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

May 24, 2017

The rate of acceleration is $= \sqrt{{e}^{2 a} - 4 {e}^{a} + 8} m {s}^{-} 2$ in the direction of $\arctan \left(\frac{2}{{e}^{a} - 2}\right)$

#### Explanation:

The acceleration is the derivative of the velocity.

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

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

Therefore,

$A \left(a\right) = \left({e}^{a} - 2 , 2\right)$

The rate of acceleration is

$| | a \left(4\right) | | = \sqrt{{\left({e}^{a} - 2\right)}^{2} + {2}^{2}}$

$= \sqrt{{e}^{2 a} - 4 {e}^{a} + 4 + 4}$

$= \sqrt{{e}^{2 a} - 4 {e}^{a} + 8} m {s}^{-} 2$

The direction is

$\theta = \arctan \left(\frac{2}{{e}^{a} - 2}\right)$