An object's two dimensional velocity is given by #v(t) = ( t^2 - 2t , 1- 3t )#. What is the object's rate and direction of acceleration at #t=7 #?

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Answer 1 · 2018-03-13T07:20:18

The rate of acceleration is #=12.37ms^-2# in the direction #-14.04^@# clockwise from the x-axis.

The acceleration is the derivative of the velocity

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

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

When #t=7#

#a(7)=(2*7-2,-3)=(12,-3)#

The rate of acceleration is

#||a(7)||=||(12,-3)||=sqrt(12^3+(-3)^2)=sqrt(144+9)=sqrt153=12.37ms^-2#

The angle is

#theta=arctan(-3/12)=arctan(-1/4)=-14.04^@# clockwise from the x-axis.