# An object's two dimensional velocity is given by v(t) = ( sqrt(3t)-t , t^2-5t). What is the object's rate and direction of acceleration at t=2 ?

Apr 14, 2017

The rate of acceleration is $= 1.07 m {s}^{-} 2$ in the direction =248.7º

#### Explanation:

The acceleration is the derivative of the velocity.

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

$v \left(t\right) = < \sqrt{3 t} - t , {t}^{2} - 5 t >$

$a \left(t\right) = < \sqrt{3} \cdot \frac{1}{2 \sqrt{t}} - 1 , 2 t - 5 >$

When $t = 2$

$a \left(2\right) = < \sqrt{3} \cdot \frac{1}{2 \sqrt{2}} - 1 , 2 \cdot 2 - 5 >$

$= < - 0.39 , - 1 >$

The rate of acceleration is

$= | | a \left(2\right) | |$

$= | | < - 0.39 , - 1 > | |$

$= \sqrt{{\left(0.39\right)}^{2} + {\left(1\right)}^{2}}$

$= 1.07 m {s}^{-} 2$

The direction is in the 3rd quadrant

$\theta = 180 + \arctan \left(\frac{1}{0.39}\right)$

=248.7º