An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #450 KJ# to #129 KJ# over #t in [0, 8 s]#. What is the average speed of the object?

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Answer 1 · 2017-03-30T10:20:57

The average speed is #=306.4ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=6kg#

The initial velocity is #=u_1#

#1/2m u_1^2=450000J#

The final velocity is #=u_2#

#1/2m u_2^2=129000J#

Therefore,

#u_1^2=2/6*450000=150000m^2s^-2#

and,

#u_2^2=2/6*129000=43000m^2s^-2#

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

The points are #(0,150000)# and #(8,43000)#

The equation of the line is

#v^2-150000=(43000-150000)/8t#

#v^2=-13375t+150000#

So,

#v=sqrt((-13375t+150000)#

We need to calculate the average value of #v# over #t in [0,8]#

#(8-0)bar v=int_0^8sqrt((-13375t+150000))dt#

#8 barv=[((-13375t+150000)^(3/2)/(-3/2*13375)]_0^8#

#=((-13375*8+150000)^(3/2)/(-20062.5))-((-13375*0+150000)^(3/2)/(-20062.5))#

#=43000^(3/2)/-20062.5-150000^(3/2)/-20062.5#

#=2451.2#

So,

#barv=22451.2/8=306.4ms^-1#