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

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Answer 1 · 2017-05-21T07:12:59

The rate of acceleration is #=2.36ms^-2# in the direction
#theta=65º# anticlockwise from the x-axis

The acceleration is the derivative of the velocity.

#v(t)=(t-2, 2cos(pi/2t)-t)#

#a(t)=v'(t)=(1, -pisin(pi/2t)-1)#

Therefore,

#a(3)=(1, pi-1)=(1,2.14)#

The rate of acceleration is

#||a(4)||=sqrt(1+2.14^2)#

#=sqrt5.59#

#=2.36ms^-2#

The direction is

#theta=arctan((pi-1))#

#theta# lies in the 1st quadrant

#theta=65º# anticlockwise from the x-axis