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

Jan 10, 2018

The rate of acceleration is $= 3.01 m {s}^{-} 2$ in the direction $= {266.6}^{\circ}$ anticlockwise from the $\text{x-axis}$

#### Explanation:

The acceleration is the derivative of the velocity.

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

Therefore,

$a \left(t\right) = v ' \left(t\right) = \left(\sin \left(\frac{\pi}{3} t\right) + \frac{\pi}{3} t \cos \left(\frac{\pi}{3} t\right) , - \pi \sin \left(\frac{\pi}{2} t\right) - 3\right)$

When $t = 2$

The acceleration is

$a \left(2\right) = v ' \left(2\right) = \left(\sin \left(\frac{\pi}{3} \cdot 2\right) + \frac{\pi}{3} \cdot 2 \cdot \cos \left(\frac{\pi}{3} \cdot 2\right) , - \pi \sin \left(\frac{\pi}{2} \cdot 2\right) - 3\right)$

$a \left(2\right) = \left(\frac{\sqrt{3}}{2} - \frac{\pi}{3} , - 3\right) = \left(- 0.18 , - 3\right)$

The rate of acceleration is

$= | | a \left(2\right) | | = \sqrt{{\left(- 0.18\right)}^{2} + {\left(- 3\right)}^{2}} = 3.01 m {s}^{-} 2$

The direction is $\theta = 180 + \arctan \left(\frac{3}{0.18}\right) = {266.6}^{\circ}$ anticlockwise from the $\text{x-axis}$