An object's two dimensional velocity is given by #v(t) = ( t^2 +2 , cospit - 3t )#. What is the object's rate and direction of acceleration at #t=1 #?

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Answer 1 · 2017-10-27T05:49:35

Rate of accelaration is #3.606# and direction is #-56.31^@#

Here velocity #v(t)=(t^2+2,cospit-3t)# is given in parametric form,

accelaration is #a(t)=(dv(t))/(dt)=(d/(dt)(t^2+2),d/(dt)(cospit-3t))#

or #(2t,-pisinpit-3)#

and at #t=1#

#a(t)=(2,-3)#

This shows that #veca(t)=2hati-3hatj#

its magnitude is #sqrt(2^2+3^2)=sqrt(4+9)=sqrt(13)=3.606#

and direction is given by #tan^(-1)((-3)/2)=tan^(-1)(-1.5)=-56.31^@#