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=7 #?

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Answer 1 · 2017-03-25T10:47:13

The rate of acceleration is #=16.28ms^-2#
in the direction #=169.4#º

The velocity is

#v(t)=(t^2+2t,cos(pit)-3t)#

Acceleration is the first derivative of the velocity

#a(t)=(dv(t))/dt=(2t+2,-pisin(pit)-3)#

When #t=7#

#a(7)=(dv(7))/dt=(2*7+2,-pisin(7pi)-3)#

#=(16,-3)#

The object rate of acceleration is

#=sqrt((16^2)+(-3)^2)#

#=16.28ms^-2#

The direction is

#=arctan(-3/16)#

The angle is in the 2nd quadrant and is

#=169.4#º