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=1 #?

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Answer 1 · 2017-06-29T13:07:44

The rate of acceleration is #=1.23ms^-2# in the direction #=54.2º# 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)#

So,

when #t=1#

#a(1)=(e-2,1)=(0.72,1)#

The rate of acceleration is

#=||a(1)||=sqrt(0.72^2+(1)^2)=sqrt1.52=1.23#

The direction is #arctan(1/0.72)=54.2º# anticlockwise from the x-axis