# An object's two dimensional velocity is given by v(t) = (3t-sqrtt , 2t-4 ). What is the object's rate and direction of acceleration at t=a ?

Mar 12, 2017

The rate of acceleration is $= \sqrt{13 - \frac{3}{\sqrt{a}} + \frac{1}{4 a}}$ in the direction $= \arctan \left(\frac{4 \sqrt{a}}{6 \sqrt{a} - 1}\right)$

#### Explanation:

$v \left(t\right) = \left(3 t - \sqrt{t} , 2 t - 4\right)$

The acceleration is the derivative of the velocity

$v ' \left(t\right) = \left(3 - \frac{1}{2 \sqrt{t}} , 2\right)$

Therefore,

$v ' \left(a\right) = \left(3 - \frac{1}{2 \sqrt{a}} , 2\right)$

The rate of acceleration is

$| | v ' \left(a\right) | | = \sqrt{{\left(3 - \frac{1}{2 \sqrt{a}}\right)}^{2} + 4}$

$= \sqrt{9 - \frac{3}{\sqrt{a}} + \frac{1}{4 a} + 4}$

$= \sqrt{13 - \frac{3}{\sqrt{a}} + \frac{1}{4 a}}$

The direction is

$= \arctan \left(\frac{2}{3 - \frac{1}{2 \sqrt{a}}}\right)$

$= \arctan \left(\frac{4 \sqrt{a}}{6 \sqrt{a} - 1}\right)$