# An object's two dimensional velocity is given by v(t) = ( e^t-2t , t-4e^2 ). What is the object's rate and direction of acceleration at t=1 ?

Jun 29, 2017

The rate of acceleration is $= 1.23 m {s}^{-} 2$ in the direction =54.2º anticlockwise from the x-axis

#### Explanation:

The acceleration is the derivative of the velocity

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

$a \left(t\right) = v ' \left(t\right) = \left({e}^{t} - 2 , 1\right)$

So,

when $t = 1$

$a \left(1\right) = \left(e - 2 , 1\right) = \left(0.72 , 1\right)$

The rate of acceleration is

$= | | a \left(1\right) | | = \sqrt{{0.72}^{2} + {\left(1\right)}^{2}} = \sqrt{1.52} = 1.23$

The direction is arctan(1/0.72)=54.2º anticlockwise from the x-axis