An object's two dimensional velocity is given by #v(t) = ( t^2 +2t , cospit - 3t )#. What is the object's rate and direction of acceleration at #t=1 #?

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Answer 1 · 2017-12-25T19:44:27

#a = 5 #
#theta = - 37^o " or " 37"^o# SE

#a(t) =(dv(t))/(dt) =d/(dt)(t^2+2t, cos pit - 3t) = ( 2t+2, -pi sin pit-3)#

a(1) = (4, -3)
The magnitude of the acceleration is
#||a(1)|| = sqrt(4^2 + (-3)^2)= 5 #

The direction of the acceleration is determined from

#tantheta = (v_y)/(v_x) #
#theta = tan^-1((v_y)/(v_x) )= tan^-1(-3/4)=-37^o#
or

# theta = 37^o# SE