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

Jan 9, 2018

The rate of acceleration is $= \sqrt{5} m {s}^{-} 2$ in the direction $= {153.4}^{\circ}$ anticlockwise from the $\text{x-axis}$

#### Explanation:

The acceleration is the derivative of the velocity

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

Therefore,

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

When $t = 1$, the acceleration is

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

The rate of acceleration is

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

The direction of acceleration is

$\theta = \arctan \left(- \frac{1}{2}\right) = {153.4}^{\circ}$ anticlockwise from the $\text{x-axis}$