# An object's two dimensional velocity is given by v(t) = ( t^2 +2t , cospit - 3t ). What is the object's rate and direction of acceleration at t=7 ?

Mar 25, 2017

The rate of acceleration is $= 16.28 m {s}^{-} 2$
in the direction $= 169.4$º

#### Explanation:

The velocity is

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

Acceleration is the first derivative of the velocity

$a \left(t\right) = \frac{\mathrm{dv} \left(t\right)}{\mathrm{dt}} = \left(2 t + 2 , - \pi \sin \left(\pi t\right) - 3\right)$

When $t = 7$

$a \left(7\right) = \frac{\mathrm{dv} \left(7\right)}{\mathrm{dt}} = \left(2 \cdot 7 + 2 , - \pi \sin \left(7 \pi\right) - 3\right)$

$= \left(16 , - 3\right)$

The object rate of acceleration is

$= \sqrt{\left({16}^{2}\right) + {\left(- 3\right)}^{2}}$

$= 16.28 m {s}^{-} 2$

The direction is

$= \arctan \left(- \frac{3}{16}\right)$

The angle is in the 2nd quadrant and is

$= 169.4$º