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

Apr 27, 2017

The rate of acceleration is $= 4 m {s}^{-} 2$ in the direction of =73.6º

#### Explanation:

Acceleration is the derivative of the velocity

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

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

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

When $t = 2$,

$a \left(2\right) = \left(2.65 , 9\right)$

The rate of acceleration is

$| | a \left(2\right) | | = \sqrt{{\left(2.65\right)}^{2} + {9}^{2}}$

$= \sqrt{16} = 4$

The direction is

theta=arctan(9/2.65)=73.6º