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 #?

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Answer 1 · 2017-01-25T07:42:00

Rate of acceleration is #1.0427# and direction is #62.364^@#

As #v(t)=(sin(pi/3t),2cos(pi/2t)-t)#,

#a(t)=d/(dt)v(t)=(pi/3cos(pi/3t),-2xxpi/2xxsin(pi/2t)-1)#

at #t=2#, #a(2)=(pi/3cos((2pi)/3),-pisin(pi)-1)=(-pi/6,-1)#

Hence object's rate of acceleration is #|a(2)|=sqrt((pi/6)^2+(-1)^2)#

= #sqrt(pi/36+1)=sqrt(1.0873)=1.0427#

and direction is given by #tan^(-1)((-1)/(-pi/6))=tan^(-1)(6/pi)=62.364^@#