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

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Answer 1 · 2016-04-21T18:39:04

#a(3)=sqrt(3/2)#

#theta=54,74^o#

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

#a_x(t)=d/(d t)v_x(t)#

#a_x(t)=d/(d t)(sqrt(t-2))#

#a_x(t)=1/(2sqrt(t-1))" "a_x(3)=1/sqrt(3-1)" "a_x(3)=1/sqrt2#

#a_y(t)=d/(d t)v_y(t)#

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

#a_y(t)=2t-5" "a_y(3)=2*3-5" "a_y(3)=6-5#

#a_y(3)=1#

#a(3)=sqrt(a_x^2+a_y^2#

#a(3)=sqrt((1/sqrt2)^2+1^2)#

#a(3)=sqrt(1/2+1)##" ; "a(3)=sqrt(3/2)#

#tan theta=(a_y(3))/(a_x(3))#

#tan theta=1/(1/sqrt2)=sqrt2#

#theta=54,74^o#