 4 \hat{j}$

$| | \vec{a} | | = \sqrt{{\left(- 0 , 85\right)}^{2} + {4}^{2}} \text{ magnitude of a}$

 a=4,09 " "("unit")/s^2

$\tan \theta = \frac{4}{- 0 , 85}$

$\theta = - {78}^{o}$