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

Oct 27, 2017

Rate of accelaration is $3.606$ and direction is $- {56.31}^{\circ}$

#### Explanation:

Here velocity $v \left(t\right) = \left({t}^{2} + 2 , \cos \pi t - 3 t\right)$ is given in parametric form,

accelaration is $a \left(t\right) = \frac{\mathrm{dv} \left(t\right)}{\mathrm{dt}} = \left(\frac{d}{\mathrm{dt}} \left({t}^{2} + 2\right) , \frac{d}{\mathrm{dt}} \left(\cos \pi t - 3 t\right)\right)$

or $\left(2 t , - \pi \sin \pi t - 3\right)$

and at $t = 1$

$a \left(t\right) = \left(2 , - 3\right)$

This shows that $\vec{a} \left(t\right) = 2 \hat{i} - 3 \hat{j}$

its magnitude is $\sqrt{{2}^{2} + {3}^{2}} = \sqrt{4 + 9} = \sqrt{13} = 3.606$

and direction is given by ${\tan}^{- 1} \left(\frac{- 3}{2}\right) = {\tan}^{- 1} \left(- 1.5\right) = - {56.31}^{\circ}$