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

Jun 16, 2017

The rate of acceleration is $= 3.47 m {s}^{-} 2$ in the direction =196.8º anticlockwise from the x-axis

#### Explanation:

The acceleration is the derivative of the velocity

$v \left(t\right) = \left(t \sin \left(\frac{1}{3} \pi t\right) , 2 \cos \left(\frac{1}{2} \pi t\right) - t\right)$

$a \left(t\right) = v ' \left(t\right) = \left(\sin \left(\frac{1}{3} \pi t\right) + \frac{1}{3} \pi t \cos \left(\frac{1}{3} \pi t\right) , - \pi \sin \left(\frac{1}{2} \pi t\right) - 1\right)$

So,

when $t = 8$

$a \left(8\right) = \left(\sin \left(\frac{8}{3} \pi\right) + \frac{8}{3} \pi \cos \left(\frac{8}{3} \pi\right) , - \pi \sin \left(4 \pi\right) - 1\right)$

$= \left(- 3.32 , - 1\right)$

The rate of acceleration is

$= | | a \left(3\right) | | = \sqrt{{3.32}^{2} + {\left(1\right)}^{2}} = \sqrt{12.04} = 3.47$

The direction is theta=180+arctan(1/3.32)=196.8º