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

Jan 18, 2018

The rate of acceleration is $= \sqrt{5} m {s}^{-} 2$ in
the direction is $= {26.6}^{\circ}$ clockwise from the x-axis.

#### Explanation:

Acceleration is the derivative of the velocity

$v \left(t\right) = \left({t}^{2} - 2 t , \cos \pi t - t\right)$

$a \left(t\right) = v ' \left(t\right) = \left(2 t - 2 , - \pi \sin \pi t - 1\right)$

When $t = 2$

$a \left(2\right) = v ' \left(2\right) = \left(2 \cdot 2 - 2 , - \pi \sin \pi \cdot 2 - 1\right)$

$= \left(2 , - 1\right)$

The rate of acceleration is

$| | a \left(2\right) | | = | | \left(2 , - 1\right) | | = \sqrt{{2}^{2} + {\left(- 1\right)}^{2}} = \sqrt{5} m {s}^{-} 2$

The direction is

$\theta = \arctan \left(- \frac{1}{2}\right) = {26.6}^{\circ}$ clockwise from the x-axis.