# An object's two dimensional velocity is given by v(t) = ( sin(pi/3t) , 2cos(pi/2t )- 3t ). What is the object's rate and direction of acceleration at t=1 ?

Sep 17, 2017

$a = \sqrt{{\pi}^{2} / 36 + 16}$

#### Explanation:

${v}_{x} \left(t\right) = \sin \left(\frac{\pi}{3} t\right)$ and ${v}_{y} \left(t\right) = 2 \cos \left(\frac{\pi}{2} t\right) - 3 t$
acceleration in X-direction
${a}_{x} \left(t\right) = \frac{d}{\mathrm{dt}} \left({v}_{x} \left(t\right)\right) = \frac{d}{\mathrm{dt}} \left(\sin \left(\frac{\pi}{3} t\right)\right) = \frac{\pi}{3} \cos \left(\frac{\pi}{3} t\right)$

acceleration in Y-direction
${a}_{y} \left(t\right) = \frac{d}{\mathrm{dt}} \left({v}_{y} \left(t\right)\right) = \frac{d}{\mathrm{dt}} \left(2 \cos \left(\frac{\pi}{2} t\right) - 3 t\right)$
${a}_{y} \left(t\right) = - 2 \left(\frac{\pi}{2}\right) \sin \left(\frac{\pi}{2} t\right) - 3 = - \pi \sin \left(\frac{\pi}{2} t\right) - 3$

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

Hence Acceleration a=sqrt((a_x(t))^2+(a_y(t))^2
$a = \sqrt{{\left(\frac{\pi}{6}\right)}^{2} + {\left(- \pi - 3\right)}^{2}} = \sqrt{{\pi}^{2} / 36 + {\pi}^{2} + 6 \pi + 9} \approx 6.1638$

direction is ${\tan}^{-} 1 \left({a}_{y} / {a}_{x}\right) = {\tan}^{-} 1 \left(\frac{- \pi - 3}{\frac{\pi}{6}}\right) \approx - {85.13}^{o}$ from Positive X- axis