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

Apr 29, 2017

The rate of acceleration is $= 5.04 m {s}^{-} 2$ in the direction $= 172.6$º

#### Explanation:

Acceleration is the derivative of the velocity

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

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

$a \left(t\right) = v ' \left(t\right) = \left(2 t - 3 , \cos t\right)$

When $t = 4$,

$a \left(4\right) = \left(2 \cdot 4 - 3 , \cos 4\right) = \left(5 , - 0.65\right)$

The rate of acceleration is

$| | a \left(4\right) | | = \sqrt{{\left(5\right)}^{2} + {\left(- 0.65\right)}^{2}}$

$= \sqrt{25.42} = 5.04$

The direction is in the 2nd quadrant

theta=180-arctan(0.65/5)=172.6º