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 #?

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Answer 1 · 2017-06-16T08:06:11

The rate of acceleration is #=3.47ms^-2# in the direction #=196.8º# anticlockwise from the x-axis

The acceleration is the derivative of the velocity

#v(t)=(tsin(1/3pit),2cos(1/2pit)-t)#

#a(t)=v'(t)=(sin(1/3pit)+1/3pitcos(1/3pit),-pisin(1/2pit)-1)#

So,

when #t=8#

#a(8)=(sin(8/3pi)+8/3picos(8/3pi),-pisin(4pi)-1)#

#=(-3.32,-1)#

The rate of acceleration is

#=||a(3)||=sqrt(3.32^2+(1)^2)=sqrt12.04=3.47#

The direction is #theta=180+arctan(1/3.32)=196.8º#