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

Apr 26, 2017

The rate of acceleration is $= 27.5$ in the direction of =146.9º

#### Explanation:

Acceleration is the derivative of the velocity

$a \left(t\right) = v ' \left(t\right)$

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

$a \left(t\right) = v ' \left(t\right) = \left(1 - 6 {t}^{2} , 8 t - 1\right)$

When $t = 2$,

$a \left(2\right) = \left(- 23 , 15\right)$

The rate of acceleration is

$| | a \left(2\right) | | = \sqrt{{\left(- 23\right)}^{2} + {15}^{2}}$

$= \sqrt{754} = 27.5$

The direction is

theta=180-arctan(15/23)=146.9º