An object's velocity is given by #v(t) = (t^2 -t +1 , t^3- 3t )#. What is the object's rate and direction of acceleration at #t=6 #?

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Answer 1 · 2016-05-06T21:35:26

#a(6)=105.57#
#theta=84.02^o#

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#a_x(t)=d/(d t)=(t^2-t+1)#

#a_x(t)=2t-1" "a_x(6)=2*6-1=11#

#a_y(t)=d/(d t)(t^3-3t)#

#a_y(t)=3t^2-3" "a_y(6)=3*6^2-3=108-3=105#

#a(6)=sqrt(a_x(6)^2+a_y(6)^2#

#a(6)=sqrt(11^2+105^2)#

#a(6)=sqrt(121+11025)#

#a(6)=105.57#

#tan theta=(a_y(6))/(a_x(6))=105/11#

#theta=84.02^o#