An object's two dimensional velocity is given by #v(t) = ( sqrt(t-2)-2t , t^2)#. What is the object's rate and direction of acceleration at #t=7 #?

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Answer 1 · 2016-05-21T10:04:54

#a(7)=14.11" "m/s^2#
#tan alpha=-7.87#

#v(t)=(sqrt(t-2)-2t,t^2)#

#v_x(t)=sqrt(t-2)-2t#

#v_y(t)=t^2#

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

#a_x(7)=1/(2sqrt(7-2))-2=1/(2sqrt5)-2=(1-2(2sqrt5))/(2sqrt5)#

#a_x(7)=(1-4sqrt5)/(2sqrt5)=(2sqrt5-40)/20=(sqrt5-20)/10#

#a_x(7)=-1.78#

#a_y(t)=d/(d t)v_y(t)=d/(d t)(t^2)=2t#

#a_y(7)=2*7=14#

#a(7)=sqrt(a_x^2+a_y^2)#

#a(7)=sqrt((-1*78)^2+14^2#

#a(7)=14.11" "m/s^2#

#tan alpha=a_y/a_x#

#tan alpha=-14/(1.78)#

#tan alpha=-7.87#