An object's two dimensional velocity is given by #v(t) = ( t^2 +2t , cospit - 3t )#. What is the object's rate and direction of acceleration at #t=5 #?

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Answer 1 · 2016-04-21T13:20:41

#a(5)=12,37#
#theta=14,04 ^o#

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#a(t)=d/ (d t)v(t)#

#a_x(t)=d/(d t)(t^2+2t)=2t+2" "a_x(5)=2*5+2=12#

#a_y(t)=d/(d t)(cos pi t-3t)=-pi sinpit-3#

#a_y(5)=-pi sin5pi-3#

#sin 5 pi=0#

#a_y(5)=-3#

#a(5)=sqrt(a_x^2+a_y^2)#

#a(5)=sqrt(12^2+(-3)^2)#

#a(5)=sqrt(144+9)#

#a(5)=sqrt153#

#a(5)=12,37#

#tan theta=-3/12#

#tan theta=-0,25#

#theta=14,04 ^o#