An object's two dimensional 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=5 #?

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Answer 1 · 2016-05-11T19:26:40

#a(5)=72.56#
#"angle="69.44^o#

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

#"for t="5 " ; "a_x(5)=2.5-1=10-1=9#

#a_x(5)=9#

#a_y(t)=d/(d t)(t^3-3t)=3t^2-3#

#"for t=5 ;"a_y(5)=3*5^2-3=75-3#

#a_y(5)=72#

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#a(5)=sqrt(72^2+9^2)" "a(5)=sqrt(5265)#

#a(5)=72.56#