# An object's velocity is given by v(t) = (2t^2 +t +1 , sin2t ). What is the object's rate and direction of acceleration at t=3 ?

Apr 7, 2017

The rate of acceleration is $= 13.04 m {s}^{-} 2$ in the direction of =4.4º

#### Explanation:

The acceleration is the derivative of the velocity.

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

$a \left(t\right) = v ' \left(t\right) = \left(4 t + 1 , 2 \cos 2 t\right)$

Therefore,

$a \left(3\right) = \left(13 , 2 \cos 6\right) = \left(13 , 0.96\right)$

The rate of acceleration is

$| | a \left(3\right) | | = \sqrt{{13}^{2} + {0.96}^{2}} = 13.04 m {s}^{-} 2$

The direction is

theta=arctan(0.96/13)=4.4º