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

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Answer 1 · 2017-06-10T11:32:30

The rate of acceleration is #=11.04ms^-2# in the direction #=5.14º# clockwise from the x-axis

The acceleration is the derivative of the velocity

#v(t)=(2t^2-t+2,sint)#

#a(t)=v'(t)=(4t-1,cost)#

So,

when #t=3#

#a(3)=(11,cos3)=(11,-0.99)#

The rate of acceleration is

#=||a(3)||=sqrt(11^2+(-0.99)^2)=sqrt121.98=11.04#

The direction is #arctan(-0.99/11)=-5.14º#