An object's two dimensional velocity is given by #v(t) = ( 3t^2 - 2t , 1- 3t )#. What is the object's rate and direction of acceleration at #t=4 #?

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Answer 1 · 2017-05-10T06:52:27

The rate of acceleration is #=22.2ms^-1# in the direction of #=352.2º#

The acceleration is the derivative of the velocity.

#v(t)=(3t^2-2t, 1-3t)#

#a(t)=v'(t)=(6t-2, -3)#

Therefore,

#a(4)=(24-2, -3)=(22,-3)#

The rate of acceleration is

#||a(4)||=sqrt(22^2+(-3)^2)#

#=sqrt493#

#=22.2ms^-2#

The direction is

#theta=arctan(-3/22)#

#theta# lies in the 4th quadrant

#theta=360-7.8=352.2º#