# An object's two dimensional velocity is given by v(t) = ( sqrt(t-2)-t , t^2-5t). What is the object's rate and direction of acceleration at t=3 ?

Apr 21, 2016

$a \left(3\right) = \sqrt{\frac{3}{2}}$

$\theta = 54 , {74}^{o}$

#### Explanation:

$a \left(t\right) = \frac{d}{d t} v \left(t\right)$

${a}_{x} \left(t\right) = \frac{d}{d t} {v}_{x} \left(t\right)$

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

${a}_{x} \left(t\right) = \frac{1}{2 \sqrt{t - 1}} \text{ "a_x(3)=1/sqrt(3-1)" } {a}_{x} \left(3\right) = \frac{1}{\sqrt{2}}$

${a}_{y} \left(t\right) = \frac{d}{d t} {v}_{y} \left(t\right)$

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

${a}_{y} \left(t\right) = 2 t - 5 \text{ "a_y(3)=2*3-5" } {a}_{y} \left(3\right) = 6 - 5$

${a}_{y} \left(3\right) = 1$

a(3)=sqrt(a_x^2+a_y^2

$a \left(3\right) = \sqrt{{\left(\frac{1}{\sqrt{2}}\right)}^{2} + {1}^{2}}$

$a \left(3\right) = \sqrt{\frac{1}{2} + 1}$$\text{ ; } a \left(3\right) = \sqrt{\frac{3}{2}}$

$\tan \theta = \frac{{a}_{y} \left(3\right)}{{a}_{x} \left(3\right)}$

$\tan \theta = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}$

$\theta = 54 , {74}^{o}$