# 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=3 ?

Sep 26, 2016

The vector is $a \left(t\right) = \left(\left(\frac{1}{2}\right) {\left(t - 2\right)}^{- \frac{1}{2}} - 1 , 2 t\right)$ Substitute 3 for t, and then compute the magnitude and direction as you would any vector.

#### Explanation:

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

$a \left(3\right) = \left(- \frac{1}{2} , 6\right)$

The rate of acceleration is the magnitude:

$| a \left(3\right) | = \sqrt{{\left(- \frac{1}{2}\right)}^{2} + {6}^{2}}$

$a \approx 6.02 \frac{u n i t s}{s} ^ 2$

$\theta = {\tan}^{-} 1 \left(\frac{6}{- \frac{1}{2}}\right)$

theta ~~ -85.2°