# 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=5 ?

Aug 21, 2017

The rate of acceleration is $= 21.07 m {s}^{-} 2$ in the direction of $= {4.6}^{\circ}$ clockwise from the x-axis

#### Explanation:

The acceleration is the derivative of the velocity

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

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

When $t = 5$

$a \left(5\right) = \left(21 , 2 \cos 10\right) = \left(21 , - 1.68\right)$

The rate of acceleration is

$| | a \left(5\right) | | = \sqrt{{21}^{2} + {\left(- 1.68\right)}^{2}} = \sqrt{443.8} = 21.07$

The direction is

$\theta = \arctan \left(- \frac{1.68}{21}\right) = {4.6}^{\circ}$ clockwise from the x-axis