# An object's two dimensional velocity is given by v(t) = ( t^2 +2 , cospit - t ). What is the object's rate and direction of acceleration at t=7 ?

Jul 3, 2018

The rate of acceleration is $= 14.04 m {s}^{-} 2$ in the direction ${4.09}^{\circ}$ clockwise from the x-axis

#### Explanation:

The acceleration is the derivative of the velocity

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

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

Therefore, when $t = 7$

$a \left(7\right) = \left(14 , - \pi \sin \left(7 \pi\right) - 1\right)$

$= \left(14 , - 1\right)$

So, the rate of acceleration is

$| | a \left(t\right) | | = \sqrt{{\left(14\right)}^{2} + {\left(- 1\right)}^{2}}$

$= 14.04 m {s}^{-} 2$

The direction is

$\theta = \arctan \left(- \frac{1}{14}\right) = {4.09}^{\circ}$ clockwise from the x-axis