# An object's two dimensional velocity is given by v(t) = ( 1/t, t^2). What is the object's rate and direction of acceleration at t=6 ?

Jun 4, 2017

THe rate of acceleration is $= 12.00003 m {s}^{-} 2$ in the direction =90.13º anticlockwise from the x-axis

#### Explanation:

We need

$\left(\frac{1}{x}\right) ' = - \frac{1}{x} ^ 2$

$\left({x}^{2}\right) ' = 2 x$

The acceleration is the derivative of the velocity

$v \left(t\right) = \left(\frac{1}{t} , {t}^{2}\right)$

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

Therefore,

When $t = 6$

$a \left(6\right) = \left(- \frac{1}{36} , 12\right)$

The rate of acceleration is $= \sqrt{{\left(- \frac{1}{36}\right)}^{2} + {\left(12\right)}^{2}}$

$= \sqrt{144.0008}$

$= 12.00003$

The direction is =arctan(-12/(1/36))=180-89.9º=90.13 anticlockwise from the x-axis

The angle is in the 2nd quadrant