# An object's two dimensional velocity is given by v(t) = ( t^2 - 2t , 1- 3t ). What is the object's rate and direction of acceleration at t=7 ?

Mar 13, 2018

The rate of acceleration is $= 12.37 m {s}^{-} 2$ in the direction $- {14.04}^{\circ}$ clockwise from the x-axis.

#### Explanation:

The acceleration is the derivative of the velocity

$v \left(t\right) = \left({t}^{2} - 2 t , 1 - 3 t\right)$

$a \left(t\right) = v ' \left(t\right) = \left(2 t - 2 , - 3\right)$

When $t = 7$

$a \left(7\right) = \left(2 \cdot 7 - 2 , - 3\right) = \left(12 , - 3\right)$

The rate of acceleration is

$| | a \left(7\right) | | = | | \left(12 , - 3\right) | | = \sqrt{{12}^{3} + {\left(- 3\right)}^{2}} = \sqrt{144 + 9} = \sqrt{153} = 12.37 m {s}^{-} 2$

The angle is

$\theta = \arctan \left(- \frac{3}{12}\right) = \arctan \left(- \frac{1}{4}\right) = - {14.04}^{\circ}$ clockwise from the x-axis.