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

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Answer 1 · 2017-11-05T07:07:26

The rate of acceleration is #=7.32ms^-2# in the direction #=10.51^@# anticlockwise from the x-axis

The acceleration is the derivative of the velocity.

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

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

Therefore,

when #t=2#

#a(2)=(e^2-2,1)#

The rate of acceleration is

#||a(t)||=sqrt((e^2-2)^2+(1)^2)=7.32ms^-2#

The direction is

#theta=arctan(1/(e^2-2))=arctan(0.18)=10.51^@# anticlockwise from the x-axis