An object's two dimensional velocity is given by #v(t) = ( e^t-2t , 2t-4e^2 )#. What is the object's rate and direction of acceleration at #t=a #?

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Answer 1 · 2017-05-24T05:45:29

The rate of acceleration is #=sqrt(e^(2a)-4e^(a)+8)ms^-2# in the direction of #arctan(2/(e^(a)-2))#

The acceleration is the derivative of the velocity.

#v(t)=(e^t-2t, 2t-4e^2)#

#A(t)=v'(t)=(e^t-2, 2)#

Therefore,

#A(a)=(e^(a)-2, 2)#

The rate of acceleration is

#||a(4)||=sqrt((e^(a)-2)^2+2^2)#

#=sqrt(e^(2a)-4e^(a)+4+4)#

#=sqrt(e^(2a)-4e^(a)+8)ms^-2#

The direction is

#theta=arctan(2/(e^(a)-2))#