An object's two dimensional velocity is given by #v(t) = ( t^2 +2 , cospit - t )#. What is the object's rate and direction of acceleration at #t=7 #?

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Answer 1 · 2018-07-03T07:23:47

The rate of acceleration is #=14.04ms^-2# in the direction #4.09^@# clockwise from the x-axis

The acceleration is the derivative of the velocity

#v(t)=(t^2+2, cos(pit)-t)#

#a(t)=v'(t)=(2t, -pisin(pit)-1)#

Therefore, when #t=7#

#a(7)=(14, -pisin(7pi)-1)#

#=(14, -1)#

So, the rate of acceleration is

#||a(t)||=sqrt((14)^2+(-1)^2)#

#=14.04ms^-2#

The direction is

#theta=arctan(-1/14)=4.09^@# clockwise from the x-axis