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

May 6, 2017

The rate of acceleration is $= 15.01 m {s}^{-} 2$ in the direction of =177.5º

#### Explanation:

The acceleration is the derivative of the velocity.

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

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

Therefore,

$a \left(4\right) = \left(15 , \cos 4\right) = \left(15 , - 0.65\right)$

The rate of acceleration is

$| | a \left(4\right) | | = \sqrt{{15}^{2} + {\left(0.65\right)}^{2}}$

$= \sqrt{225.4}$

$= 15.01 m {s}^{-} 2$

The direction is

$\theta = \arctan \left(- \frac{0.65}{15}\right)$

$\theta$ lies in the 2nd quadrant

theta=180-2.5=177.5º