# 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=5 ?

Mar 7, 2018

#### Answer:

The rate of acceleration is $= 146.4 m {s}^{-} 1$ in the direction of ${0.4}^{\circ}$ anticlockwise from the x-axis

#### Explanation:

The acceleration is the derivative of the velocity

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

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

When $t = 5$

$a \left(5\right) = \left({e}^{5} - 2 , 1\right)$

$a \left(5\right) = \left(146.4 , 1\right)$

The rate of acceleration is

$| | a \left(5\right) | | = \sqrt{{146.4}^{2} + {1}^{2}} = 146.4 m {s}^{-} 1$

The angle is

$\theta = \arctan \left(\frac{1}{146.4}\right) = {0.4}^{\circ}$ anticlockwise from the x-axis

Mar 7, 2018

#### Answer:

Please see below.

#### Explanation:

As velocity is given by $v \left(t\right) = \left({e}^{t} - 2 t , t - 4 {e}^{2}\right)$

accelaration is $\frac{\mathrm{dv}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left({e}^{t} - 2 t\right) \hat{i} + \frac{d}{\mathrm{dt}} \left(t - 4 {e}^{2}\right) \hat{j}$

or $\left({e}^{t} - 2\right) \hat{i} + \hat{j}$

Note that $- 4 {e}^{2}$ is a constant hence its differential is $0$.

and at $t = 5$, accelaration is $\left({e}^{5} - 2\right) \hat{i} + \hat{j}$

Its magnitude is $\sqrt{{\left({e}^{5} - 2\right)}^{2} + 1} = \sqrt{{\left(148.4132 - 2\right)}^{2} + 1}$

= $146.417$

and direction is ${\tan}^{- 1} \left(\frac{1}{{e}^{5} - 2}\right) = {\tan}^{- 1} \left(\frac{1}{146.4132}\right)$

= ${0.39}^{\circ}$