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=9 #?

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Answer 1 · 2017-02-11T12:18:22

#dot(v)(9)=(49,-239) ms^(-2)#

in direction #" "tan^(-1)(-239/49)#

acceleration is given by the first derivative of its velocity

so #v(t)=(3t^2-5t, -t^3+4t)#

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

#dot(v)(t)=(6t-5,-3t^2+4)#

#dot(v)(9)=(6xx9-5,-3xx9^2+4)#

#dot(v)(9)=(49,-239) ms^(-2)#

direction given by#" "tan^(-1)(y/x)#

#" "tan^(-1)(-239/49)#