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

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Answer 1 · 2018-01-18T07:32:51

The rate of acceleration is #=sqrt5ms^-2# in
the direction is #=26.6^@# clockwise from the x-axis.

Acceleration is the derivative of the velocity

#v(t)=(t^2-2t, cospit-t)#

#a(t)=v'(t)=(2t-2, -pisinpit-1)#

When #t=2#

#a(2)=v'(2)=(2*2-2, -pisinpi*2-1)#

#=(2,-1)#

The rate of acceleration is

#||a(2)||=||(2,-1)||=sqrt(2^2+(-1)^2)=sqrt5 ms^-2#

The direction is

#theta=arctan(-1/2)=26.6^@# clockwise from the x-axis.