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 1, 2017

The rate of acceleration is $= 1.71 m {s}^{-} 2$ in the direction 35.7º clockwise from the x-axis.

Explanation:

The acceleration is the derivative of the velocity.

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

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

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

Therefore,

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

$= \left(1.39 , - 1\right)$

The rate of acceleration is

$| | a \left(8\right) | | = \sqrt{{1.39}^{2} + {1}^{2}}$

$= \sqrt{2.93}$

$= 1.71 m {s}^{-} 2$

The direction is

theta=arctan(-1/1.39)=-35.7º