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

Jul 23, 2017

The rate of acceleration is $= 24.52 m {s}^{-} 2$ in the dorection $= {78.2}^{\circ}$ anticlockwise from the x-axis

#### Explanation:

The acceleration is the derivative of the velocity

$v \left(t\right) = \left({t}^{2} - t + 1 , {t}^{3} - 3 t\right)$

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

At $t = 3$, the acceleration is

$a \left(3\right) = \left(5 , 24\right)$

The rate of acceleration is

$= | | a \left(3\right) | | = \sqrt{{5}^{2} + {24}^{2}} = \sqrt{601} = 24.52 m {s}^{-} 2$

The direction is

$\theta = \arctan \left(\frac{24}{5}\right) = {78.2}^{\circ}$