An object's two dimensional velocity is given by #v(t) = ( 3t^2 - 5t , -t^3 +4t )#. What is the object's rate and direction of acceleration at #t=4 #?

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Answer 1 · 2016-05-30T18:55:55

#a(4)=33.84" " m/s^2#

#a_x(t)=d/(d t) (3t^2-5t)=6t-5#

#a_x(4)=6*4-5=24-5=19#

#a_y(t)=d/(d t)(-t^3+4t)=-3t^2+4#

#a_y(4)=-2*4^2+4=-32+4=-28#

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#a(4)=sqrt(a_x(4)^2+a_y(4)^2#

#a(4)=sqrt(19^2+(-28)^2)#

#a(4)=sqrt(361+784)#

#a(4)=sqrt 1145#

#a(4)=33.84" " m/s^2#