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

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Answer 1 · 2017-06-21T12:04:53

The rate of acceleration is #=39.1ms^-2# in the direction #9.4º# clockwise from the x-axis

The acceleration is the derivative of the velocity

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

#a(t)=v'(t)=(e^t-4t,1-e^2)#

So,

when #t=4#

#a(4)=(e^4-16,1-e^2)=(38.6,-6.4)#

The rate of acceleration is

#=||a(4)||=sqrt(38.6^2+(-6.4)^2)=sqrt1530.8=39.1#

The direction is #arctan(6.4/38.6)=9.4º# clockwise from the x-axis