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

May 21, 2016

$a \left(7\right) = 14.11 \text{ } \frac{m}{s} ^ 2$
$\tan \alpha = - 7.87$

Explanation:

$v \left(t\right) = \left(\sqrt{t - 2} - 2 t , {t}^{2}\right)$

${v}_{x} \left(t\right) = \sqrt{t - 2} - 2 t$

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

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

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

${a}_{x} \left(7\right) = \frac{1 - 4 \sqrt{5}}{2 \sqrt{5}} = \frac{2 \sqrt{5} - 40}{20} = \frac{\sqrt{5} - 20}{10}$

${a}_{x} \left(7\right) = - 1.78$

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

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

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

a(7)=sqrt((-1*78)^2+14^2

$a \left(7\right) = 14.11 \text{ } \frac{m}{s} ^ 2$

$\tan \alpha = {a}_{y} / {a}_{x}$

$\tan \alpha = - \frac{14}{1.78}$

$\tan \alpha = - 7.87$