An object has a mass of #8 kg#. The object's kinetic energy uniformly changes from #150 KJ# to # 640KJ# over #t in [0, 3 s]#. What is the average speed of the object?

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Answer 1 · 2017-03-26T13:04:16

The average speed is #=308.8ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=8kg#

The initial velocity is #=u_1#

#1/2m u_1^2=150000J#

The final velocity is #=u_2#

#1/2m u_2^2=640000J#

Therefore,

#u_1^2=2/8*150000=37500m^2s^-2#

and,

#u_2^2=2/8*640000=160000m^2s^-2#

The graph of #v^2=f(t)# is a straight line

The points are #(0,37500)# and #(3,160000)#

The equation of the line is

#v^2-37500=(160000-37500)/3t#

#v^2=40833.3t+37500#

So,

#v=sqrt((40833.3t+37500)#

We need to calculate the average value of #v# over #t in [0,3]#

#(3-0)bar v=int_0^3sqrt((40833.3t+37500)dt#

#3 barv=[((40833.3t+37500)^(3/2)/(3/2*40833.3)]_0^3#

#=((40833.3*3+37500)^(3/2)/(61250))-((40833.3*0+37500)^(3/2)/(61250))#

#=160000^(3/2)/61250-37500^(3/2)/61250#

#=926.3#

So,

#barv=926.3/3=308.8ms^-1#