# An object's two dimensional velocity is given by v(t) = ( sqrt(t-2)-t , t^2). What is the object's rate and direction of acceleration at t=7 ?

Apr 27, 2016

$a {\left(7\right)}_{x} = - 0 , 76$
$a {\left(7\right)}_{y} = 2 \cdot 7 = 14$
$a \left(7\right) = 14 , 02$
$\alpha = 86 , {89}^{o}$

#### Explanation:

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

$a {\left(7\right)}_{x} = \frac{1}{2 \cdot \sqrt{7 - 2}} - 1$

$a {\left(7\right)}_{x} = \frac{1}{2 \cdot \sqrt{5}} - 1$

$a {\left(7\right)}_{x} = \frac{1 - 2 \cdot \sqrt{5}}{2 \cdot \sqrt{5}}$

$a {\left(7\right)}_{x} = \frac{\left(1 - 2 \cdot \sqrt{5}\right) \cdot \left(2 \cdot \sqrt{5}\right)}{\left(2 \cdot \sqrt{5}\right) \cdot \left(2 \cdot \sqrt{5}\right)}$

$a {\left(7\right)}_{x} = \frac{2 \cdot \sqrt{5} - 20}{20}$

$a {\left(7\right)}_{x} = \frac{\sqrt{5} - 10}{10} = \frac{2 , 37 - 10}{10} = - 0 , 76$

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

$a {\left(7\right)}_{y} = 2 \cdot 7 = 14$

$a \left(7\right) = \sqrt{{\left(a {\left(x\right)}_{7}\right)}^{2} + {\left(a {\left(7\right)}_{y}\right)}^{2}}$

a(7)=sqrt(((-0,76)^2+14^2)

$a \left(7\right) = \sqrt{0 , 58 + 196}$

$a \left(7\right) = 14 , 02$

$\tan \alpha = \frac{a {\left(7\right)}_{y}}{a {\left(7\right)}_{x}}$

$\tan \alpha = \frac{14}{- 0 , 76}$

$\tan \alpha = - 18 , 42$

$\alpha = 86 , {89}^{o}$