# An object's two dimensional velocity is given by v(t) = ( cost , -t^3 +4t ). What is the object's rate and direction of acceleration at t=3 ?

Nov 21, 2017

The rate of acceleration is $= = 23 m {s}^{-} 2$ in the direction $= {269.7}^{\circ}$anticlockwise from the x-axis.

#### Explanation:

The acceleration is the derivative of the velocity

$v \left(t\right) = \left(\cos t , - {t}^{3} + 4 t\right)$

$a \left(t\right) = v ' \left(t\right) = \left(- \sin t , - 3 {t}^{2} + 4\right)$

When $t = 3$

$a \left(3\right) = \left(- \sin \left(3\right) , - {3}^{3} + 4\right) = \left(- 0.14 , - 23\right)$

The rate of acceleration is

$= | | a \left(3\right) | | = \sqrt{{\left(- 0.14\right)}^{2} + {\left(- 23\right)}^{2}} = 23 m {s}^{-} 2$

The direction is

$\theta = \arctan \left(- \frac{23}{-} 0.14\right) = {269.7}^{\circ}$ anticlockwise from the x-axis.

Nov 21, 2017

$23$ units and $4.71$ radian

#### Explanation:

As two dimensional velocity of object is given by $v \left(t\right) = \left(\cos t , - {t}^{3} + 4 t\right)$

accelaration is given by $\frac{\mathrm{dv}}{\mathrm{dt}}$ i.e.

$a \left(t\right) = \left(- \sin t , - 3 {t}^{2} + 4\right)$

and at $t = 3$, this is $- \sin 3 \hat{i} + \left(- 3 \cdot {3}^{2} + 4\right) \hat{j}$

or $a \left(t = 3\right) = - 0.14112 \hat{i} - 23 \hat{j}$

and its magnitude is $\sqrt{{\left(0.14112\right)}^{2} + {23}^{2}} = \sqrt{529.02} = 23$

and direction is $\pi + {\tan}^{- 1} \frac{23}{0.14112} = \pi + 1.569 = 4.71$ radian