# An object's two dimensional velocity is given by v(t) = ( 5t-t^3 , 4t^2-t). What is the object's rate and direction of acceleration at t=4 ?

Aug 22, 2017

The rate of acceleration is $= 53 m {s}^{-} 2$ in the direction $= {144.2}^{\circ}$ anticlockwise from the x-axis

#### Explanation:

The acceleration is the derivative of the velocity

$v \left(t\right) = \left(5 t - {t}^{3} , 4 {t}^{2} - t\right)$

$a \left(t\right) = v ' \left(t\right) = \left(5 - 3 {t}^{2} , 8 t - 1\right)$

When $t = 4$

$a \left(4\right) = \left(- 43 , 31\right)$

The rate of acceleration is

$| | a \left(4\right) | | = \sqrt{{\left(- 43\right)}^{2} + {\left(31\right)}^{2}} = \sqrt{2810} = 53$

The direction is

$\theta = \arctan \left(- \frac{31}{43}\right) = {144.2}^{\circ}$ anticlockwise from the x-axis