# An object's two dimensional velocity is given by v(t) = ( t^3, t-t^2sin(pi/8)t). What is the object's rate and direction of acceleration at t=3 ?

Apr 1, 2017

The rate of acceleration is $= 27.63 m {s}^{-} 1$ in the direction of $= 167.69$º

#### Explanation:

The velocity is

$v \left(t\right) = \left({t}^{3} , t - {t}^{2} \sin \left(\left(\frac{\pi}{8}\right) t\right)\right)$

Acceleration is the first derivative of the velocity

$a \left(t\right) = \frac{\mathrm{dv} \left(t\right)}{\mathrm{dt}} = \left(3 {t}^{2} , 1 - 2 t \sin \left(\frac{\pi}{8}\right) t - {t}^{2} \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right) t\right)$

When $t = 3$

$a \left(3\right) = \frac{\mathrm{dv} \left(3\right)}{\mathrm{dt}} = \left(27 , 1 - 6 \sin \left(\frac{3}{8} \pi\right) - \left(\frac{9}{8} \pi\right) \cos \left(\frac{3}{8} \pi\right)\right)$

$= \left(27 , 1 - 5.54 - 1.35\right)$

$= \left(27 , - 5.89\right)$

The object rate of acceleration is

$= \sqrt{\left({27}^{2}\right) + {\left(- 5.89\right)}^{2}}$

$= 27.63 m {s}^{-} 2$

The direction is

$= \arctan \left(- \frac{5.89}{27}\right)$

The angle is in the 2nd quadrant and is

$= 167.69$º