An object's velocity is given by #v(t) = (2t^2 +t +1 , sin2t )#. What is the object's rate and direction of acceleration at #t=3 #?

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Answer 1 · 2017-04-07T10:07:06

The rate of acceleration is #=13.04ms^-2# in the direction of #=4.4º#

The acceleration is the derivative of the velocity.

#v(t)=(2t^2+t+1,sin2t)#

#a(t)=v'(t)=(4t+1,2cos2t)#

Therefore,

#a(3)=(13,2cos6)=(13,0.96)#

The rate of acceleration is

#||a(3)||=sqrt(13^2+0.96^2)=13.04ms^-2#

The direction is

#theta=arctan(0.96/13)=4.4º#