An object's two dimensional velocity is given by #v(t) = (3t-sqrtt , 2t-4 )#. What is the object's rate and direction of acceleration at #t=3 #?

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Answer 1 · 2017-06-11T05:31:38

The rate of acceleration is #=3.14ms^-2# in the direction #=39.6º# anticlockwise from the x-axis.

The acceleration is the derivative of the velocity

#v(t)=(3t-sqrtt,2t-4)#

#a(t)=v'(t)=(3-1/(sqrtt),2)#

So,

when #t=3#

#a(3)=(3-1/sqrt3,2)=(2.42,2)#

The rate of acceleration is

#=||a(3)||=sqrt(2.42^2+(2)^2)=sqrt9.87=3.14#

The direction is #arctan(2/2.42)=39.6º#