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

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Answer 1 · 2017-04-26T05:45:46

The rate of acceleration is #=27.5# in the direction of #=146.9º#

Acceleration is the derivative of the velocity

#a(t)=v'(t)#

#v(t)=(t-2t^3,4t^2-t)#

#a(t)=v'(t)=(1-6t^2,8t-1)#

When #t=2#,

#a(2)=(-23,15)#

The rate of acceleration is

#||a(2)||=sqrt((-23)^2+15^2)#

#=sqrt754=27.5#

The direction is

in the second quadrant

#theta=180-arctan(15/23)=146.9º#