An object's two dimensional velocity is given by #v(t) = ( cost , -t^3 +4t )#. What is the object's rate and direction of acceleration at #t=1 #?

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Answer 1 · 2018-04-13T07:28:07

The rate of acceleration is #=1.31ms^-1# in the direction #=130^@# anticlockwise from the x-axis

The acceleration is the derivative of the velocity

#v(t)=(cost, -t^3+4t)#

#a(t)=v'(t)=(-sint, -3t^2+4)#

When #t=1#

#a(1)=(-sin1, -3+4)=<-0.84, 1>#

The rate of acceleration of the object is

#||a(1)||= ||<-0.84, 1>||= sqrt((-0.84)^2+1^2)=sqrt1.71=1.31ms^-2#

The direction is

#theta=arctan(-1/0.84)=arctan(-1.19)=130^@# anticlockwise from the x-axis