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

Jan 25, 2017

Rate of acceleration is $1.0427$ and direction is ${62.364}^{\circ}$

#### Explanation:

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

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

at $t = 2$, $a \left(2\right) = \left(\frac{\pi}{3} \cos \left(\frac{2 \pi}{3}\right) , - \pi \sin \left(\pi\right) - 1\right) = \left(- \frac{\pi}{6} , - 1\right)$

Hence object's rate of acceleration is $| a \left(2\right) | = \sqrt{{\left(\frac{\pi}{6}\right)}^{2} + {\left(- 1\right)}^{2}}$

= $\sqrt{\frac{\pi}{36} + 1} = \sqrt{1.0873} = 1.0427$

and direction is given by ${\tan}^{- 1} \left(\frac{- 1}{- \frac{\pi}{6}}\right) = {\tan}^{- 1} \left(\frac{6}{\pi}\right) = {62.364}^{\circ}$