An object's two dimensional velocity is given by #v(t) = ( 1/t, t^2)#. What is the object's rate and direction of acceleration at #t=6 #?

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Answer 1 · 2017-06-04T18:45:38

THe rate of acceleration is #=12.00003ms^-2# in the direction #=90.13º# anticlockwise from the x-axis

We need

#(1/x)'=-1/x^2#

#(x^2)'=2x#

The acceleration is the derivative of the velocity

#v(t)=(1/t,t^2)#

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

Therefore,

When #t=6#

#a(6)=(-1/36,12)#

The rate of acceleration is #=sqrt((-1/36)^2+(12)^2)#

#=sqrt(144.0008)#

#=12.00003#

The direction is #=arctan(-12/(1/36))=180-89.9º=90.13# anticlockwise from the x-axis

The angle is in the 2nd quadrant