# An object's two dimensional velocity is given by v(t) = ( e^t-2t^2 , t^3 - 4t ). What is the object's rate and direction of acceleration at t=7 ?

Oct 16, 2017

The rate of acceleration is $= 1078.1 m {s}^{-} 2$ in the direction of ${7.62}^{\circ}$ anticlockwise from the x-axis

#### Explanation:

Acceleration is the derivative of the velocity

The velocity is

$v \left(t\right) = \left({e}^{t} - 2 {t}^{2} , {t}^{3} - 4 t\right)$

The acceleration is

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

Therefore, when $t = 7$

$a \left(7\right) = \left({e}^{7} - 28 , 3 \cdot 49 - 4\right) = \left(1068.6 , 143\right)$

The rate of acceleration is

$| | a \left(7\right) | | = \sqrt{{1068.6}^{2} + {143}^{2}} = 1078.1 m {s}^{-} 2$

The direction is

$\theta = \arctan \left(\frac{143}{1068.6}\right) = {7.62}^{\circ}$ anticlockwise from the x-axis