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 $" "tan^(-1)(y/x)$

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

An object's two dimensional velocity is given by v(t) = ( 3t^2 - 5t , -t^3 +4t )v(t) = ( 3t^2 - 5t , -t^3 +4t ). What is the object's rate and direction of acceleration at t=9 t=9 ?