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

Jan 25, 2017

Magnitude is $15.8115$ and direction is $\left(- {18.435}^{\circ}\right)$

#### Explanation:

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

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

= $\left(9 + 6 , 2 \times \left(- 1\right) - 3\right)$ = $\left(15 , - 5\right)$

Hence object's rate is $| v \left(3\right) | = \sqrt{{15}^{2} + {5}^{2}}$

= $\sqrt{250} = 5 \sqrt{10} = 5 \times 3.1623 = 15.8115$

and direction is given by ${\tan}^{- 1} \left(\frac{- 5}{15}\right) = {\tan}^{- 1} \left(\frac{- 1}{3}\right) = - {18.435}^{\circ}$