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

Apr 20, 2017

The rate of acceleration is $= 5.1 m {s}^{-} 2$ in the direction of =11.3º

#### Explanation:

The acceleration is the derivative of the velocity.

$a \left(t\right) = v ' \left(t\right)$

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

$a \left(t\right) = < 1 + \frac{1}{t} ^ 2 , 2 t >$

When $t = \frac{1}{2}$

$a \left(\frac{1}{2}\right) = < 1 + \frac{1}{\frac{1}{2}} ^ 2 , 2 \cdot \frac{1}{2} >$

$= < 5 , 1 >$

The rate of acceleration is

$= | | a \left(\frac{1}{2}\right) | |$

$= | | < 5 , 1 > | |$

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

$= 5.1 m {s}^{-} 2$

The direction is in the 1st quadrant

$\theta = \arctan \left(\frac{1}{5}\right)$

=11.3º