# 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=5 ?

Mar 13, 2017

Given that object's two dimensional velocity is
v(t)=(tsin(π/3t),2cos(π/2t)−3t).

Acceleration a(t)=dotv(t)=d/dt(tsin(π/3t),2cos(π/2t)−3t)
=>a(t)=(tpi/3cos(π/3t)+sin(pi/3t),-2pi/2sin(π/2t)−3)

Required acceleration at $t = 5$ is
a(5)=((5pi)/3cos((5π)/3 )+sin((5pi)/3 ),-pisin((5π)/2 )−3)
$\implies a \left(5\right) = \left(\frac{5 \pi}{3} \times \frac{1}{2} - \frac{\sqrt{3}}{2} , - \left(\pi + 3\right)\right)$
$\implies a \left(5\right) = \left(1.75197 , - 6.14159\right)$

$| a \left(5\right) | = \sqrt{{\left(1.75197\right)}^{2} + {\left(- 6.14159\right)}^{2}}$
$| a \left(5\right) | = 6.3866 \text{ units}$

Direction is given by
$\alpha = {\tan}^{-} 1 \left(- \frac{6.14159}{1.75197}\right) = - {74.08}^{\circ}$