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

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Answer 1 · 2016-04-20T10:44:20

#v_x(t) =3t - sqrt t#

#=>a_x(t) = dotv_x(t) = 3-1/(2sqrtt)#

#:. a_x(3) = 3-1/(2sqrt3)#

#v_y(t) =t^3 - 3 t#

#=>a_y(t) = dotv_y(t) = 3t^2-3#

#:. a_y(3) = 24#

hence, the object's acceleration is given as #sqrt(a_x^2+a_y^2)#

and its direction is given as #tantheta = v_y/v_x#