An object's two dimensional velocity is given by #v(t) = ( t-1/t, t^2)#. What is the object's rate and direction of acceleration at #t=1/2 #?

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Answer 1 · 2017-04-20T12:15:54

The rate of acceleration is #=5.1ms^-2# in the direction of #=11.3º#

The acceleration is the derivative of the velocity.

#a(t)=v'(t)#

#v(t) = < t-1/t , t^2> #

#a(t)= <1+1/t^2,2t>#

When #t=1/2#

#a(1/2) = <1+1/(1/2)^2,2*1/2 >#

#=<5,1 >#

The rate of acceleration is

#=||a(1/2)||#

#=||< 5,1 >||#

#=sqrt((5)^2+(1)^2)#

#=5.1ms^-2#

The direction is in the 1st quadrant

#theta=arctan(1/5)#

#=11.3º#