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 · 2017-06-01T12:36:53

The rate of acceleration is #=1.71ms^-2# in the direction #35.7º# clockwise 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)=(1*sin(1/3pit)+t*1/3picos(1/3pit), -2*1/2pisin(1/2pit)-1)#

#=(sin(1/3pit)+1/3pitcos(1/3pit),-pisin(1/2pit)-1)#

Therefore,

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

#=(1.39,-1)#

The rate of acceleration is

#||a(8)||=sqrt(1.39^2+1^2)#

#=sqrt(2.93)#

#=1.71ms^-2#

The direction is

#theta=arctan(-1/1.39)=-35.7º#