An object's two dimensional velocity is given by #v(t) = ( sqrt(t^2-1)-2t , t^2-5)#. What is the object's rate and direction of acceleration at #t=6 #?

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Answer 1 · 2016-06-30T09:24:20

#a(6)=11.96#

#theta=-85.33 ^o#

#"Horizontal component of acceleration :"#
#a_x(t)=d/(d t)(sqrt(t^2-1)-2t)#

#a_x(t)=(2t)/(2*sqrt(t^2-1))-2#

#a_x(6)=(2*6)/(2*sqrt(6^2-1))-2#

#a_x(6)=6/sqrt35-2#

#a_x(6)=(6-2*sqrt35)/sqrt35=(-5.83)/5.92=-0.98#

#"Vertical component of acceleration :"#

#a_y(t)=d/(d t) (t^2-5)#

#a_y(t)=2t#

#a_y(6)=2.6=12#

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

#a(6)sqrt((-0.98)^2+12^2)#

#a(6)=sqrt(0.96+144)#

#a(6)=11.96#

#tan theta=12/(-0.98)#

#Tan theta=-12.24#

#theta=-85.33 ^o#